text:
\begin{document}

\begin{center}
{\bf \Large On Thompson's conjecture for finite simple groups}

{\bf I. B. Gorshkov}
\medskip
\footnote{The work was supported by
RFBR 18-31-00257.}

\end{center}
{\it Abstract: Let $G$ be a finite group, $N(G)$ be the set of conjugacy classes of the group $G$. In the present paper it is proved $G\simeq L$ if $N(G)=N(L)$, where $G$ is a finite group with trivial center and $L$ is a finite simple group.
\smallskip

Keywords: finite group, simple group, group of Lie type, conjugacy classes, Thompson conjecture. \smallskip
}

\section*{Introduction}

Consider a finite group $G$. For $g\in G$, denote by $g^G$ the conjugacy class of $G$ containing $g$, and by $|g^G|$ the size of $g^G$. The centralizer of $g$ in $G$ is denoted by $C_G(g)$. Put $N(G)=\{n\  |\  \exists g\in G$ such that $|g^G|=n\}\setminus\{1\}$. In
1987 Thompson posted the following conjecture concerning $N(G)$.
\medskip

\textbf{Thompson's Conjecture (see \cite{Kour}, Question 12.38)}. {\it If $L$ is a finite
simple non-abelian group, $G$ is a finite group with trivial
center, and $N(G)=N(L)$, then $G\simeq L$.}

\medskip

A more general question can also be formulated. When the property of equality sets of conjugacy classes sizes of two groups with a trivial center implies an isomorphism.

We say that the group $L$ is recognizable by the set of conjugacy classes size among finite groups with a trivial center (briefly recognizable) if the equality $N(L)=N(G)$, where $G$ is a group with trivial center, implies an isomorphism $L\simeq G$. Obviously, any group having a non-trivial center is not recognizable. However, it is easy to show that $S_3$ is recognizable. Thus, the condition of solvability is not a necessary condition for recognizability. As an example of an non-recognizeble group, we can take a Frobenius group of order $18$. There exists two non-isomorphic Frobenius groups of order $18$ and they have the same sets of conjugacy classes sizes. The question of the existence of the group $L$ such that there exists an infinite set of groups $\{G_i, i \in\mathbb{N} \} $ with trivial center such that $N(L)=N(G_i)$, where $i \in \mathbb{N}$, is open. The set $N(L)$ is closely connected with the order of the group $L$, and sometimes precisely determines it, which essentially limits the possibilities for constructing an infinite series of groups with a trivial center and the same set of conjugacy classes size.

We denote by $\omega(G)$ and $\pi(G)$ the set of elements orders in $G$ and the set of all prime divisors of orders of $G$ respectively. The set $\omega(G)$ defines a prime graph $GK(G)$, whose vertex set is $\pi(G)$ and two distinct primes $p, q\in\pi(G)$ are adjacent if $pq\in\omega(G)$. The greatest power of a prime $p$ dividing the natural number $n$ will be denoted by $n_p$.
The number $p^n$ such that there exists $\alpha \in N(G)$ is a multiple of $p^n$ and there is no $\beta \in N(G)$ that is a multiple of $p^{n + 1}$ is denoted by $|G||_p $. If $\pi\subseteq\pi(G)$ then $|G||_{\pi}=\prod_{p\in \pi}|G||_p $. To shorten the notation, we introduce the following notation $|G||=|G||_{\pi (G)}$. Note that the definitions of the number $|G||$ are correct and $(|G||)_p=|G||_p$. We have $|G||_p$ divides $|G|_p $ for any $ p\in\pi(G)$.

Take a group $L$ with trivial center and disconnected prime graph $GK(L)$. In \cite{Ch96} it is proved that if $N(L)=N(G)$ for any group $G$ with trivial center then $|G|=|G||=|L|$. However, in the general case the equality $|L|=|L||$ may not hold. An example of a group $G$ with a trivial center such that $|G|>|G||$ is constructed in \cite {camin11}.

The question of recognizability is most interesting for simple groups. One of the first groups for which the proof of recognizability was the sporadic group whose prime graph has more than two connected components, see \cite{Chen92}. In \cite{Shi}, using previous results, it is proved that any sporadic group is recognizable. In \cite{Shi2} it is proved that almost sporadic groups are also recognizable.
In \cite{Wil} end \cite{Kon} obtained the classification of finite simple groups with disconnected prime graph. Using this deep result Chen \cite{Ch96,Ch99} established the Thompson's conjecture for all finite simple groups with prime graph have more then two connected components. In the last 10 years, the recognition of a large number of simple groups. However, the results were of a partial nature. Until 2009, the prime graph of any group for which the recognition was proven was disconnected. In \cite {VasT} was shown the recognition of groups with a connected prime graph, in particular $Alt_ {10}$ and $L_4 (4)$ are recognizable. In \cite {Ah11}, the proof of the validity of the Thompson's conjecture for all simple Lie groups of type $A_n (q)$ is completed.
In article \cite{Xu14} showed that Thompson's conjecture holds for finite simple exceptional groups of type $E_7(q)$. In \cite{GorandCo} showed that Thompson's conjecture holds for $F_4(q)$ for odd $q, E_6(q), ^2E_6(q)$. As corollary of thats resaults we obtain that Thompson's conjecture has been proved valid for all exceptional group of lie type. In seryes articles \cite{AD, VasT, Gor0, GorA, GorA2, GorA3, GorA4, Iran, Kh1, Kh2} it was proved that Thompson's conjecture holds for all simple alternating groups. Recognizability of simple groups $^2A_n(q)$ is proved in \cite{Ah17}. In \cite {Ah16} and \cite{Ah12}, the recognizability of the groups $B_n(q)$ and $C_n(q)$ is proved. The prime graph$\ ^2B_2$ has $4$ connected components and recognition of that's groups was proved in \cite{Ch96}. In \cite{Darafsheh}, the recognizability of the groups $D_{p + 1}(2)$ and $D_{p + 1}(3)$ is proved, where $ p $ is a prime. In \cite{Ah132}, the Thompson's conjecture was proved for groups of type $D_n(q)$ where $n \not \in \{4, 8\}$. The validity of the conjecture for the groups $^2D_n(q)$ was proved in \cite{Ah13}. The graph of prime numbers of the groups $^3D_4(q) $ is disconnected, the validity of the conjecture for these groups was proved in \cite{Ch01}. Thus, to complete the proof of the validity of Thompson's conjecture, it is necessary to prove the recognizability of the groups $D_4(q)$ and $D_8(q)$ where $q>3$.

\begin{theorem}
The groups $D_4(q)$ and $D_8(q) $ are recognizable.
\end{theorem}

As a corollary of the theorem and the results obtained earlier, we obtain the following assertion.

\begin{cor}
Thompson's conjecture is hold.
\end{cor}

\section{Notation and preliminary results}

If $n$ is a nonzero integer and $r$ is an odd prime with $(r,n)=1$, then $e(r, n)$ denotes the multiplicative order of $n$ modulo $r$.
Fix an integer $a$ with $|a|>1$. A prime $r$ is said to be a primitive prime divisor of $a^i-1$ if $e(r,a)=i$. We write $r_i(a)$ to denote some primitive prime divisor of $a^i-1$, if such a prime exists, and $R_i(a)$ to denote the set of all such divisors. Zsigmondy \cite{zs} proved that primitive prime divisors exist for almost all pairs $(a, i)$.

\begin{lem}(Zsigmondy).
  Let $a$ be an integer and $|a|>1$. For every natural number $i$ the set $R_i(a)$ is nonempty, except for the pairs $(a, i)\in\{(2, 1), (2, 6), (-2, 2), (-2, 3), (3, 1), (-3, 2)\}$.
\end{lem}

For $i\neq 2$ the product of all primitive divisors of $a^i-1$ taken with multiplicities is denoted by $k_i(a)$. Put $k_2(a)=k_1(-a)$. The number $k_i(a)$ is said to be the greatest primitive divisor of $a^i-1$.

It follows from \cite{A1919} that for $i>2$,

\begin{center}

$k_i(a)=\frac{|\Phi_i(a)|}{(r, \Phi_{i_{\{r\}'}}(a))}$
\end{center}
where $\Phi_i(x)$ is the $i$-th cyclotomic polynomial and $r$ is the largest prime dividing $i$; moreover, if $i_{\{r\}'}$ does not divide $r-1$ then $(r,\Phi_{i_{\{r\}'}}(a))=1$.

As a special case of this assertion, we obtain the following lemma.

\begin{lem}\label{kqi}
If $q$ is a power of prime then:
\begin{enumerate}
\item $k_3(q)=(q^2+q+1)/(3,q-1)$

\item $k_4(q)=(q^2+1)/(2,q+1)$

\item $k_6(q)=(q^2-q+1)/(3,q+1)$

\item $k_7(q)=(q^7-1)/(q-1)(7,q-1)$

\item $k_{14}(q)=(q^7+1)/(q+1)(7,q+1)$
\end{enumerate}
\end{lem}

\begin{lem}[{\rm \cite[Lemma 1.4]{GorA2}}]\label{factorKh}
Let $G$ be a finite group, $K\unlhd G$ and put $\overline{G}= G/K$. Take $x\in G$ and $\overline{x}=xK\in G/K$.
The following claims hold

(i) $|x^K|$ and $|\overline{x}^{\overline{G}}|$ divide $|x^G|$.

(ii) If $L$ and $M$ are neigboring members of a composition series of $G$, $L<M$, $S=M/L$, $x\in M$  and
$\widetilde{x}=xL$ is an image of $x$, then $|\widetilde{x}^S|$ divides $|x^G|$.

(iii) If $y\in G, xy=yx$, and $(|x|,|y|)=1$, then $C_G(xy)=C_G(x)\cap C_G(y)$.

(iv) If $(|x|, |K|) = 1$, then $C_{\overline{G}}(\overline{x}) = C_G(x)K/K$.

(v) $\overline{C_G(x)}\leq C_{\overline{G}}(\overline{x})$.
\end{lem}

\begin{lem}[{\rm \cite[Theorem 5.2.3]{Gore}}]\label{Gore5hzOcomutantePstavtomor}
Let $A$ be a $\pi(G)'$-group of automorphisms of
an abelian group $G$. Then $G=C_G(A)\times[G,A]$.
\end{lem}

If $\pi$ is a set of primes, then $n_{\pi}$ denotes the $\pi$-part of $n$, that is, the largest divisor $k$ of $n$ with $\pi(k)\subseteq\pi$; and $n_{\pi'}$ denotes the $\pi'$-part of $n$, that is, the ratio $|n|/n_{\pi}$.

Let $G$ be a group, $a\in G$. We denote by $a^G$ the conjugacy class of the group $G$ containing the element $a$, $Ind_G(a)$ is the index of $C_G(a)$ in the group $G$.
Let $p,q$ be distinct numbers. We say that the group $G$ satisfies the condition $\{p,q\}^* $ (for brevity we will write $G\in\{p, q\}^*$) if for any $\alpha \in N(G) $ we have $\alpha_{\{p, q\}}\in\{|G||_p, |G||_q, |G||_{\{p, q\}}\} $.

\begin{lem}[{\rm \cite[Corollary]{GorBig}}]\label{GorBig}
Let $G\in\{p,q\}^*$ be a group with trivial center, where $p,q\in\pi(G)$ and $p>q>5$. Then $|G|_{\{p,q\}}=|G||_{\{p, q\}} $ and
$C_G(g)\cap C_G(h)=1$ for any $p$- and $q$-elements $g$ and $h$, respectively.
\end{lem}

Define the following function on positive integers:
 \begin{equation*}
\eta(k) =
 \begin{cases}
   k &\text{ if $k$ is odd}\\
   k/2 &\text{if k is even}
 \end{cases}
\end{equation*}

Now we introduce a new function in order to unify further arguments. Namely, given a simple classical group $L$ over a field of order $q$ and a prime $r$ coprime to $q$, we put

\begin{equation*}
\varphi(r,L)=
 \begin{cases}
   e(r,\varepsilon q) &\text{ if $L= A^{\varepsilon}_n(q)$}\\
   \eta(e(r,q))       &\text{if $L$ is symplectic or orthogonal}
 \end{cases}
\end{equation*}

For a classical group $L$, we put $prk(L)$ to denote its dimension if $L$ is a linear or unitary group, and its Lie rank if $L$ is a symplectic or orthogonal group.

\begin{lem}[{\rm\cite[Lemma 2.12]{VasBig}}]\label{vas}
Let $L$ be a simple classical group over a field of order $q$ and characteristic $p$, and let $prk(L)=n\geq4$. If $r\in\pi(L)\setminus\{p\}, i=e(r,q)$, and $n/2<\varphi(r, L)\leq n$, then $L$ includes a cyclic Hall subgroup of order $k_i(q)$.

\end{lem}

\begin{lem}[{\rm\cite[Section 5.1.7]{Carter}}]\label{Regular}
Let $L$ be a simple group of Lie type in characteristic $p$, $L\leq G\leq Aut(L)$. The group $G$ contains an element $h$ such that $(Ind_G (h))_{p'}=|G|_{p'}$.
\end{lem}

\begin{lem}\label{Order}
If $S\leq A\leq Aut(S)$, where $S$ is a non abelian simple group, then $|A|=|A||$.
\end{lem}
\begin{proof}
We show that $|A|_p =|A||_p$ for any $p\in \pi(A)$. To do this, we need to show that there exists $\alpha \in N(A)$ such that $\alpha_p=|A|_p$. Assume that $S$ is a simple alternating group of degree $n$. If $n<100$, then the statement is checked with \cite{GAP}. Suppose that $n\geq100$. Lemma \cite[Lemma 1.11]{GorA2} implies that the interval $\Omega=[n/2..3n/4]$ contains two prime numbers. Hence $\Omega$ contains a prime number $t$ different from $p$. Let $g\in A$ be an element of order $t$. Then $C_A(g)\simeq\langle g \rangle \times R$, where $R$ is an alternating or symmetric group of degree $n-t$. Therefor $R$ contains an element $h$ such that $(Ind_R(h))_p =|R|_p$. Thus, $(Ind_A(gh))_p=|A|_p$.

Assume that $S$ is a group of Lie type over  field in characteristic $t$. Lemma \ref{Regular} implies that $S$ contains a unipotent element $g$ such that $\pi(C_A(g))=\{t\}$. Therefore, if $p\neq t$ then $|Ind_A (g)|_p=|A|_p$. Assume that $p=t$. The set $\pi(S)$ contains a number $r$ that is not adjacent in $GK(S)$ with $p$ (see \cite{VV}). Let $h\in A$ is an element of order $r$. We note that cetraliser of every external automorphism of the group $S$ no contain an element of order $r$. Hence $(Ind_A (h))_p=|A|_p$.

\end{proof}

\begin{lem}\cite[Theorem 1]{vse}\label{pat}
Let $G$ be a finite group, and let $p$ and $q$ be different primes. Then
some Sylow $p$-subgroup of $G$ commutes with some Sylow $q$-subgroup of $G$ if and only
if the class sizes of the $q$-elements of $G$ are not divisible by $p$ and the class sizes of
the $p$-elements of $G$ are not divisible by $q$.
\end{lem}

\begin{lem}\cite{navarro}\label{navarro}
Let $G$ be a finite group and $p$ a prime, $p\not\in\{3, 5\}$. Then $G$ has abelian Sylow $p$-subgroups if and only if $|x^G|_p=1$ for all $p$-elements $x$ of $G$.
\end{lem}

\begin{lem}\label{hz3}
Let $S\leq A\leq Aut(S)$, where $S$ is a simple group of Lie type, $H\in Syl_p(A)$, where $p>5$ is a prime number. If for every $\alpha \in N(A)$ we have $|\alpha|_p \in \{1,|H|\}$, then $|H|$ divides $S$ or $|H|$ divides $A/S$.
\end{lem}
\begin{proof}
Lemma \ref{Order} imply that $|A|=|A||$. In particular $|A||_p=|A|_p$.
Therefor $(Ind_A(x))_p=1$ for any $p$-element $x$. It follows from Lemma \ref{navarro} that the Sylow $p$-subgroups of $A$ is abelian.
Assume that $p$ divides $|S|$ and $|A/S|$. Take a $p$-element $x\in A$ such that $x$ acts on $S$ as an outer automorphism. We have $C_S(x)$ includes a some Sylow $p$-subgroup of $S$. The graph $GK(C_S(x))$ contains a vertex $t$ such that $t$ is not adjacent to $p$ see \cite{VV} and a description of the centralizers of outer automorphisms of groups of Lie type. Hence, there is an element $y\in C_S(x)$ of order $t$ such that $ (Ind_{C_S(x)}(y))_ p=|C_S(x)|_p$. Thus, $|S|_p\leq(Ind_A(xy))_p<|A|_p$; a contradiction.

\end{proof}

\begin{lem}\label{hz5}
Let $S\leq A\leq Aut(S)$, where $S\simeq S(q)$ is a simple group of Lie type over a field of order $q=p^n$ where $p$ is prime, $H$ is a Hall $\pi$-subgroup of $A$, where $\pi\subseteq\pi(A)$ and $\{2,3,5\}\cap\pi=\varnothing$. If for every $\alpha \in N(A)$ we have $|\alpha|_{\pi} \in \{1,|H|\}$, then $|H|$ divides $S$ or $|H|$ divides $A/S$.

\end{lem}
\begin{proof}
It follows from Lemma \ref{hz3} that for any $t\in\pi$, $|A|_t$ divide one of the numbers $|S|$ or $|A/S|$. Assume that there exists numbers $a, b\in\pi$ such that $a$ divides $|S|$ and $b$ divides $|A/S|$. Take a $b$-element $x\in A$. Since $b$ does not divide $|S|$, we see that $x$ acted on $S$ as a fields automorphism of $S$. We have $C_S(x)\simeq S(p^m)$, where $m$ is a non trivial divisor of $n$. The graph $GK(C_S(x))$ contains a vertex $r$ that is not adjacent to $a$ see \cite {VV}. Hence, there exists an element $y\in C_S(x)$ such that $(Ind_{C_S (x)} (y))_t =|C_S (x)|_t$ and $1<(Ind_A (xy))_{\pi}<|H|$; a contradiction.
\end{proof}

\begin{lem}\label{hz2}
Let $L$ be a group of Lie type over a field of order $q$ and characteristic $p$, $H\in Syl_t(L)$, where $t>5$ is prime. If for any $\alpha\in N(L)$ we have $|\alpha|_t\in \{1,|H|\}$ then $H$ is a cyclic or $L\simeq A_1(t^n)$ where $n>1$.
\end{lem}
\begin{proof}
Lemma \ref{Order} implies that $|L|=|L||$. Therefor $(Ind_L(x))_t=1$, for any $t$-element $x$. It follows from \ref{navarro} that a Sylow $t$-subgroup of $L$ is abelian. A Sylow $p$-subgroup of $L$ is abelian iff $L\simeq A_1(q)$. Note that if $t\neq p\in\pi(A_1(q))\setminus\{2\}$, then the Sylow $t$-subgroup of $A_1(q)$ is cyclic. We assume that $L\not\simeq A_1(q)$, and hence $p\neq t$.

Assume that $L$ is a classical group. From the description of the cyclic structure of tors of the finite simple groups see \cite{GrBu}, it follows that $L$ contains an element of order $k_{\varphi(t, L)}(q)$. If $\varphi(t,L)>n/2$, then from Lemma \ref{vas} it follows that the $t$-subgroup of $L$ is cyclic. Assume that $\varphi(t, L)\leq n/2$. Then $|L|_t>|k_{\varphi(t,L)}(q)|_t$. Let $r=|k_{\varphi(t,L)}(q)|_t$, $k=|k_{\varphi(a,L)}(q)|_a$, where $a$ is a prime number such that $\varphi(a, L)=n/2-1$ if $\varphi(t, L)=n/2$ and $\varphi(a,L)= n-\varphi(t, L)$ otherwise.
From the description of the spectra of finite simple groups it follows that $L$ contains an element $h$ of order $r.k$. Therefor $(Ind_L(h^r))_t<|L|_t$, and hence $(Ind_L(h^r))_t=1$. Thus, the centralizer of any $t$-elements contains an element conjugate to $h^r$. Let $T<L$ be a tors having the cyclic structure $\varphi(t, L) + \varphi (t, L)$, $x\in T$ is a $t$-element, $y \in C_L(x)\cap(h^r)^L$. Since $t, a\neq p$, then in $L$ concludes a maximal torus $X$ such that $xy\in X$. Hence $X$ has the cyclic structure $\varphi(t,L) + \varphi(t, L) + \varphi(a, L)>n$. It follows from \cite{GrBu} that $L$ contains no torus with such a cyclic structure; a contradiction.

If $L$ is an exceptional group of Lie type, then the lemma follows from \cite{Der}.

\end{proof}

\begin{lem}\label{hz4}
Let $L$ be a simple group, $L<G\leq Aut(L)$, $H\in Syl_t(L)$, where $t>5$ is prime. If $|H|$ divide $|G/L|$ then $H$ is cyclic.
\end{lem}
\begin{proof}
Since $t>5$, we see that $L$ is group of Lie type. From the fact that $t$ does not divide $|L|$, it follows that $H$ is a subgroup of the group of field automorphisms. The group of field automorphisms of the group $L$ is cyclic and hence $H$ is cyclic.
\end{proof}

\begin{lem}\cite[Corollary 13]{Ah132}\label{Neda} Let $r \in R_{2(n-1)}(q), r_1\in R_n(q), r_2\in R_{n-1}(q)$ and $r_3 \in R_{2(n-2)}(q)$, $L=D_n(q)$, $d=(q^n-1,4)$.

(i) If $x\in L\setminus \{1\}$, then either $(Ind_L(x))_r = |L|_r$ or $(Ind_L(x))_r<|L|_r$
and one of the following holds:

\begin{itemize}

           \item{$Ind_L(x)=|L|d/(|GU_{(n-1)/m}(q^m)|(q+1))$, where $(n-1)/m$
is an odd number and if $q=3$ and $n$ is an even number, then $m\neq 1$;}
           \item{$Ind_L(x) = |L|d/(\alpha|GU_n(q)|)$ , where $n$ is an even number and, if $q=3$, then $\alpha=2$,
if $q\neq3$ and $(q+1)_2>2$, then $\alpha\in{1, 2}$ and otherwise, $\alpha=1$;}
           \item{$(q, n)\neq(3, 2u)$ $(u\in \mathbb{N})$ and $Ind_L(x) = 2|L|d/(|GO^-_{2(n-1)}(q)|(q+1))$;}

            \item{$q\equiv-1(mod 4)$ and $Ind_L(x)=|L|d/(|GO^-_{2(n-1)}(q)|(q+1))$;}

         \end{itemize}

(ii) if $n$ is odd and $x_1\in L\setminus\{1\}$, then $(Ind_L(x_1))_{r_1}=|L|_{r_1}$ or $(Ind_L(x_1))_{r_1}<|L|_{r_1}$ and there exists a divisor $m_1$ of $n$ such that $Ind_L(x_1)=|L|d/(|GL_{n/(m_1)}(q^{m_1})|$.
In particular, if $q\in\{2, 3, 5\}$, then $m_1\neq1$;

(iii) if $n$ is even and $x_2\in L\setminus\{1\}$, then $(Ind_L(x_2))_{r_2}=|L|_{r_2}$ or $(Ind_L(x_2))_{r_2}<|L|_{r_2}$ and one of the following holds:
\begin{itemize}
\item{ $Ind_L(x_2)=|L|d/(|GL_{(n-1)/m_2}(q^{m_2})|(q-1))$, where $m_2$ is a divisor of $n-1$. In particular, if $q\in\{2, 3, 5\}$, then $m_2\neq1$;}
\item{ $q\not\in\{2, 3\}$ and $Ind_L(x_2)=|L|d/(\alpha|GL_n(q)|)$, where if $(q-1)_2 > 2$ and $q\neq5$, then $\alpha\in\{1, 2\}$, if $q=5$, then $\alpha=2$ and otherwise, $\alpha = 1$;}
\item{ $q\not\in\{2, 3, 5\}$ and $Ind_L(x_2)=2|L|d/(|GO^+_{2(n-1)}(q)|(q-1))$;}

\item{ $q\equiv 1(mod 4)$ and $Ind_L(x_2)|=|L|d/(|GO^+_{2(n-1)}(q)|(q-1))$;}
\end{itemize}

(iv) if $n\geq6$ is even and $x_3\in G\setminus\{1\}$, then either $(Ind_L(x_3))_{r_3} = |L|_{r_3}$ or $Ind_L(x_3)|_{r_3}<|L|_{r_3}$ and one of the following holds:
\begin{itemize}
\item{$Ind_L(x_3)=\alpha|L|d/(|GO^{\varepsilon}_{2(n-1)}(q)|(q-\varepsilon1))$, where $\alpha | 2$ under conditions mentioned in $(i,iii)$ and $\varepsilon\in\{+, -\}$;}
    \item{ $Ind_L(x_3)=|L|d/(\alpha\beta)$, for some divisor $\alpha$ of $|\Omega^-_4(q)|$ and $\beta\in\{|GU_{(n-2)/m_3}(q^{m_3})|,|GO^-_{2(n-2)}(q)|,|GO^-_{2(n-2)}(q)|/2\}$, where $(n-2)/m_3$
is an odd number;}
\item{$Ind_L(x_3) = |L|d/q^{2(n-1)}|SP_{2(n-2)}(q)|$;}
\item{$Ind_L(x_3)= |L|d/|GU_{n/2}(q^2)|$, where $n/2$ is an even number.}
\end{itemize}
\end{lem}
\begin{lem}\cite[Lemma 2.6]{Ah11}\label{Lieq3}
Let $G(q)$ be a simple group of Lie type in characteristic $p$, then $|G(q)|<(|G(q)|_p)^3$.
\end{lem}
\begin{lem}\label{Lieq3Aut}
If $G(q)$ be a simple group of Lie type in characteristic $p$, then $|Aut(G(q))|<(|G(q)|_p)^{3,5}$.
\end{lem}
\begin{proof}
The assertion of the lemma follows from the description of the orders of automorphism groups of finite simple groups of Lie type and Lemma \ref{Lieq3}.
\end{proof}

\begin{lem}\label{pchast}
Let $G(q)$ be a simple group of Lie type in characteristic $p$, $t\in \pi(G(q))\setminus\{p\}$. The group $G(q)$ contains an element $h$ such that $(Ind_{G(q)}(h))_p=|G(q)|_p$ and $(Ind_{G(q)}(h))_t<|G(q)|_t$.
\end{lem}
\begin{proof}
It follows from the description of the spectra of finite groups of Lie type see \cite{Bu1, Bu2, Bu3, Bu4} that $G(q)$ contain an element $h$ such that $t\in\pi(|h|)$ and $G(q)$ no contain en element of order $|h|p$. Hence $|C_{G(q)}(h)|$ is not divisible by $p$ and $(Ind_{G(q)}(h))_p=|G(q)|_p$.
\end{proof}

\section{Proof the Theorem for groups $D_4(q)$}

\begin{prop}
Let $L=D_4(q)$, where $q=p^n$. If $G$ is a group with trivial centre and $N(G)=N(L)$ then $G\simeq L$.
\end{prop}

Proposition is proved for $q=2$ and $q=3$ see \cite{Darafsheh}. We will assume that $q>3$.

\begin{lem}\label{good}
$G\in\{r_3(q), r_6(q)\}^*$
\end{lem}
\begin{proof}
Take $h\in L$ such that $(Ind_L(h))_{r_i(q)}<|G|_{r_i(q)}$ where $i\in\{3, 6\}$. Since $r_i(q)$ and $p$ are not adjacent in $GK(L)$, see \cite{VV}, then $h$ is a semisimple element. Hence $h$ lies in some maximal torus $T$ of $L$. It follows from the description of the orders of maximal torus of $L$ see \cite{Car1} that $(Ind_L(h))_{R_i(q)}=1$ and $(Ind_L(h))_{R_{i'}(q)}=|L|_{R_{i}(q)}$, where $i'\in\{3, 6\}\setminus\{i\}$. We have $L\in\{r_3(q), r_6(q)\}^*$. Since $N(G)=N(L)$, we obtain $G\in\{r_3(q), r_6(q)\}^*$.
\end{proof}

\begin{lem}\label{HollD4R3R6}
A Hall $R_3(q)$-subgroup and $R_6(q)$-subgroup of $G$ exists and abelian.
\end{lem}
\begin{proof}
Note that $2,3,5\not\in R_3(q)\cup R_6(q) $. It follows from Lemmas \ref{good} and \ref{GorBig} that for any $r\in R_3(q)\cup R_6(q)$ a Sylow $r$-subgroup of $G$ is abelian. Hence for any $r$-element $h$ we have $(Ind_G(h))_r=1$. It follows from Lemma \ref{Neda} that $(Ind_G (h))_{R_i(q)}=1$, where $i$ such that $r\in R_i(q)$. It follows from Lemma \ref{pat} that the Hall $R_i(q)$-subgroup exists and abelian.

\end{proof}

\begin{lem}\label{factorD4}
The group $G$ includes a normal subgroup $K$ such that $S\leq \overline{G}=G/K\leq Aut(S)$, where $S$ is non abelian simple group, and $|\overline{G}|$ is a multiple of $k_3(q)k_6(q)$.
\end{lem}
\begin{proof}
Let $\widetilde{G}=G/O_{(R_3(q)\cup R_6(q))'} $.
Suppose that $\widetilde{G}$ include non-trivial solvable minimal normal subgroup $X$. We have $\pi(X)\subseteq R_i (q) $, where $i\in \{3, 6\}$. Take a Hall $R_j(q)$-subgroup $H$ of $\widetilde{G}$, where $j\in \{3, 6\}\setminus\{i\}$. Therefor $XH$ is a Frobenius group with kernel $X$. Hence $N(\widetilde{G})$ contains a number is multiple of $|X|$. It follows that $|X|$ divides $k_i(q)$. Since $X$ is a Frobenius kernel of $XH$, we obtain $|X|-1$ is multiple of $|H|$. It follows from $|H|=|G|_{R_j(q)}\geq |L|_{R_j(q)}=k_j(q)$ that $k_j(q)$ divides $k_i(q)$; a contradiction with Lemma \ref{kqi}.

Let $R$ be the socle of $\widetilde {G}$. We get that $R=S_1\times...\times S_k$, where $S_1,...,S_k$ are non abelian simple groups.
Assume that $k> 1$. It follows from the definition that the order of $S_h$ is multiple of some number of $R_3(q)\cup R_6(q)$, for every $1\leq h\leq k$. Note that $\omega(\widetilde{G})\subseteq \omega(G)$. Hence $R$ contains no element of order $tr$, where $t\in R_3(q)$ and $r\in R_6(q)$. Thus, $\pi(R)\cap(R_3(q)\cup R_6(q))\subseteq R_i(q)$, where $i\in\{3,6\}$. The $S_1$ contains an element $g$ such that $(Ind_{R}(g))_{R_3 (q)\cup R_6(q)} =|S_1|_{R_3(q)\cup R_6(q )}$. We have $1<(Ind_{S_1}(g))_{R_3 (q)\cup R_6(q)}<|R|_{R_3(q)\cup R_6(q)}$. Similarly, as in Lemma \ref{Order} we can shows that $|R|=|R||$. In particular $|R|_{R_3(q)\cup R_6(q)}>(Ind_R(g))_{R_3(q)\cup R_6(q)}$; using Lemma \ref {factorKh} we get a contradiction.

Thus $k=1$ and $R\leq \widetilde{G}\leq Aut (R)$, where $R$ is a simple group. Since $|G|_{R_3(q)\cup R_6(q)}=|\widetilde{G}|_{R_3(q)\cup R_6(q )}$ we get $k_3(q)k_6(q)$ divides $|\widetilde{G}|$.

\end{proof}

Let $S$, $\overline{G}$ and $K$ be similar as in Lemma \ref{factorD4}.

\begin{lem}\label{AutD4}
$|Out(S)|$ is not a multiple of $k_3(q)$ and $k_6(q)$.
\end{lem}
\begin{proof}
Assume that there exists $i\in\{3,6\}$ such that $k_i(q)$ divides $|Out(S)|$.
Since $r_i(q)>5$, we get $S$ can not be isomorphic to a sporadic group or an alternating group. Hence $S$ is a group of Lie type over a field of order $l=t^r$ of characteristic $t$. From Lemmas \ref{Order} and \ref{factorKh} it follows that $|S|$ divides $|G||$. Since $|G||=|L||$, we obtain $|S|$ divides $|L|$. In particular, for any $k\in \pi (S) $, we have $|S|_k\leq|L|_k$.

Suppose that $S$ is a group of exceptional Lie type. From $2,3\not\in R_i (q) $, it follows that $k_i(q)$ divides order of the field automorphisms group of $S$. Hence $l\geq t^{k_i (q)}$. It follows from Lemma \ref{kqi} that $k_i(q)\in\{(q^2-q+1)/(3, q+1), (q^2+q+1)/(3, q-1)$. Note that for any $k\in\pi(L) $, we have $|L|_k<|L|_p$ and $|L|_p=q^{12}$.
If $S\not\simeq\ ^2G_2(l)$ then $|S|_t> l^4 $. Since $q>3$, we have $|S|_t\geq t^{4k_i (q)}>q^{12}=|L|_p$; a contradiction. If $S\simeq\ ^2G_2(l)$, and $l>2$ or $q>4$, then we obtain $|S|_t>|L|_p$. Assume that $l= 2, q=4$. Then $43 \in\pi(S)\setminus\pi(L)$; a contradiction.

Assume that $S$ is a classical group of Lie type and $k_i(q)$ divides order of the fields automorphism group. If $|S|_t>l^3$, then similarly as above we get a contradiction. Assume that $|S|_t=l^3$, then $S$ is isomorphic to one of the groups $A_2(l)$ or$\ ^2A_2(l)$. If $t>2$ or $q>4$, then $|S|_t>|L|_p$; a contradiction. We get $t=2, q=4$. In this case $k_3(4)=13, k_6(4)=7$. If $l \geq t^{13}$, then we obtain $|S|_t>|L|_p$; a contradiction. Hence $l=t^7$ and $127\in \pi(S) \setminus \pi (L)$; a contradiction.

Assume that $|S|_t=l^2$. Therefor $S$ is isomorphic to$\ ^2B_2(r)$. If $q>5$ or $t>3 $, then $|S|_t>|L|_p$; a contradiction. Assume that $t=3$ and $|S|_t\leq |L|_p $. Then $1093 \in\pi(S)\setminus \pi (L) $; a contradiction. Assume that $t=2$, then $127\in \pi(S)\setminus\pi(L)$; a contradiction.

Assume that $|S|_t=l$. Hence $S$ is isomorphic to $A_1(r)$. From the fact that $|S|_t\leq|L|_p$ it follows that $q<11$. Analyzing all possible variants similarly as above we arrive at a contradiction.

Thus, there exists $r_i(q)$, such that $r_i(q)$ divides order of the diagonal automorphisms group of $S$. In this case, $S$ is isomorphic to one of the groups $A^{\varepsilon}_m(l)$ where $\varepsilon \in \{+, -\} $. We have $r_i(q)$ divides $(n+1, l-(\varepsilon1))$. It is easy to shows that in this case $|S|_t\geq|L|_p$; a contradiction.

\end{proof}

\begin{lem}\label{Aut2D4}
The number $|\overline{G}|/|S|$ is not a multiple of $r_i(q)$, where $i\in\{3,6\}$.
\end{lem}
\begin{proof}
Assume that there exists $r_i(q)\in\pi(|\overline{G}|/|S|)$, where $i\in\{3,6\}$. Take a Hall $R_i(q)$-subgroup $H$ of $\overline{G}$. We have for every $h\in\overline{G}$, $(Ind_{\overline{G}}(h))_{R_i(q)}=1$ or $(Ind_{\overline{G}}( h))_{R_i(q)}=k_i(q)$. Using Lemma \ref{hz5}, we obtain $|H|$ divides $|S|$ or $|\overline{G}|/|S|$. Since $r_i(q)$ divides $|\overline{G}|/|S|$, we see that $k_i(q)$ divides $|\overline{G}|/|S|$; a contradiction with Lemma \ref{AutD4}.
\end{proof}

\begin{lem}\label{R4}
$R_4(q)\cap\pi(K)=\varnothing$.
\end{lem}
\begin{proof}
Assume that $R_4(q)\cap\pi(K)\neq\varnothing$. Let $\widetilde{\ }:G\rightarrow G/O_{R_4 (q)'}(K)$ be a natural homomorphism, $R<\widetilde{G}$ be a minimal normal subgroup. Take $h\in G$ such that $\pi(|h|)\subseteq R_6(q)$. It follows from Lemma \ref{Aut2D4} that $\overline{h}\in S$, where $\overline{h}\in\overline{G}$ is the image of the element $h$. By the definition we have $|\pi(R)\cap R_4 (q)|\neq\varnothing$. If $R$ is not solvable, then it is easy to show that $R$ contains an $R_4(q)$-element $x$ such that $(Ind_{\widetilde{G}}(x))_{R_3(q)}<|G|_{R_3(q)}$ and $(Ind_{\widetilde{G}}(x))_{R_6(q)}<|G|_{R_6(q)}$. We have $\pi(|x|)\cap\pi(O_{R_4 (q)'}(K))=\varnothing$. It follows from Lemma \ref{factorKh} that $G$ contains $y$ such that $(Ind_G(y))_{R_3 (q)\cup R_6 (q)}=1$; a contradiction with the fact that $G\in\{r_3(q), r_6(q)\}^*$ and Lemma \ref{GorBig}. Hence, $R$ is an elementary Abelian $r$-group, where $r\in R_4(q)$. Assume that $C_R(h)>1$. Let $x\in C_R(h)$. Then $(Ind_{\widetilde{G}}(x))_{R_6 (q)}=1$. It follows from Lemma \ref{Neda} that for any $\alpha\in N(G)$ such that $\alpha_{R_6(q)}=1$, we have $\alpha_{R_4 (q)}=|L|_{R_4(q)}$. Therefor $(Ind_{\widetilde{G}}(hx))_{R_4(q)}=(Ind_{\widetilde{G}}(x))_{R_4(q)}$. Thus $C_{\widetilde{G}}(h)$ contains a some Sylow $r$-subgroup of $C_{\widetilde {G}}(x)$. In particular, $h$ acts trivially on $R$. Since $R$ is normal subgroup of $\widetilde {G} $, we obtain any element conjugate with $h$ acts trivially on $R$. In particular, the minimal preimage $H<\widetilde{G}$ of $S$ acts trivially on $R$. If $|S|_{R_3(q)}>1$, then $G$ contains an element $y$ such that $Ind_G(y)_{R_3(q)\cup R_6(q)}=1$; a contradiction. Thus, $|\overline{G}/S|$ is a multiple of $k_3(q) $; a contradiction with Lemma \ref{AutD4}.

Therefor $C_{R}(h)=1$. Thus, $|R|$ divides $Ind_{\widetilde{G}}(h)$. Since $(Ind_{\widetilde {G}}(h))_r\leq|L|_{r}$, we obtain $|R|\leq(k_4(q))^2$. The group $H$ acts faithfully on $R$. Hence $|R|-1\geq|H|$. From Lemma \ref{Aut2D4} we get that $|H|\geq 4k_3(q)k_6(q)>(k_4(q))^2$; a contradiction.

\end{proof}

\begin{lem}\label{D4Alt}
The group $S$ is not isomorphic to an alternating group.
\end{lem}
\begin{proof}
Assume that $S\simeq Alt_m$ for some $m>5$. From Lemmas \ref{factorD4} and \ref{R4} it follows that $|\overline{G}|$ is a multiple of $(k_4 (q))^2k_3(q)k_6(q)$. Since $|Out(S)|\leq4$ and $|S|$ is a multiple to $3$ and $4$, we obtain $|S|$ is divisible by $12(k_4(q))^2k_3(q)k_6(q)=(q^2+1)^2(q^4+q^2+1)$.
From Lemma \ref{HollD4R3R6}, we have that a Halls $R_3(q)$-subgroup and $R_6(q)$-subgroup of $S$ are abelian. Lemmas \cite{Hall} and \cite{Tom66} implies that a Hall $\pi$-subgroup of $Alt_m$ is abelian only if $|\pi\cap\pi(Alt_m)|=1$. Hence $|R_3(q)|=|R_6(q)|=1$, and $m\geq max((q^2+q+1)/(3, q-1), (q^2-q+1)/(3, q+1))$. If $q\geq 5$, then $|S|>|L| $. Hence $S$ contains an element $h$ such that $Ind_S(h)$ does not divide any number from $N(L)$; a contradiction. If $q=4$ then $R_4(q)=\{17\}$. Hence $m>34$ and $|S|>|L|$; a contradiction.

\end{proof}

\begin{lem}\label{Sporadic}
The group $S$ is not isomorphic to any of the sporadic groups.
\end{lem}
\begin{proof}
The assertion of the lemma follows from the fact that $(q^2+1)^2(q^4+q^2+1)$ divides $|\overline{G}|$, $|\overline{G}|$ divides $|L|$ and \cite{Atlas}.
\end{proof}
\begin{lem}\label{R1R2}
$(R_1(q)\cup R_2(q))\cap\pi(K)=\varnothing$.
\end{lem}
\begin{proof}
Assume that there exists $r\in(R_1(q)\cup R_2(q))\cap\pi(K)$. Let $R=O_{r'}(G)$, $X\unlhd K/R $ be a minimal normal subgroup. Assume that $S\simeq A_1(t^m)$, where $t\in R_3(q)\cup R_6(q)$. We have $|\overline{G}|\leq2mt^m(t^m-1)^2$, $|S|_t=k_j(q)$ where $j\in\{3,6\}$. It follows from Lemmas \ref{factorD4} and \ref{R4} that $|\overline{G}|$ is divisible by $(k_4(q))^2k_3(q)k_6 (q)$; a contradiction. Hence $S\not\simeq A_1(t^m)$ where $t\in R_3(q)\cup R_6(q)$.

It follows from Lemmas \ref{hz2} and \ref{hz4} that for any $t\in R_3(q)\cup R_6(q)$ a Sylow $t$-subgroup of $\overline{G}$ is cyclic. It follows from Lemma \ref{HollD4R3R6} that a Hall $R_3(q)$-subgroup and $R_6(q)$-subgroup of $\overline{G}$ are cyclic. Let $i=3$ if $r\in R_1(q)$ and $i=6$ if $r\in R_2(q)$, $H\in Hall_{R_i(q)}(G/R)$. It follows from Lemma \ref{Neda} that if $r\neq 2$ then for any $x\in H $ we have $(Ind_{G/R}(x))_{r}\leq 1/4(q +(-1)^e)^2$, and $(Ind_G(x))_{r}\leq 4(q+(-1)^e)^2$ if $r=2$, where $e\in\{1,2\}$. Let $h$ be the generating element of $H$, and let $x\in H$ be an element of order $t\in R_i(q)$. From Lemma \ref{Gore5hzOcomutantePstavtomor} it follows that $X=C_X(H)\times[X, H]$. Hence $(Ind_X(h))_r\geq|[X, H]|_r$. For any $x\in H$, we have that $X\neq C_X(x)$. Therefor $|[X, H]|>|H|$. If $r\neq 2$ then $|Ind_X(h)|_{r}|\geq|[X,H]|>|Ind_{G/R}(h)|_{r}$; a contradiction. If $r=2$, then $(Ind_S (\overline{h}))_2\geq 4$, hence $(Ind_{G/R}(h))_2\geq 4|[X, H]|$; a contradiction.
\end{proof}

From Lemmas \ref{factorD4}, \ref{R4} and \ref{R1R2} it follows that $\pi(K)\in\{p\}$.

\begin{lem}\label{LieType}
$S$ is a group of Lie type over the field of characteristic $p$.
\end{lem}
\begin{proof}
From Lemmas \ref{D4Alt} and \ref{Sporadic} it follows that $S$ is a group of Lie type. Let $t$ be the characteristic of the field over which $S$ is given. From Lemma \ref {Lieq3}, it follows that $|S|<|S|^3_t$. It follows from Lemmas \ref{AutD4}, \ref{R4} and \ref{R1R2} that $|\overline{G}|$ is divisible by $(k_1(q)k_2(q))^ 4(k_4(q))^2k_3(q)k_6(q)$. Lemma \ref{Lieq3Aut} implies that $t\not\in R_3(q)\cup R_6(q)$. Assume that $t\in R_1(q) \cup R_2(q)$. If $t\in R_1(q)$ put $r\in R_3(q)\cup\pi(S)$. If $t\in R_2(q)$ put $r\in R_6(q)\cup\pi(S)$. From Lemma \ref{pchast}  it follows that $S$ contains an element $h$ such that $(Ind_S(h))_t=|S|_t$ and $(Ind_S(h))_r<|S|_r$. Since $G\in \{r_3(q), r_6(q)\}^*$, we obtain $(Ind_S(h))_r=1$. It follows from Lemma \ref{Neda} that $(Ind_S(h))_t\leq 4(k_i(q))^2_t$, where $i$ is such that $t\in R_i(q)$; a contradiction with Lemma \ref{Lieq3Aut}.

 Assume that $t\in R_4(q)$. From Lemma \ref{Regular} it follows that $S$ contains an element $h$ such that $(Ind_{\overline{G}}(h))_{t'}=|\overline{G}|_ {t'}$. Using Lemma \ref {Neda} we obtain $\alpha_2<|\overline{G}|_2$ for any $\alpha\in N(\overline{G})$ such that $\alpha_{r_4(q)}<|\overline{G}|_{r_4(q)}$; a contradiction.
Thus, $t=p$.

\end{proof}

\begin{lem}\label{Liner}
$S\simeq D_4(q)$.
\end{lem}
\begin{proof}
It follows from Lemmas \ref{factorD4}, \ref{R4}, and \ref{R1R2} that $|\overline{G}|$ is a multiple of $|L|_{p'}$, in particular $|L|_p\geq |S|_p$. Lemma \ref{LieType} implies that $S$ is a group of Lie type over the field of characteristic $p$. From Lemmas \ref{hz2} and \ref{hz4}, it follows that any Sylow $t$-subgroup of $\overline{G}$ is cyclic, where $t\in R_3(q)\cup R_6(q)$. It follows from Lemma \ref{HollD4R3R6} that the Hall $R_3(q)$-subgroup and $R_6(q)$-subgroup of $\overline{G}$ are cyclic. It follows from Lemma \ref{AutD4} that $k_3(q)k_6(q)$ divides $|S|$. Hence $k_i(q)$ divides the order of some maximal torus, where $i\in\{3,6\}$. Let $a$ be a maximal number such that the set $\pi(S)\cap R_{a}(p)$ is not empty. From Lemma \ref{factorD4} and the fact that $\pi(S)\leq \pi(L)$ it follows that $a=6n$. We have $|S|_p\leq|L|_p=p^{2a}$.

Assume that $S\simeq A_m(p^l)$. The maximal number $b$ such that $R_b(p)\subseteq\pi(S)$ is equal to $l(m+1)$. We have $b=a=l(m+1)$, $|S|_p=p^{lm(m+1)/2}=p^{am/2}\leq p^{2a}$. Thus, $m\leq4$ and $l=6n/m$. From the fact that $|S|_{p'}\leq |L|_{p'}\leq|Aut(S)|_{p'}$, it is easy to get a contradiction.

Assume that $S\simeq\ ^2A_m(p^l)$. We have $a=2lm$ if $m$ is even and $a=2l(m+1)$ if $m$ is odd, $|S|_p=p^{lm(m+1)/2}$. Thus, $m\leq7$ and $l=3n/m$ if $m$ is even and $l=3n/(m+1)$ if $m$ is odd. From the fact that $|S|_{p'}\leq|L|_{p'}\leq|Aut(S)|_{p'}$, it is easy to get a contradiction.

Assume that $S\simeq B_m(p^l)$ or $S\simeq C_m(p^l)$. We have $a=2lm$, $|S|_p=p^{lm^2}=p^{am/2}\leq p^{2a}$. Thus, $m\leq4$ and $l=3n/m$. From the fact that $ |S|_{p'}\leq|L|_{p'}\leq|Aut(S)|_{p'}$, it is easy to get a contradiction.

Assume that $S\simeq\ ^2B_2(p^l)$. We have $a=4l$, $|S|_p=p^{2l}$. Hence $l=3n/2$. Thus, $|Aut(S)|_{p'}<q(q-1)(q^3+1)$; a contradiction with the fact that $q^{12}<|L|_{p'}\leq|Aut(S)|_{p'}$.

Similarly show that $S$ can not be isomorphic to$\ ^3D_4(p^l), G_2(p^l),\ ^2G_2(p^l), F_4(p^l),\ ^2F_4(p^l)$.

Assume that $S\simeq D_m(p^l)$, where $m\geq4$. We have $a=2l(m-1)$, $|S|_p=p^{lm(m-1)}=p^{am/2}\leq p^{2a}$. Thus, $m\leq4$. Hence, $l=n$ and $S\simeq L$.

Assume that $S\simeq\ ^2D_m(p^l)$. We have $a=2lm$, $|S|_p=p^{lm(m-1)}=p^{a(m-1)/2} \leq p^{2a}$. Thus, $m\leq5$. From the fact that $|S|_{p'}\leq |L|_{p'}\leq|Aut(S)|_{p'}$, it is easy to get a contradiction.

Assume that $S\simeq\ ^{\varepsilon}E_{\alpha}(p^l)$, where $\alpha \in \{6,7,8\}$, if $\alpha=6$ then $\varepsilon \in \{+,-\}$ otherwise $\alpha$ is the empty symbol. In this case it is easy to show that $|S|_p>|L|_p$; a contradiction.

\end{proof}

\begin{lem}
$G\simeq L$
\end{lem}
\begin{proof}
From Lemma \ref{Liner} it follows that $S\simeq L$. Assume that $K$ is not trivial. From Lemmas \ref {factorD4}, \ref {R4}, and \ref{R1R2} it follows that $K$ is a $p$-group. Take $h\in S$ is an element of order $k_6(q)$. Since $r_6(q)$ is not adjacent to $p$ in the graph $GK(S)$, we obtain $(Ind_S(h))_p=|S|_p=|L|_p$. Let $h'\in G$ be the preimage of the element $h$. If $h'$ acts non trivially on $K$ then $(Ind_G(h'))_p>|L|_p$; a contradiction. Thus $h'$ acts trivially on $K$. Hence any element conjugate to $h'$ acts trivially on $K$. Since $S$ is a simple group, the minimal preimage $H$ of $ S $ is contained in $C_G(K)$. Hence for any $x\in K$ we have $(Ind_G(x))_{R_3(q)\cup R_6(q)}=1$; a contradiction with Lemma \ref {GorBig}. Thus $K$ is trivial.

From Lemma \ref{Order} we obtain $S=G$.
\end{proof}

The proposition $1$ is proved.

\section{Proof the Theorem for groups $D_8(q)$}

\begin{prop}
Let $L=D_8(q)$, where $q=p^n$. If $G$ is a group with trivial centre and $N(G)=N(L)$ then $G\simeq L$.
\end{prop}

\begin{lem}\label{good2}
$G\in\{r_7(q), r_{14}(q)\}^*$
\end{lem}
\begin{proof}
The assertion of the lemma follows from Lemma \ref{Neda}.
\end{proof}
\begin{lem}\label{HollD8R3R6}
A Hall $R_7(q)$-subgroup and $R_{14}(q)$-subgroup of $G$ exists and is abelian.
\end{lem}
\begin{proof}
Note that $2,3,5\not\in R_7(q)\cup R_{14}(q)$. It follows from Lemmas \ref{good2} and \ref{GorBig} that for any $r\in R_7(q)\cup R_{14}(q)$ a Sylow $r$-subgroup of $G$ is abelian. Hence for any $r$-element $h$ we have $(Ind_G(h))_r=1$. It follows from Lemma \ref{Neda} that $(Ind_G (h))_{R_i(q)}=1$, where $i$ such that $r\in R_i(q)$. It follows from Lemma \ref{pat} that the Hall $R_i(q)$-subgroup exists and is abelian.
\end{proof}

\begin{lem}\label{factorD8}
The group $G$ includes a normal subgroup $K$ such that $S\leq \overline{G}=G/K\leq Aut(S)$, where $S$ is non abelian simple group, and $|\overline{G}|$ is a multiple of $k_7(q)k_{14}(q)$.
\end{lem}
\begin{proof}
Let $\widetilde{G}=G/O_{(R_7(q)\cup R_{14}(q))'} $.
Suppose that $\widetilde{G}$ include non-trivial solvable minimal normal subgroup $X$. We have $\pi(X)\subseteq R_i(q)$, where $i\in \{7, 14\}$. Take a Hall $R_j(q)$-subgroup $H$ of $\widetilde{G}$, where $j\in \{7, 14\}\setminus\{i\}$. Therefor $XH$ is a Frobenius group with kernel $X$. Hence $N(\widetilde{G})$ contains a number is multiple of $|X|$. It follows that $|X|$ divides $k_i(q)$. Since $X$ is a Frobenius kernel of $XH$, we obtain $|X|-1$ is multiple of $|H|$. It follows from $|H|=|G|_{R_j(q)}\geq |L|_{R_j(q)}=k_j(q)$ that $k_j(q)$ divides $k_i(q)$; a contradiction with Lemma \ref{kqi}.

Let $R$ be the socle of $\widetilde {G}$. We get that $R=S_1\times...\times S_k$, where $S_1,...,S_k$ are non abelian simple groups.
Assume that $k> 1$. It follows from the definition that the order of $S_h$ is multiple of some number of $R_7(q)\cup R_{14}(q)$, for every $1\leq h\leq k$. Note that $\omega(\widetilde{G})\subseteq \omega(G)$. Hence $R$ contains no element of order $tr$, where $t\in R_7(q)$ and $r\in R_{14}(q)$. Thus, $\pi(R)\cap(R_7(q)\cup R_{14}(q))\subseteq R_i(q)$, where $i\in\{7, 14\}$. The $S_1$ contains an element $g$ such that $(Ind_{R}(g))_{R_7 (q)\cup R_{14}(q)} =|S_1|_{R_{7}(q)\cup R_{14}(q )}$. We have $1<(Ind_{S_1}(g))_{R_7 (q)\cup R_{14}(q)}<|R|_{R_7(q)\cup R_{14}(q)}$. Similarly, as in Lemma \ref{Order} we can shows that $|R|=|R||$. In particular $|R|_{R_7(q)\cup R_{14}(q)}>(Ind_R(g))_{R_7(q)\cup R_{14}(q)}$; using Lemma \ref{factorKh} we get a contradiction.

Thus $k=1$ and $R\leq \widetilde{G}\leq Aut (R)$, where $R$ is a simple group. Since $|G|_{R_7(q)\cup R_{14}(q)}=|\widetilde{G}|_{R_7(q)\cup R_{14}(q )}$ we get $k_7(q)k_{14}(q)$ divides $|\widetilde{G}$.
\end{proof}

Let $S$, $\overline{G}$ and $K$ be similar as in Lemma \ref{factorD8}.

\begin{lem}\label{AutD8}
$|Out(S)|$ is not a multiple of $k_7(q)$ and $k_{14}(q)$.
\end{lem}
\begin{proof}
Assume that there exists $i\in\{7, 14\}$ such that $k_i(q)$ divides $|Out(S)|$.
Since $r_i(q)>13$, we get $S$ can not be isomorphic to a sporadic group or an alternating group. Hence $S$ is a group of Lie type over a field of order $l=t^r$ of characteristic $t$. From Lemmas \ref{Order} and \ref{factorKh} it follows that $|S|$ divides $|G||$. Since $|G||=|L||$, we obtain $|S|$ divides $|L|$. In particular, for any $k\in \pi (S) $, we have $|S|_k\leq|L|_k$.

From $2,3\not\in R_i (q) $, it follows that $k_i(q)$ divides order of the field automorphisms group of $S$. Hence $l\geq t^{k_i (q)}$. It follows from Lemma \ref{kqi} that $k_i(q)\in\{(q^7-1)/((7, q-1)(q-1)), (q^7+1)/((7, q+1)(q+1)\}$. Note that for any $k\in\pi(L)$, we have $|L|_k<|L|_p$ and $|L|_p=q^{56}$.
Since $q>3$, we have $|S|_t\geq t^{k_i (q)}>q^{56}=|L|_p$; a contradiction.
\end{proof}

\begin{lem}\label{Aut2D8}
The number $|\overline{G}|/|S|$ is not a multiple of $r_i(q)$, where $i\in\{7,14\}$.
\end{lem}
\begin{proof}
Assume that there exists $r_i(q)\in\pi(|\overline{G}|/|S|)$, where $i\in\{7,14\}$. Take a Hall $R_i(q)$-subgroup $H$ of $\overline{G}$. We have for every $h\in\overline{G}$, $(Ind_{\overline{G}}(h))_{R_i(q)}=1$ or $(Ind_{\overline{G}}( h))_{R_i(q)}=k_i(q)$. Using Lemma \ref{hz5}, we obtain $|H|$ divides $|S|$ or $|\overline{G}|/|S|$. Since $r_i(q)$ divides $|\overline{G}|/|S|$, we see that $k_i(q)$ divides $|\overline{G}|/|S|$; a contradiction with Lemma \ref{AutD8}.
\end{proof}

\begin{lem}\label{R8}
$R_8(q)\cap\pi(K)=\varnothing$.
\end{lem}
\begin{proof}
Assume that $R_8(q)\cap\pi(K)\neq\varnothing$. Let $\widetilde{\ }:G\rightarrow G/O_{R_8(q)'}(K)$ be a natural homomorphism, $R<\widetilde{G}$ be a minimal normal subgroup. Take $h\in G$ such that $\pi(|h|)\subseteq R_{14}(q)$. It follows from Lemma \ref{AutD8} that $\overline{h}\in S$, where $\overline{h}\in\overline{G}$ is the image of the element $h$. By the definition we have $|\pi(R)\cap R_8 (q)|\neq\varnothing$. If $R$ is not solvable, then it is easy to show that $R$ contains an $R_8(q)$-element $x$ such that $(Ind_{\widetilde{G}}(x))_{R_7(q)}<|G|_{R_7(q)}$ and $(Ind_{\widetilde{G}}(x))_{R_{14}(q)}<|G|_{R_{14}(q)}$. We have $\pi(|x|)\cap\pi(O_{R_8 (q)'}(K))=\varnothing$. It follows from Lemma \ref{factorKh} that $G$ contains $y$ such that $(Ind_G(y))_{R_7 (q)\cup R_{14}(q)}=1$; a contradiction with the fact that $G\in\{r_7(q), r_{14}(q)\}^*$ and Lemma \ref{GorBig}. Hence, $R$ is an elementary Abelian $r$-group, where $r\in R_8(q)$. Assume that $C_R(h)>1$. Let $x\in C_R(h)$. Then $(Ind_{\widetilde{G}}(x))_{R_{14}(q)}=1$. It follows from Lemma \ref{Neda} that for any $\alpha\in N(G)$ such that $\alpha_{R_{14}(q)}=1$, we have $\alpha_{R_8 (q)}=|L|_{R_8(q)}$. Therefor $(Ind_{\widetilde{G}}(hx))_{R_8(q)}=(Ind_{\widetilde{G}}(x))_{R_8(q)}$. Thus $C_{\widetilde{G}}(h)$ contains a some Sylow $r$-subgroup of $C_{\widetilde{G}}(x)$. In particular, $h$ acts trivially on $R$. Since $R$ is normal subgroup of $\widetilde{G} $, we obtain any element conjugate with $h$ acts trivially on $R$. In particular, the minimal preimage $H<\widetilde{G}$ of $S$ acts trivially on $R$. If $|S|_{R_7(q)}>1$, then $G$ contains an element $y$ such that $Ind_G(y)_{R_7(q)\cup R_{14}(q)}=1$; a contradiction. Thus, $|\overline{G}/S|$ is a multiple of $k_7(q) $; a contradiction with Lemma \ref{AutD8}.

Therefor $C_{R}(h)=1$. Thus, $|R|$ divides $Ind_{\widetilde{G}}(h)$. Since $(Ind_{\widetilde {G}}(h))_r\leq|L|_{r}$, we obtain $|R|\leq(k_8(q))^2$. The group $H$ acts faithfully on $R$. Hence $|R|-1\geq|H|$. From Lemma \ref{Aut2D8} we get that $|H|\geq 4k_7(q)k_{14}(q)>(k_8(q))^2$; a contradiction.
\end{proof}

\begin{lem}\label{D8Alt}
The group $S$ is not isomorphic to an alternating group.
\end{lem}
\begin{proof}
Assume that $S\simeq Alt_m$ for some $m>5$. Since $S$ is divided by $r_{14}(q)>13$, we get $m>13$. Therefor $|Out(S)|=2$.
From Lemma \ref {HollD8R3R6}, it follows that the Hall $R_7(q)$- and $R_{14}(q)$-subgroups are abelian. It follows from Lemmas \cite{Hall} and \cite{Tom66} that the Hall $\pi$-subgroup of $Alt_m$ is abelian only if $|\pi\cap\pi(Alt_m)|=1$. Thus, $m\geq k_i(q)$, where $i\in\{7, 14\} $. Hence $|S|>|L|$; a contradiction.
\end{proof}

\begin{lem}\label{Sporadic2}
The group $S$ is not isomorphic to any of the sporadic groups.
\end{lem}
\begin{proof}
The assertion of the lemma follows from the fact that $k^2_8(q)k_7(q)k_{14}(q)$ divides $|\overline{G}|$, $|\overline{G}|$ divides $|L|$ and \cite{Atlas}.
\end{proof}

\begin{lem}\label{MnogoR}
$\pi(K)\subseteq\{p,2\}\cup R_4(q)$.
\end{lem}
\begin{proof}
Assume that $r_j(q)\in\pi(K)$, where $j\neq 4$ and $r_j(q)\neq2$. From Lemmas \ref{factorD8} and \ref{R8} it follows that $j\not\in\{7,8,14\}$. Let $R=O_{r_j'}(G)$, $X\unlhd K/R$ be the minimal normal subgroup. Similarly, as in the Lemma \ref{AutD4}, we can show that $X$ is an elementary abelian group. Let $i=14$ if $j=2$ in the another case $i=7$. It follows from Lemmas \ref{hz2} and \ref{hz4} that every $s$-subgroup of $\overline{G}$ is cyclic, where $t\in R_7(q) \cup R_{14}(q)$. It follows from Lemma \ref{HollD8R3R6} that the Hall $R_7(q)$- and $R_{14}(q)$-subgroups of $\overline{G}$ are cyclic. Let $H\in Hall_{R_i(q)}(G/R)$, and $h\in H$ be the generating element. Then $h$ acts regularly on $[X,H]$. Hence $(Ind_{G/R}(h))_{r_j(q)}>|[X,H]|\geq k_i(q)+ 1$. if $j\in\{3,6,5,10,12\}$ then $k_i(q)>|L|_{r_l(q)}$; a contradiction.
If $j\in\{1, 2\}$ and $r_j(q)\neq\{2 \}$, then $(Ind_{G/R}(h))_{r_j(q)}\leq(k_j(q))^6<k_i(q) $; a contradiction.

\end{proof}

\begin{lem}\label{LieType8}
The $S$ is a group of Lie type over the field of characteristic $p$.
\end{lem}
\begin{proof}
From Lemmas \ref{D8Alt} and \ref{Sporadic2} it follows that $S$ is a group of Lie type over a field of order $t^m$ for some prime $t$. The assertion of the lemma follows from the fact that $|\overline{G}|<|S|^3_tt^m$ and Lemmas \ref{factorD8}, \ref{R8}, \ref{MnogoR}.
\end{proof}
\begin{lem}\label{Liner8}
$S\simeq D_8(q)$.
\end{lem}
\begin{proof}
It follows from Lemmas \ref{factorD8}, \ref{R8}, and \ref{MnogoR} that $|\overline{G}|$ is a multiple of $|L|_{(\{p,2\}\cup R_4(q))'}$. Lemma \ref{LieType8} implies that $S$ is a group of Lie type over the field of characteristic $p$. From Lemmas \ref{hz2} and \ref{hz4}, it follows that any Sylow $t$-subgroup of $\overline{G}$ is cyclic, where $t\in R_7(q)\cup R_{14}(q)$. It follows from Lemma \ref{HollD8R3R6} that the Hall $R_7(q)$-subgroup and $R_{14}(q)$-subgroup of $\overline{G}$ are cyclic. It follows from Lemma \ref{Aut2D8} that $k_7(q)k_{14}(q)$ divides $|S|$. Hence $k_i(q)$ divides the order of some maximal torus, where $i\in\{7, 14\}$. Let $a$ be a maximal number such that the set $\pi(S)\cap R_{a}(p)$ is not empty. From Lemma \ref{factorD4} and the fact that $\pi(S)\leq \pi(L)$ it follows that $a=14n$. We have $|S|_p\leq|L|_p=p^{4a}$.

Assume that $S\simeq A_m(p^l)$. We have $a=l(m+1)$, $|S|_p=p^{lm(m+1)/2}=p^{am/2}\leq p^{4a}$. Thus, $m\leq8$ and $l=(14n-1)/m$. Thus $S$ contains a torus of order $(q^{14}-1)/(m+1, p^l-1)$. In particular $S$ contains an element of order $r_{n7}(p)r_{n14}(p)$; a contradiction with Lemma \ref{good2}.

Assume that $S\simeq\ ^2A_m(p^l)$. We have $a=2l(m+1)$ if $m$ is even and $a=2lm$ if $m$ is odd, $|S|_p=p^{lm(m+1)/2}$. Thus, $m\leq8$ and $l=3n/m$ if $m$ is even and $l=3n/(m + 1) $ if $ m $ is odd. It is easy to obtain a contradiction from the fact that $|Aut(S)|_{p'}>|L|_{(\{p,2\}\cup R_4(q))'}$.

Assume that $S\simeq B_m(p^l)$ or $S\simeq B_m(p^l)$. We have $a=2lm$, $|S|_p=p^{lm^2}=p^{am/2}\leq p^{2a}$. Thus, $m\leq4$ and $l=3n/m$. From the fact that $|S|_{p'}\leq |L|_{(\{2, p\}\cup R_4(q))'}\leq|Aut(S)|_{p'}$, it is easy to obtain a contradiction.

Assume that $S\simeq\ ^2B_2(p^l)$. We have $l\leq a<4l$, $|S|_p=p^{2l}$. Hence $14n\leq l<7n/2$. Thus, $|Aut(S)|_{p'}<q(q-1)(q^2+1)$; a contradiction with the fact that $q^{12}<|L|_{(\{2,p\}\cup R_4(q))'}\leq|Aut(S)|_{p'}$.

Similarly show that $S$ can not be isomorphic to$\ ^3D_4(p^l), G_2(p^l),\ ^2G_2(p^l), F_4(p^l),\ ^2F_4(p^l)$.

Assume that $S\simeq D_m(p^l)$. We have $a=2l(m-1)$, $|S|_p=p^{lm(m-1)}=p^{am/2}\leq p^{4a}$. Thus, $m\leq8$. If $m=8$ then $l=n$ and consequently $S\simeq L$. Assume that $m<8$. If $m=7$ or $m=5$ then from Lemma \ref{Neda} we yet that $S\not\in\{r_7(q),r_{14}(q)\}^*$. Therefor $G\not\in \{r_7(q),r_{14}(q)\}^*$, a contradiction with Lemma \ref{good2}. If $m=6$ or $m=4$ then $|Aut(S)|_{(\{2,p\}\cup R_4(q))'}<|L|_{(\{2,p\}\cup R_4(q))'}$, a contradiction.

Assume that $S\simeq\ ^2D_m(p^l)$. We have $a=2lm$, $|S|_p=p^{lm(m-1)}=p^{a(m-1)/2} \leq p^{4a}$. Thus, $m\leq9$. From the facts that $|S|\leq |L|$ and $|L|_{(\{2,p\}\cup R_4(q))'}\leq|Aut(S)|_{(\{2,p\}\cup R_4(q))'}$, it is easy to get a contradiction.

Assume that $S\simeq\ ^{\varepsilon}E_{\alpha}(p^l)$, where $\alpha \in \{6,7,8\}$, if $\alpha=6$ then $\varepsilon \in \{+,-\}$ otherwise $\alpha$ is the empty symbol. In this case it is easy to show that $|S|_p>|L|_p$; a contradiction.
\end{proof}

\begin{lem}
$G\simeq L$
\end{lem}
\begin{proof}
From Lemma \ref{Liner8} it follows that $S\simeq L$. Assume that $K$ is not trivial. From Lemmas \ref{factorD8}, \ref{R8}, and \ref{MnogoR} it follows that $K$ is a $\{2,p\}\cup R_4(q)$-group. Take $h\in S$ is an element of order $k_{14}(q)$. Since $r_{14}(q)$ is not adjacent to $p$ in the graph $GK(S)$, we obtain $(Ind_S(h))_p=|S|_p=|L|_p$. Assume that $p\in\pi(K)$. Let $\widetilde{G}=G/O_{p'}(G)$, $h'\in \widetilde{G}$ be the preimage of the element $h$. If $h'$ acts non trivially on $O_p(\widetilde{G})$ then $(Ind_{\widetilde{G}}(h'))_p>|L|_p$; a contradiction. Thus $h'$ acts trivially on $O_p(\widetilde{G})$. Hence any element conjugate to $h'$ acts trivially on $O_p(\widetilde{G})$. Since $S$ is a simple group, the minimal preimage $H<\widetilde{G}$ of $S$ is contained in $C_G(O_p(\widetilde{G}))$. Hence for any $x\in O_p(\widetilde{G})$ we have $(Ind_{\widetilde{G}}(x))_{R_3(q)\cup R_6(q)}=1$. Since $p$ is not divisible $O_{p'}(G)$, we obtain that $G$ contain $x'$ such that $(Ind_{G}(x'))_{R_3(q)\cup R_6(q)}=1$; a contradiction with Lemma \ref{GorBig}. Thus $p\not\in\pi(K)$. Similarly, we can show that $K$ is trivial.

From Lemma \ref{Order} we obtain $S=G$.
\end{proof}

config: arxiv


TITLE: What property is being used?
QUESTION [1 upvotes]: I'm looking over a stats problem and there is a jump which I don't understand. It is:
$$\theta_1^{-n}(\prod_{i=1}^{n}x_i)^{1-\theta_1}=\theta_1^{-n}exp((\theta_1-1)(-\sum_{i=1}^{n}logx_i))$$
This question is about likelihood functions. We also know that $f(x|\theta)=\theta x^{\theta-1}$ for $x\epsilon(0,1)$ with $\theta>0$

REPLY [0 votes]: Use the fact that for $x \in \mathbb{R}_{++},$ $$x = e^{\log{x}}.$$ Also $$\log{(ab)} = \log{(a)} + \log{(b)}$$ for $a,b \in \mathbb{R}_{++}$. Here $\mathbb{R}_{++}$ indicates strictly positive numbers.


TITLE: Show that $S$ has elements of order $36$, $39$ but does not have an element of order $37$ or $38$.
QUESTION [1 upvotes]: This question was asked in my quiz and I  was unable to solve one part , so I am asking it here.

Let S denote the group of all those permutations of the english alphabet that fix the letters $T,E,N,D,U,L,K,A,R$ .Other letters may or may not be fixed. Show that $S$ has elements of order $36 ,39$ but does not have an element of order $37$ or $38$.

Attempt : Order of Group is $17!$. So , Group can have element of order $3$ and $13$ by Cauchy theorem (call $a$ and $b$ respectively) and so $ab$ will have order $39$. Also ,as order of element divides order of group so order $37$ or $38$ is not possible.

But I am not able to think about element of order $36$.
Kindly help with that !

REPLY [4 votes]: Take any $4$ letters which might not be fixed, call them $a_1,...,a_4$. Pick other $9$ letters which might not be fixed and are distinct from the $a_i$'s, call them $b_1,...,b_9$. Then the permutation $(a_1\ a_2\ a_3\ a_4)(b_1\ b_2\ ...\ b_9)$ has order lcm$(4,9)=36$.
By the way, it is in general not true that the order of $ab$ is the product of orders of $a$ and $b$, even if these orders are coprime. So in order to prove there is an element of order $39$ just repeat my argument for order $36$, just this time pick $3$ elements $a_1, a_2, a_3$ and $13$ elements $b_1,...,b_{13}$.


TITLE: Existence of a function extending "even/odd" to infinite sets
QUESTION [1 upvotes]: Let $F$ be the function that sends a finite subset of naturals to the set $\{\text{even}, \text{odd}\}$, according to the parity of its size. Can $F$ be extended to a $G$ on all subsets, so that for any (potentially infinite) set $A$ and element $a$ of $A$, we have the following?
$$G(A-\{a\})=\begin{cases}
\text{even}, & \text{if}\; G(A)=\text{odd} \\
\text{odd},  & \text{if}\; G(A)=\text{even}
\end{cases}$$
Transfinite induction doesn't seem to work.

REPLY [1 votes]: Here is a (non-canonical) construction:
If $A,B$ are infinite subsets of $\mathbb N$, we say that they are equivalent if they differ by a finite set.  That is, if the set of natural numbers which are in one but not both is finite.  If we think of such subsets as identifying real numbers in $(0,1]$ then equivalence means that the two decimals coincide from some point on.
Choose a representative of each equivalence class and assign its parity arbitrarily (for example, say that all the representatives are Even).  Then extend the notion of parity to the whole class via the given rule.
Note that this notion of parity does not appear to extend to infinite subsets.  That is to say, if $B\subset A$ and $A,B$ are both infinite and the parity of both are known, then we can't infer the parity of the complement, $A-B$.  My sense is that there is no notion of parity which works in this sense, but I am not sure.


text:
\begin{document}
\title{Moment-angle manifolds and connected sums of sphere products}
\author[F.~Fan, L.~Chen, J.~Ma \& X.~Wang]{Feifei Fan, Liman Chen, Jun Ma and Xiangjun Wang}
\thanks{The authors are supported by NSFC grant No. 11261062 and SRFDP No.20120031110025}
\address{Feifei Fan, School of Mathematical Sciences and LPMC, Nankai University, Tianjin 300071, P.~R.~China}
\email{fanfeifei@mail.nankai.edu.cn}
\address{Liman Chen, School of Mathematical Sciences and LPMC, Nankai University, Tianjin 300071, P.~R.~China}
\email{chenlimanstar1@163.com}
\address{Jun Ma, School of Mathematical Sciences, Fudan University, Shanghai 200433, P.~R.~China}
\email{tonglunlun@gmail.com}
\address{Xiangjun Wang, School of Mathematical Sciences and LPMC, Nankai University, Tianjin 300071, P.~R.~China}
\email{xjwang@nankai.edu.cn}
\subjclass[2010]{Primary 13F55, 14M25, 55U10; Secondary 57R18, 57R19}
\date{}
\maketitle
\begin{abstract}
Corresponding to every finite simplicial complex $K$, there is a
¡°moment-angle¡± complex $\mathcal {Z}_{K}$; if $K$ is a triangulation of a sphere, $\mathcal {Z}_{K}$ is a compact manifold.
The question of whether $\mathcal {Z}_{K}$ is a connected sum of sphere products was considered in {\cite[\S 11]{BM06}}.
So far, all known examples of moment-angle manifolds which are homeomorphic to connected sums of sphere products have the property that every product is of exactly two spheres.
In this paper, we give a example whose cohomology ring is isomorphic to that of a connected sum of sphere products with one product of three spheres. We also give some general properties of this kind of moment-angle manifolds.
\end{abstract}

\section{Introduction}
Throughout this paper, we assume that $m$ is a positive integer and $[m]=\{1,2,\dots,m\}$.
For an abstract simplicial complex $K$ with $m$ vertices labeled by $[m]$ and a sequence
$I=(i_1,\dots,i_k)\subseteq[m]$ with $1\leq i_1\leq\cdots\leq i_k\leq m$,
we denote by $K_I$ the \emph{full subcomplex} of $K$ on $I$,
and $\wh I=[m]\setminus I$.
\subsection{Moment-angle complex}
Given a simple polytope $P$ with $m$ fecets, Davis and Januszkiewicz \cite{DJ91} constructed a manifold  $\mathcal{Z}_P$ with an action of a real torus $T^m$. After that Buchstaber and Panov  \cite{BP00} generalized this definition to any simplicial complex $K$, that is
\[\mathcal {Z}_{K}=\bigcup_{\sigma\in K}(D^2)^{\sigma}\times (S^1)^{[m]\setminus \sigma},\]
and named it the \emph{moment-angle complex} associated to $K$, whose study connects algebraic geometry, topology, combinatorics, and commutative algebra.
This cellular complex is always $2$-connected and
has dimension $m+n+1$, where $n$ is the dimension of $K$.

It turns out that the algebraic topology of a moment-angle complex $\mathcal {Z}_{K}$,
such as the cohomology ring and the homotopy groups is intimately related to the combinatorics of the underlying simplicial complex $K$.

\subsection{Moment-angle manifold}
Now suppose that $K$ is an $n$-dimentional simplicial sphere (a triangulation of a sphere) with $m$ vertices. Then, as shown by Buchstaber and Panov \cite{BP00},
the moment-angle complex $\mathcal {Z}_{K}$ is a manifold of dimension $n+m+1$, referred to as a \emph{momnet-angle manifold}.
In particular, if $K$ is a polytopal sphere (see Definition \ref{def:1}), or more generally a starshaped sphere (see Definition \ref{def:0}), then $\mathcal {Z}_{K}$ admits a smooth structure.
\begin{Def}\label{def:1}
A \emph{polytopal sphere} is a triangulated sphere isomorphic to the boundary complex of a simplicial polytope.
\end{Def}

\begin{Def}\label{def:0}
a simplicial sphere $K$ of dimension $n$ is said to be \emph{sastarshaped} if there is a geometric realization $|K|$ of $K$ in $\mathbb{R}^n$ and a point $p\in \mathbb{R}^n$
with the property that each ray emanating from $p$ meets $|K|$ in exactly one point.
\end{Def}

\begin{rem}
A polytopal sphere is apparently a starshaped sphere, but for $n\geq 3$, there are examples that are starshaped and not polytopal. The easiest such example is given by the \emph{Br\"uckner sphere} (see \cite{GS67}).
\end{rem}

The topology of a moment-angle manifold can be quite complicated.
The complexity increases when the dimension $n$ of the associated simplicial sphere $K$ increases.
for $n=0$, $\mathcal {Z}_{K}$ is $S^3$. For $n=1$, $K$ is the boundary of a polygon,
and $\mathcal {Z}_{K}$ is a connected sum of sphere products. In higher dimensions, the situation becomes much more complicated.
On the other hand, McGavran \cite{M79} showed that, for any $n>0$, there are infinitely
many $n$-dimensional polytopal spheres whose corresponding moment-angle manifolds are connected sums of sphere products.

\begin{thm}[McGavran, see {\cite[Theorem 6.3]{BM06}}]\label{thm:2}
Let $K$ be a polytopal sphere dual to the simple polytope obtained from the $k$-simplex by cutting off vertices for $l$ times.
Then the corresponding moment-angle manifold is homeomorphic to a connected sum of sphere products
\[\mathcal {Z}_{K}\cong \overset{l}{\underset{j=1}\#}j\binom{l+1}{j+1}S^{j+2}\times S^{2k+l-j-1}.\]
\end{thm}

For $k=2$ or $3$, the above theorem gives all moment-angle manifolds which are homomorphic to connected sums of sphere products
(see {\cite[Proposition 11.6]{BM06}}).
Nevertheless, in higher dimension they are not the only ones whose cohomology ring is isomorphic
to that of a connected sum of sphere products. Bosio and Meersseman {\cite[\S 11]{BM06}} gave many other examples of moment-angle manifolds
whose cohomology rings have this property.
We notice that all examples of connected sums of sphere products given in \cite{BM06} have the property that every product is of two spheres, this leads to a question:
\begin{que}
If $\mathcal {Z}_{K}$ is a connected sum of sphere products, is it ture that every product is of exactly two spheres?
\end{que}
In this paper (Proposition \ref{prop:1}),
we give a negative answer to this question at the aspect of cohomology rings, by constructing a $3$-dimentional polytopal
sphere, so that the cohomology ring of the corresponding moment-angle manifold is isomorphic to the cohomology ring of the connected sum of sphere products
\[S^3\times S^3\times S^6\#(8)S^5\times S^7\#(8)S^6\times S^6.\]

\section{cohomology ring of moment-angle complex}\label{sec:2}
\begin{Def}\label{def:2}
Let $K$ be a simplicial complex with vertex set $[m]$.
A \emph{missing face} of $K$ is a sequence $(i_1,\dots,i_k)\subseteq[m]$ such that $(i_1,\dots,i_k)\not\in K$,
but every proper subsequence of $(i_1,\dots,i_k)$ is a simplex of $K$. Denote by $MF(K)$ the set of all missing faces of $K$.
\end{Def}

From definition \ref{def:2}, it is easy to see that if $K_I$ is a full subcomplex of $K$, then $MF(K_I)$ is a subset of $MF(K)$. Concretely,
\[MF(K_I)=\{\sigma\in MF(K):\sigma\subseteq I\}.\]

Let $R[m]=R[v_1,\dots,v_m]$ denote the graded polynomial algebra over $R$, where $R$ is a field or $\mathbb{Z}$, $\mathrm{deg}v_i=2$.
The \emph{face ring} (also known as the \emph{Stanley-Reisner ring}) of a simplicial complex $K$ on the vertex set [m] is the quotient ring
\[R(K)=R[m]/\mathcal {I}_K,\]
where $\mathcal {I}_K$ is the ideal generated by all square free monomials $v_{i_1}v_{i_2}\cdots v_{i_s}$
such that $(i_1,\dots,i_s)\in MF(K)$.

The following result is used to calculate the cohomology ring of $\mathcal {Z}_{K}$,
which is proved by Buchstaber and Panov \cite[Theorems 7.6]{BP02} for the case over a field, \cite{BBP04} for the general case;
see also \cite[Theorem 4.7]{P08}. Another proof of Theorem \ref{thm:1} for the case over $\mathbb{Z}$ was given by Franz \cite{M06}.

\begin{thm}[Buchstaber-Panov, {\cite[Theorem 4.7]{P08}}]\label{thm:1}
Let $K$ be a abstract simplicial complex with $m$ vertices. Then the cohomology
ring of the moment-angle complex $\mathcal {Z}_{K}$ is given by the isomorphisms
\begin{align*}
H^*(\mathcal {Z}_{K};R)\cong \mathrm{Tor}_{R[m]}^{*,*}(R(K),R)\cong\bigoplus_{I\subseteq [m]} \w {H}^*(K_I;R)
\end{align*}
where
\[H^p(\mathcal {Z}_{K};R)\cong \bigoplus_{\substack{J\subseteq[m]\\-i+2|J|=p}}\mathrm{Tor}^{-i,\,2|J|}_{R[m]}(R(K),R)\]
and
\[\mathrm{Tor}^{-i,\,2|J|}_{R[m]}(R(K),R)\cong \w {H}^{|J|-i-1}(K_J;R).\]
\end{thm}

\begin{rem}\label{rem:1}
There is a canonical ring structure on $\bigoplus_{I\subseteq [m]} \w {H}^*(K_I)$
(called the \emph{Hochster ring} and denoted by $\mathcal {H}^{*,*}(K)$, where $\mathcal {H}^{i,J}(K)=\w H^i(K_J)$) given by the maps
\[\eta:\w H^{p-1}(K_I)\otimes H^{q-1}(K_J)\to H^{p+q-1}(K_{I\cup J}),\]
which are induced by the canonical simplicial inclusions $K_{I\cup J}\to K_I*K_J$ (join
of simplicial complexes) for $I\cap J=\emptyset$ and zero otherwise.
Precisely, Let $\w C^q(K)$ be the $q$th reduced simplicial cochain group of $K$.
For a oriented simplex $\sigma=(i_1,\dots,i_p)$ of $K$ (the orientation is given by the order of vertices of $\sigma$), denote by
$\sigma^*\in\w C^{p-1}(K)$ the basis cochain corresponding to $\sigma$; it takes value $1$ on $\sigma$ and vanishes on all other simplices.
Then for $I,J\in[m]$ with $I\cap J=\emptyset$, we have isomorphisms of reduced simplicial cochains
\begin{align*}
\mu :\w C^{p-1}(K_I)\otimes \w C^{q-1}(K_J)&\to \w C^{p+q-1}(K_I*K_J), \quad p,q\geq 0\\
\sigma^*\otimes\tau^*&\mapsto (\sigma\sqcup \tau)^*
\end{align*}
where $\sigma\sqcup \tau$ means the juxtaposition of $\sigma$ and $\tau$.
Given two cohomology classes $[c_1]\in \w H^{p-1}(K_I)$ and $[c_2]\in\w H^{q-1}(K_J)$, which are represented by the cocycles
$\sum_i \sigma_i^*$ and $\sum_j\tau_j^*$ respectively. Then
\[\eta([c_1]\otimes[c_2])=\varphi^*([\mu(\sum_{i,j}\sigma_i^*\otimes\tau^*_j)]),\]
where $\varphi:K_{I\cup J}\to K_I*K_J$ is the simplicial inclusion.

We denote by $\psi([c])$ the inverse image of a class $[c]\in \bigoplus_{I\subseteq [m]} \w {H}^*(K_I)$
by the composition of the two isomorphisms in Theorem \ref{thm:1}.
Given two cohomology classes $[c_1]\in \w H^{p}(K_I)$ and $[c_2]\in\w H^{q}(K_J)$, define
\[[c_1]*[c_2]=\eta([c_1]\otimes[c_2]).\]
Bosio and Meersseman proved in \cite{BM06} (see also \cite[Proposition 3.2.10]{BP13}) that, up to sign
\[\psi([c_1])\smile \psi([c_2])=\psi([c_1]*[c_2]).\]
\end{rem}

\begin{rem}\label{rem:2}
Baskakov showed in \cite{B02} (see {\cite[Theorem 5.1]{P08}}) that the isomorphisms in Theorem \ref{thm:1} are functorial with respect to simplicial maps (here we only consider simplicial inclusions). That is, for a simplicial inclusion $i:K'\hookrightarrow  K$
(suppose the vertex sets of $K'$ and $K$ are $[m']$ and $[m]$ respectively) which
induces natural inclusions
\[\phi: \mathcal{Z}_{K'}\hookrightarrow  \mathcal {Z}_{K}\]
and
\[i|_{K'_I}: K'_I\hookrightarrow  K_I,\ \text{ for each } I\subseteq [m'],\]
there is a commutative diagram of algebraic homomorphisms
\[\begin{CD}
H^*(\mathcal{Z}_{K})@>\phi^*  >>H^*(\mathcal {Z}_{K'})\\
@V\cong VV @VV\cong V\\
\underset{I\subseteq [m]}\bigoplus \w {H}^*(K_I)@>\bigoplus_I(i|_{K'_I})^* >>\underset{I\subseteq [m']}\bigoplus\w {H}^*(K'_I)
\end{CD}\]
\end{rem}
Actually, there are three ways to calculate the integral cohomology ring of a moment-angle complex $\mathcal {Z}_{K}$.

(1) The first is to calculate the Hochster ring $\mathcal {H}^{*,*}(K)$ of $K$ and apply the isomorphisms in Theorem \ref{thm:1}.

(2) The second is to calculate $\mathrm{Tor}^{*,*}_{\mathbb{Z}[m]}(\mathbb{Z}(K),\mathbb{Z})$ by means of the Koszul resolution
(\cite[Theorem 7.6 abd Theorem 7.7]{BP02}), that is
\[\mathrm{Tor}_{\mathbb{Z}[m]}^{*,*}(\mathbb{Z}(K),\mathbb{Z})\cong H(\Lambda[u_1,\dots,u_m]\otimes \mathbb{Z}(K),d),\]
where $\Lambda[u_1,\dots,u_m]$ is the exterior algebra over $\mathbb{Z}$ generated by $m$ generators. On the right side, we have
\[\mathrm{bideg}u_i=(-1,2),\quad \mathrm{bideg}v_i=(0,2),\quad du_i=v_i,\quad dv_i=0.\]
In fact, there is a simpler way to calculate the cohomology of this differential graded algebra by applying the following result
\begin{prop}[{\cite[Lemma 3.2.6]{BP13}}]
The projection homomorphism
\[\varrho: \Lambda[u_1,\dots,u_m]\otimes \mathbb{Z}(K)\to A(K)\]
induces an isomorphism in cohomology, where $A(K)$ is the quotient algebra
\[A(K)=\Lambda[u_1,\dots,u_m]\otimes \mathbb{Z}(K)/(v_i^2=u_iv_i=0,\ 1\leq i\leq m).\]
\end{prop}

(3) The third way is to use the Taylor resolution for $\mathbb{Z}(K)$ to calculate $\mathrm{Tor}_{\mathbb{Z}[m]}(\mathbb{Z}(K),\mathbb{Z})$.
This was introduced first by Yuzvinsky in \cite{Y99}. Wang and Zheng \cite{WZ13} applied this method to toric topology. Concretely, let
$\mathbb{P}=MF(K)$, and let $\Lambda[\mathbb{P}]$ be the exterior algebra generated by $\mathbb{P}$.
Given a monomial $u=\sigma_{k_1}\sigma_{k_2}\cdots\sigma_{k_r}$ in $\Lambda[\mathbb{P}]$, let
\[S_u=\sigma_{k_1}\cup\sigma_{k_2}\cdots\cup\sigma_{k_r}.\]
Define $\mathrm{bideg}\,u=(-r,2|S_u|)$, and define
\[\partial_i(u)=\sigma_{k_1}\cdots\wh{\sigma_{k_i}}\cdots\sigma_{k_r}=\sigma_{k_1}\cdots\sigma_{k_{i-1}}\sigma_{k_{i+1}}\cdots\sigma_{k_r}.\]

Let $(\Lambda^{*,*}[\mathbb{P}],d)$ be the cochain complex (with a different product structure from $\Lambda[\mathbb{P}]$) induced from the bi-graded exterior algebra on $\mathbb{P}$.
The differential $d: \Lambda^{-q,*}[\mathbb{P}]\to \Lambda^{-(q-1),*}[\mathbb{P}]$ is given by
\[d(u)=\sum_{i=1}^q (-1)^i\partial_i(u)\delta_i,\]
where $\delta_i=1$ if $S_u=S_{\partial_i(u)}$ and zero otherwise. The product structure in $(\Lambda^{*,*}[\mathbb{P}],d)$ is given by
\[u\times v=
\begin{cases}
u\cdot v \quad &\text{if } S_u\cap S_v=\emptyset,\\
0 &\text{otherwise,}
\end{cases}\]
where {\LARGE$\cdot$} denote the ordinary product in the exterior algebra $\Lambda[\mathbb{P}]$.
\begin{prop}[see {\cite[Theorem 2.6 and Theorem 3.2]{WZ13}}]
There is a algebraic isomorphism
\[\mathrm{Tor}_{\mathbb{Z}[m]}^{*,*}(\mathbb{Z}(K),\mathbb{Z})\cong H(\Lambda^{*,*}[\mathbb{P}],d)\]
\end{prop}

\section{Construction of a polytopal $3$-sphere with eight vertices}
In this section, we construct a $3$-dimensional polytopal sphere $K$ with eight vertices,
such that the cohomology ring of the correponding moment-angle manifold $\mathcal {Z}_{K}$
is isomorphic to the the cohomology ring of a connected sum of sphere products with one product of three spheres.
\begin{cons}\label{cons:1}
We construct $K$ by three steps. First give a $2$-dimensional simplicial complex $K_0$ with $4$ vertices shown in Figure \ref{fig:1}.
\[MF(K_0)=\{(1,2,3), (1,3,4)\}.\] It has two subcomplex
$K_1$ and $K_2$ also shown in Figure \ref{fig:1}. Next let $L_1=K_0\cup \mathrm{cone}(K_1)$ with a new vertex $5$.
(i.e., $L_1$ is the mapping cone of the inclusion map $K_1\hookrightarrow K_0$), and let $L_2=K_0\cup \mathrm{cone}(K_2)$ with a new vertex $6$.
Let $K_0'=L_1\cup L_2$ be a simplicial complex obtained by gluing $L_1$ and $L_2$ along $K_0$ (see Figure \ref{fig:2}). Then
\[MF(K_0')=\{(1,2,3), (1,3,4), (2,3,5), (3,4,6), (5,6)\}.\]
Note that $K_0'$ can be viewed as a "thick" $2$-sphere with two $3$-simplices $(1,2,4,5)$ and $(1,2,4,6)$,
shown shaded in Figure \ref{fig:2}.
$K_0'$ has two subcomplexes $K_1'$ and $K_2'$ (see Figure \ref{fig:2}),
which are all triangulations of $S^2$. Let $\mathrm{cone}(K_2')$ be the cone of $K_2'$ with a new vertex $8$.
Then it is easy to see that $K'=K_0'\cup \mathrm{cone}(K_2')$ is a triangulation of $D^3$ and its boundary
is $K_1'$. Finally, let $K=K'\cup \mathrm{cone}(K_1')$ with a new vertex $7$.
Clearly, $K$ is a triangulation of $S^3$, and the missing faces of $K$ are
\begin{equation}\label{eq:0}
\begin{split}
MF(K)=\{&(1,2,3),(1,3,4),(2,3,5),(3,4,6),(5,6),\\
&(1,4,7),(4,6,7),(1,2,8),(2,5,8),(7,8)\}.
\end{split}
\end{equation}
\begin{figure}[!ht]
\captionsetup[subfloat]{labelformat=empty,textfont=normalsize}
\centering
\subfloat[$K_0$]{\includegraphics[scale=0.35]{K0.pdf}}
\subfloat[$K_1$]{\includegraphics[scale=0.35]{K1.pdf}}
\subfloat[$K_2$]{\includegraphics[scale=0.35]{K2.pdf}}
\caption{$K_0$, $K_1$ and $K_2$.}\label{fig:1}
\end{figure}
\begin{figure}[!ht]
\captionsetup[subfloat]{labelformat=empty,textfont=normalsize}
\centering
\subfloat[$K_0'$]{\includegraphics[scale=0.4]{P0.pdf}}
\subfloat[$K'_1$]{\includegraphics[scale=0.4]{P1.pdf}}
\subfloat[$K'_2$]{\includegraphics[scale=0.4]{P2.pdf}}
\caption{$K_0'$, $K'_1$ and $K'_2$.}\label{fig:2}
\end{figure}
\end{cons}

Gr\"unbaum and Sreedharan \cite{GS67}
gave a complete enumeration of the simplicial 4-polytopes with 8 vertices. A direct verification shows that $K$ we construct above
is isomorphic to the boundary of $P_{28}^8$ (a 4-polytope with $18$ facets) in \cite{GS67}.
Then $K$ is actually a polytopal sphere. From the construction we know that all $3$-simplices of $K$ are
\[\begin{split}
&(1,2,4,5),\ (1,2,4,6),\ (1,2,5,7),\ (1,2,6,7),\ (1,3,5,7),\ (1,3,6,7)\\
&(2,3,4,7),\ (2,3,6,7),\ (2,4,5,7),\ (3,4,5,7),\ (1,4,5,8),\ (1,4,6,8)\\
&(1,3,5,8),\ (1,3,6,8),\ (2,3,4,8),\ (2,3,6,8),\ (2,4,6,8),\ (3,4,5,8).
\end{split}\]

\section{connected sums of sphere products}
In the first part of this section, we calculate the cohomology ring of $\mathcal {Z}_K$ corresponding to the polytopal sphere $K$
constructed in the last section. In the second part, we give some general properties for the moment-angle manifolds whose cohomology ring is isomorphic to
that of a connected sum of sphere product.
\labelformat{subfigure}{#1}
\begin{figure}
\captionsetup[subfloat]{labelformat=simple,textfont=normalsize,labelfont=normalsize}
\centering
\subfloat[\label{subf:1}]{\includegraphics[scale=0.22]{1.pdf}}\
\subfloat[\label{subf:4}]{\includegraphics[scale=0.32]{4.pdf}}\
\subfloat[\label{subf:5}]{\includegraphics[scale=0.32]{5.pdf}}
\caption{}\label{fig:3}
\end{figure}

\begin{prop}\label{prop:1}
For the polytopal sphere $K$ defined in Construction \ref{cons:1}, the cohomology ring of the corresponding moment-angle manifold
$\mathcal {Z}_{K}$ is isomorphic to the cohomology ring of \[S^3\times S^3\times S^6\#(8)S^5\times S^7\#(8)S^6\times S^6\].
\end{prop}

We will calculate $H^*(\mathcal {Z}_{K})$ in the first way introduced in section \ref{sec:2}. Therefore we need first to calculate the reduced cohomology rings of
all full subcomplexes of $K$. Note first the following obvious fact: Let $\Gamma$ be a simplicial complex with vertex set $[m]$. Define
$\mathcal {I}=\bigcup_{\sigma\in MF(\Gamma)}\sigma.$
If $\mathcal {I}\neq [m]$, then $\Gamma=K_{\mathcal {I}}*\Delta^{m-|\mathcal {I}|-1}$, and therefore $\Gamma$ is contractible.

Now we do this work in $6$ cases according to the cardinality of $I$ for $K_I$.
\begin{enumerate}[(1)]
\item Since the case $|I|=1$ is trivial, we start with the case $|I|=2$.
In this case, from \eqref{eq:0}, it is easy to see that $\w {H}^*(K_I)\neq 0$ if and only if $I=(5,6)$ or $(7,8)$,
and if so, $\w {H}^*(K_I)\cong \w {H}^0(K_I)\cong \mathbb{Z}$. Denote by $a_1$ (respectively $a_2$) a generator of $\w {H}^0(K_I)$
for $I=(5,6)$ (respectively $(7,8)$).

\item \label{case:2} $|I|=3$. It is easy to see that the union of any two missing faces of $K$ contains at least four vertices.
Combining the preceding argument we have that $\w {H}^*(K_I)\neq 0$
if and only if $I$ is one of the eight missing faces with three vertices in $MF(K)$,
and if so, $\w {H}^*(K_I)\cong \w {H}^1(K_I)\cong \mathbb{Z}$, whose generator we denote by $b_i$ ($1\leq i\leq 8$).

\item $|I|=4$. An easy observtion shows that the union of any three missing faces of $K$ contains at least five vertices,
and $K$ has no missing face with four vertices.
So if $K_I$ is not contractible, then it has exactly two missing faces.
Thus from \eqref{eq:0}, the form of $MF(K_I)$ is one of $\{(v_1,v_2),(v_3,v_4)\}$, $\{(v_1,v_2,v_3),(v_2,v_3,v_4)\}$ and $\{(v_1,v_2),(v_1,v_3,v_4)\}$ , for which
the corresponding simplicial complexes are respectively $A$, $B$ and $C$ shown in Figure \ref{fig:3}.
It is easy to see that they are all homotopic to $S^1$. In Table \ref{tab:1} we list all non-contractible full subcomplexes $K_I$ of $K$ for $|I|=4$
(each $I_j$ contains vertex 1). Denote by $\alpha_j$ (respectively $\alpha_j'$) a generator of
$\w {H}^*(K_{I_j})\cong \w {H}^1(K_{I_j})\cong \mathbb{Z}$ (respectively $\w {H}^*(K_{\wh{I_j}})$) for $0\leq j\leq 8$.

\begin{table}\renewcommand{\arraystretch}{1.5}
\begin{tabular}{c|c|c|c|c|c|c}\hline
&$K_{I_0}$&$K_{\wh{I_0}}$&$K_{I_1}$&$K_{\wh{I_1}}$&$K_{I_2}$&$K_{\wh{I_2}}$\\ \hline
vertex set&$\{1,2,3,4\}$&$\{5,6,7,8\}$&$\{1,2,3,5\}$&$\{4,6,7,8\}$&$\{1,2,3,8\}$&$\{4,5,6,7\}$\\ \hline
\shortstack{missing\\faces}&$\shortstack{(1,2,3)\\(1,3,4)}$&$\shortstack{(5,6)\\(7,8)}$&$\shortstack{(1,2,3)\\(2,3,5)}$
&$\shortstack{(4,6,7)\\(7,8)}$&$\shortstack{(1,2,3)\\(1,2,8)}$&$\shortstack{(4,6,7)\\(5,6)}$\\ \hline
\end{tabular}\vspace{10pt}
\begin{tabular}{c|c|c|c|c|c}\hline
$K_{I_3}$&$K_{\wh{I_3}}$&$K_{I_4}$&$K_{\wh{I_4}}$&$K_{I_5}$&$K_{\wh{I_5}}$\\ \hline
$\{1,2,5,8\}$&$\{3,4,6,7\}$&$\{1,2,7,8\}$&$\{3,4,5,6\}$&$\{1,3,4,6\}$&$\{2,5,7,8\}$\\ \hline
$\shortstack{(1,2,8)\\(2,5,8)}$&$\shortstack{(3,4,6)\\(4,6,7)}$&$\shortstack{(1,2,8)\\(7,8)}$
&$\shortstack{(3,4,6)\\(5,6)}$&$\shortstack{(1,3,4)\\(3,4,6)}$&$\shortstack{(2,5,8)\\(7,8)}$\\ \hline
\end{tabular}\vspace{10pt}
\begin{tabular}{c|c|c|c|c|c}\hline
$K_{I_6}$&$K_{\wh{I_6}}$&$K_{I_7}$&$K_{\wh{I_7}}$&$K_{I_8}$&$K_{\wh{I_8}}$\\ \hline
$\{1,3,4,7\}$&$\{2,5,6,8\}$&$\{1,4,6,7\}$&$\{2,3,5,8\}$&$\{1,4,7,8\}$&$\{2,3,5,6\}$\\ \hline
$\shortstack{(1,3,4)\\(1,4,7)}$&$\shortstack{(2,5,8)\\(5,6)}$&$\shortstack{(1,4,7)\\(4,6,7)}$
&$\shortstack{(2,3,5)\\(2,5,8)}$&$\shortstack{(1,4,7)\\(7,8)}$&$\shortstack{(2,3,5)\\(5,6)}$\\ \hline
\end{tabular}\vspace{5pt}\caption{Non-contractible full subcomplexes of $K$ with four vertices.}\label{tab:1}
\end{table}

\item\label{case:4} $|I|=5$. We need to use the following well known fact: Let $\Gamma$ be a simplicial complex on [m],
$\Gamma_J$ a full subcomplex on $J\subseteq [m]$. Then $\Gamma_{\wh{J}}$ is a deformation retract of $\Gamma\setminus\Gamma_J$.
From this and Alexander duality on $K$ we have that $\w {H}^j(K_I)\cong \w {H}_{2-j}(K_{\wh{I}})$. Since $|I|=5$, $|\wh I|=3$.
From the arguments in case \eqref{case:2}, $H_*(K_{\wh{I}})$ are all torsion free, so $\w H^*(K_{\wh I})\cong \w H_*(K_{\wh I})$.
Thus $\w H^*(K_I)$ is non-trivial if and only if $\wh I$ is one of the eight missing faces with three vertice,
and if so, $\w {H}^*(K_I)\cong \w {H}^1(K_I)\cong \mathbb{Z}$, whose generator we denote by $\beta_i$ ($1\leq i\leq 8$).

\item\label{case:5} $|I|=6$.
The same argument as in \eqref{case:4} shows that $\w H^*(K_I)$ is non-trivial if and only if $\wh I$ is $(5,6)$ or $(7,8)$,
and if so, $\w {H}^*(K_I)\cong \w {H}^2(K_I)\cong \mathbb{Z}$. Denote by $\lambda_1$ (respectively $\lambda_2$)
a generator of $\w {H}^2(K_I)$ for $\wh I=(5,6)$ (respectively $(7,8)$).

\item $|I|\geq 7$. If $|I|=7$, $\w H^*(K_I)=0$ is clear.  If $|I|=8$, $K_I=K$, so $\w H^*(K)\cong H^3(K)\cong\mathbb{Z}$. Denote by
$\xi$ a generator of it.
\end{enumerate}

\begin{proof}[Proof of Proposition \ref{prop:1}]
Theorem \ref{thm:1} and the preceding arguments give the cohomology group of $\mathcal {Z}_{K}$
\begin{center}
\renewcommand{\arraystretch}{1.5}
\begin{tabular}{|c|c|}\hline
$i$&$\w H^i(\mathcal {Z}_{K})\cong$\\ \hline
$1,2,4,8,10,11$&$0$\\ \hline
$3$&$\mathbb{Z}\cdot\psi(a_1)\oplus\mathbb{Z}\cdot\psi(a_2)$\\ \hline
$5$&$\bigoplus_{1\leq i\leq 8}\mathbb{Z}\cdot\psi(b_i)$\\ \hline
$6$&$\bigoplus_{0\leq i\leq 8}\big(\mathbb{Z}\cdot\psi(\alpha_i)\oplus \mathbb{Z}\cdot\psi(\alpha_i')\big)$\\ \hline
$7$&$\bigoplus_{1\leq i\leq 8}\mathbb{Z}\cdot\psi(\beta_i)$\\ \hline
$9$&$\mathbb{Z}\cdot\psi(\lambda_1)\oplus\mathbb{Z}\cdot\psi(\lambda_2)$\\ \hline
$12$&$\mathbb{Z}\cdot\psi(\xi)$\\ \hline
\end{tabular}
\end{center}

Now we give the cup product structure of $H^*(\mathcal {Z}_{K})$. First by Poincar\'e duality on $\mathcal {Z}_{K}$ and Remark \ref{rem:1}, up to sign
\begin{align}
\psi(a_i)\smile \psi(\lambda_i)&=\psi(\xi),\quad i=1,2;\label{eq:2}\\
\psi(b_i)\smile \psi(\beta_i)&=\psi(\xi),\quad i=1,\dots,8;\\
\psi(\alpha_i')\smile \psi(\alpha_i)&=\psi(\xi),\quad i=0,\dots,8.
\end{align}
Note that $K_{\wh {I_0}}=K_{(5,6)}*K_{(7,8)}$,
so up to sign $\psi(a_1)\smile \psi(a_2)=\psi(\alpha_0')$ (see Remark \ref{rem:1}), and so
\begin{equation}\label{eq:1}
\psi(a_1)\smile \psi(a_2)\smile \psi(\alpha_0)=\psi(\xi)
\end{equation}
Since $a_2*\alpha_0\in\w H^*(K_{\wh{(5,6)}})$,
$\psi(a_2)\smile \psi(\alpha_0)=\psi(a_2*\alpha_0)=p\cdot\psi(\lambda_1)$ for some $p\in \mathbb{Z}$.
From fomulae \eqref{eq:2} and \eqref{eq:1} we have $p=1$. Similarly,
$\psi(a_1)\smile \psi(\alpha_0)=-\psi(\lambda_2)$. Moreover from the arguments in case \eqref{case:5}, we have that
$\psi(a_i)\smile \psi(\alpha_j)=0$ for $1\leq j\leq 8$, and $\psi(a_i)\smile \psi(\alpha_j')=0$ for $0\leq j\leq 8$; $i=1,2$.
By an observation on the dimension of the non-trivial cohomology groups of $\mathcal {Z}_{K}$,
it is easy to verify that any other products between these generators are trivial.
Combining all the product relations above we get the desired result.
\end{proof}
There are other two different polytopal spheres from $K$
(corresponding to the two $4$-polytopes $P_{27}^8$ and $P_{29}^8$ in \cite{GS67}), so that
the corresponding moment-angle manifolds have the same cohomology rings as $\mathcal {Z}_{K}$.
The proof of this is the same as Proposition \ref{prop:1}.

For a moment-angle manifold corresponding to a simplicial $2$-sphere, if its cohomology ring is isomorphic to the one of a connected sum of
sphere products, then it is actually diffeomorphic to this connected sum of sphere products (\cite{BM06}, Proposition 11.6).
This leads to the following conjecture:
\begin{conj}
$\mathcal {Z}_{K}$ is diffeomorphic to the connected sum of sphere products in Proposition \ref{prop:1}.
\end{conj}
Note that the connected sum of sphere products in Proposition \ref{prop:1} only has one product of three spheres, we then ask:
Is there a moment-angle manifold (corresponding to a simplicial $3$-sphere) whose cohomology ring is isomorphic to the one of
a connected sum of sphere products with more than one product of three spheres? The following Theorem gives a negative answer to this question.
\begin{thm}\label{thm:a}
Let $K$ be a $n$-dimensional simplicial sphere ($n\geq 2$) satisfies $H^*(\mathcal {Z}_{K})\cong H^*(M)$,
where $M\cong M_1\#\cdots\#M_k$, and each $M_i$ is a product of spheres. Let $q_i$ be the number of sphere factors of $M_i$. Then
\begin{enumerate}[(a)]
\item If $q_i=n+1$ for some $i$, then $k=1$, and $\mathcal {Z}_{K}\cong M\cong S^3\times S^3\cdots\times S^3$.
\item Let $I=\{i:q_i\geq [\frac{n}{2}]+2\}$ (where $[\cdot]$ denotes the integer part). Then $|I|\leq 1$.
\end{enumerate}
\end{thm}

\begin{lem}\label{lem:1}
Let $K$ be a simplicial complex on $[m]$. Given two classes $[a],\,[b]\in \mathcal {H}^{0,*}(K)$, if $[a]*[b]\neq 0$, then there must
be a full subcomplex $K_I$ ($|I|\geq 4$) which is isomorphic to the boundary of a polygon, and satisfying $[a]*[b]\in \w H^1(K_I)$.
\end{lem}
\begin{proof}
Let $\mathcal {M}=\{I\in MF(K):|I|\neq 3\}$, and let $K'$ be a simplicial complex on $[m]$ so that $MF(K')=\mathcal {M}$.
Clearly, $K$ is a subcomplex of $K'$. Note that $K'$ and $K$ have the same $1$-skeleton, so if we can prove that for some $I\in [m]$,
$K'_I$ is isomorphic to the boundary of a polygon ($K'_I$ can not be the boundary of a triangle by the definition of $\mathcal {M}$),
then the result holds. From Remark \ref{rem:2}, there is a ring homomorphism $i^*:\mathcal {H}^{*,*}(K')\to \mathcal {H}^{*,*}(K)$
induced by the simplicial inclusion $i:K\hookrightarrow K'$.
It is easy to see that $i^*$ is a isomorphism when restricted to $\mathcal {H}^{0,*}(K')$.
Suppose $i^*([a'])=[a]$ and $i^*([b'])=[b]$. By assumption, $i^*([a']*[b'])\neq 0$, so $[a']*[b']\in \mathcal {H}^{1,*}(K')\neq 0$. Without loss of generality,
we can assume $[a']*[b']\in \w H^1(K'_J)$ for some $J\subset [m]$.
The lemma follows once we prove the following assertion:
\begin{ass}
For any simplicial complex $\Gamma$ satisfies $\w H^1(\Gamma)\neq 0$,
there must be a full subcomplex $\Gamma_I$ which is isomorphic to the boundary of a polygon, satisfying that
\[j^*:\w H^1(\Gamma)\to \w H^1(\Gamma_I)\]
is an epimorphism, where $j:\Gamma_I\to \Gamma$ is the inclusion map.
\end{ass}
Now we prove this. Since $H^1(\Gamma)\neq 0$, $H_1(\Gamma)\neq 0$, then there is a nonzero homology class $[c]\in H_1(\Gamma)$
represented by the $1$-cycle
\[c=(v_1,v_2)+(v_2,v_3)\cdots+(v_{k-1},v_k)+(v_{k},v_1),\]
where $v_i$ is a vertex of $\Gamma$. Without loss of generality, we assume $v_1,v_2,\dots,v_k$ are all different
and the vertex number $k$ is minimal among all $[c]$'s and their representations. Let $I=(v_1,\dots,v_k)$, we claim that $\Gamma_I$ is isomorphic to the
boundary of a polygon. If this is not true, then there must be a $1$-simplex, say $(v_1,v_j)\in \Gamma_I$ such that $j\neq 2,k$. Let
\[c_1=(v_1,v_2)+(v_2,v_3)\cdots+(v_j,v_1);\quad c_2=(v_1,v_j)+(v_j,v_{j+1})\cdots+(v_k,v_1).\]
Then $c=c_1+c_2$, and therefore $[c_1]\neq 0$ or $[c_2]\neq 0$. In either case, the vertex number of $c_i$ ($i=1,2$) is less than $k$, a contradiction.
Apparently, $j_*([c])$ is the fundamental class of $\Gamma_I$.
\end{proof}

\begin{lem}\label{lem:2}
Let $K$ be a simplicial sphere satisfies $H^*(\mathcal {Z}_{K})$ is isomorphic to the cohomology ring
of a connected sum of sphere products. If a proper
full subcomplex is isomorphic to the boundary of a $m$-gon, then $m\leq 4$.
\end{lem}

\begin{proof}
Suppose on the contrary that there is a proper full subcomplex $K_I$ isomorphic to the boundary of a $m$-gon with $m\geq 5$.
Then $H^*(\mathcal {Z}_{K_I})$ is a proper subring and a direct summand of $H^*(\mathcal {Z}_{K})$.
By Theorem \ref{thm:2} we can find five elements $a_1,a_2,b_1,b_2,c$ of $H^*(\mathcal {Z}_{K})$,
where $\mathrm{dim}(a_1)=3$, $\mathrm{dim}(a_2)=4$,
$\mathrm{dim}(b_1)=m-1$, $\mathrm{dim}(b_2)=m-2$ and $\mathrm{dim}(c)=m+2$, such that each of them is a generator of a $\mathbb{Z}$
summand of $H^*(\mathcal {Z}_{K})$, and the cup product relations between them are given by:
\[a_1\smile b_1=a_2\smile b_2=c,\]
all other products are zero. Clearly, $\mathrm{dim}(c)$ is not equal to the top dimension of $H^*(\mathcal {Z}_{K})$.
Suppose $H^*(\mathcal {Z}_{K})$ is isomorphic to the cohomology ring of
\[S^{f(1,1)}_{1,1}\times S^{f(1,2)}_{1,2}\times\cdots\times S^{f(1,k(1))}_{1,k(1)}\#\cdots\#
S^{f(n,1)}_{n,1}\times S^{f(n,2)}_{n,2}\times\cdots\times S^{f(n,k(n))}_{n,k(n)},\]
where $f$ is a function of $\mathbb{(Z^+)}^2\to\mathbb{Z^+}$ ($f(i,j)\geq3$ for all $i,j$), $S^{f(i,j)}_{i,j}=S^{f(i,j)}$,
$k(i)\in \mathbb{Z}^+$ denote the number of spheres in the $i$th summand of sphere product.
Denote by $e_{ij}^{(k)}$ a generator of $H^k(\mathcal {Z}_{K})$ corresponding to $S^k_{i,j}$ ($f(i,j)=k$). Then we can write
\[a_1=\sum_{f(i,j)=3}\lambda_{ij}e^{(3)}_{ij},\quad a_2=\sum_{f(i,j)=4}\lambda_{ij}'e^{(4)}_{ij}\]
where $\lambda_{ij},\,\lambda_{ij}'\in \mathbb{Z}$. It is easy to see that $e_{ij}^{(k)}\smile e^{(t)}_{rs}\neq0$
if and only if $i=r$ and $j\neq s$. Since $a_1\smile a_2=0$, we have that if $\lambda_{ij},\lambda_{i'j'}'\neq0$, then $i\neq i'$.
However this implies that $a_1\smile b_1\neq a_2\smile b_2$ since
$\mathrm{dim}(c)$ is not equal to the top dimension of $H^*(\mathcal {Z}_{K})$, a contradiction.
\end{proof}

\begin{lem}\label{lem:3}
Let $K$ be a $n$-dimensional simplicial sphere satisfies $H^*(\mathcal {Z}_{K})$ is isomorphic to the cohomology ring
of a connected sum of sphere products. If there is a
full subcomplex isomorphic to the boundary of a quadrangle, then for any full subcomplex $K_I$ saisfies $\w H^0(K_I)\neq0$,
we have $|I|=2$. Moreover, if $I_1,I_2$ are two different such sequences, then $K_{I_1\cup I_2}$ is isomorphic to the boundary of a quadrangle.
\end{lem}

\begin{proof}
The case $n=1$ are trivial, so we assume $n>1$. If we can prove the statement that for any two different missing faces $\sigma_1,\,\sigma_2\in MF(K)$, which contain two vertices, we have $\sigma_1\cap\sigma_2=\emptyset$, then the lemma holds.

Suppose $J=(1,2,3,4)$, and $MF(K_J)=\{(1,3),(2,4)\}$ (i.e., $K_J$ is isomorphic to the boundary of a quadrangle) by assumption.
First we will prove that for any vertex $v\not\in J$ and any $j\in J$, $(j,v)$ is a simplex of $K$.
Without loss of generality, suppose on the contrary that
$(1,5)\in MF(K)$. Let $\Gamma$ be a simplicial complex with vertex set $\{1,3,5\}$ such that $MF(\Gamma)=\{(1,3),(1,5)\}$.
Then $K_{(1,3,5)}$ is a subcomplex of $\Gamma$.
Clearly, $\w H^0(\Gamma)\cong \mathbb{Z}$, denote by $c_1$ a generator of it. Let $L=\Gamma*K_{(2,4)}$. Denote by $c_2$ a generator of
$\w H^0(K_{(2,4)})\cong \mathbb{Z}$, then an easy calculation shows that (see Remark \ref{rem:1}) $c_1*c_2$ is a generator of
$\w H^1(L)\cong \mathbb{Z}$. Let $J'=(1,2,3,4,5)$. Then $K_{J'}$ is a subcomplex of $L$, and the inclusion map induces
a monomorphism $\w H^1(L)\xrightarrow{\mu}\w H^1(K_{J'})$ (actually, $\mu(H^1(L))$ is a direct summand of $H^1(K_{J'})$).
There is a commutative diagram
\[\begin{CD}
\w H^0(\Gamma)\otimes\w H^0(K_{(2,4)})@>\eta >>\w H^1(L)\\
@V\phi\otimes\mathrm{id} VV @VV\mu V\\
\w H^0(K_{(1,3,5)})\otimes\w H^0(K_{(2,4)})@>\eta >>\w H^1(K_{J'}),
\end{CD}\]
where $\phi$ is induced by the inclusion map. So $\mu(c_1*c_2)=\phi(c_1)*c_2$ is a generator of $\w H^1(K_{J'})$.
Thus by Poincar\'e duality on $\mathcal {Z}_{K}$,
there is an element $c_0$ of $\w H^{n-2}(K_{\wh{J'}})$ such that $c_0*\phi(c_1)*c_2$ is a generator of
$\w H^n(K)\cong \mathbb{Z}$. On the other hand, let $e_1$ be a generator of $\w H^0(K_{(1,3)})$. Clearly $e_1*c_2$ is a generator
of $\w H^1(K_J)\cong \mathbb{Z}$, so there is a element  $e_0$ of $\w H^{n-2}(K_{\wh J})$ such that $e_0*e_1*c_2=c_0*\phi(c_1)*c_2$.
Since $e_0*e_1,\,c_0*\phi(c_1)\in \w H^{n-1}(K_{\wh{(2,4)}})\cong \mathbb{Z}$, we have $e_0*e_1=c_0*\phi(c_1)$. Since $\mathrm{dim}(\psi(e_1))=3$,
$\mathrm{dim}(\psi\phi(c_1))=4$, and $e_1*\phi(c_1)=0$,
then we get a contradiction by applying the arguments as in the proof of Lemma \ref{lem:2}.

Now suppose $v_1,v_2,v_3\in \wh J$ such that $(v_1,v_2),\,(v_1,v_3)\in MF(K)$. Let $J_0=(v_1,v_2,1,3)$. Then
from the result in the last paragraph we have $K_{J_0}$ is isomorphic to the boundary of a quadrangle. Thus by applying the
same arguments as in the last paragraph, we have that $(v_1,v_3)$ is a simplex of $K$, a contradiction.
\end{proof}
Now let us use the preceding results to complete the proof of Theorem \ref{thm:a}
\begin{proof}[Proof of Theorem \ref{thm:a}]
(a) From the assumption and Theorem \ref{thm:1}, we have that there are $n+1$ elements $c_i\in \w H^{k_i}(K_{J_i})$, $1\leq i\leq n+1$,
such that $\prod_{i=1}^{n+1} c_i\neq0\in \w H^n(K)$
(clearly, $J_i\cap J_k=\emptyset$ for $i\neq k$ and $\bigcup_{i=1}^{n+1}J_i$ is the vertex set of $K$).
From Remark \ref{rem:1}, the cohomology dimention of the class $\prod_{i=1}^{n+1} c_i$ is $n+\sum_{i=1}^{n+1}k_i$. Thus $k_i=0$ for all $1\leq i\leq n+1$.
Combine all the preceding lemmas, we have that $|J_i|=2$ for all $1\leq i\leq n+1$, so $J_i\in MF(K)$, and so $K$ is a subcomplex of
$K_{J_1}*\cdots *K_{J_{n+1}}$. Since $K_{J_1}*\cdots *K_{J_{n+1}}$ is a triangulation of $S^n$ itself,
then $K\cong K_{J_1}*\cdots *K_{J_{n+1}}$, and then the conclusion follows.

(b) Suppose there is a $M_u$ with $q_u\geq [\frac{n}{2}]+2$,
then as in (a) there are $q_u$ elements $c_i\in \w H^{k_i}(K_{J_i})$, $1\leq i\leq q_u$, such that $\prod_{i=1}^{q_u} c_i\neq0\in \w H^n(K)$.
The cohomology dimention of the class $\prod_{i=1}^{q_u} c_i$ is $q_u-1+\sum_{i=1}^{q_u}k_i$,
then from the inequality $q_u\geq [\frac{n}{2}]+2$, there are at least two $k_i$'s with $k_i=0$.
Then $K$ satisfies the conditions in all of the three Lemmas above.
From the first statement of Lemma \ref{lem:3}, we have that for any $a\in \mathcal {H}^{0,*}(K)$, $\mathrm{dim}(\psi(a))=3$.
So there are at least two $S^3$ factors in $M_u$.
From the second statement of Lemma \ref{lem:3}, we have that for any two linear independent element $a_1,a_2\in\mathcal {H}^{0,*}(K)$,
$a_1*a_2\neq 0$. This implies that all $S^3$ factors in the expression of $M$ are in $M_u$. Then there can not be another
$M_v$ with $q_v\geq [\frac{n}{2}]+2$. The conclusion holds.
\end{proof}

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TITLE: Prove $4\sin(40^{\circ})-\sqrt{3}=\cot(50^{\circ})$
QUESTION [0 upvotes]: Prove

$$4\sin(40^{\circ})-\sqrt{3}=\cot(50^{\circ})$$

I started like this:
Let $\:$ $\tan(20^{\circ})=x$
$$4\sin(40^{\circ})-\sqrt{3}=\frac{8\tan(20^{\circ})}{1+\tan^2(20^{\circ})}-\left(\frac{3\tan(20^{\circ})-\tan^3(20^{\circ})}{1-3\tan^2(20^{\circ})}\right)=\frac{8x}{1+x^2}-\frac{3x-x^3}{1-3x^2}$$
Any way to continue from here?

REPLY [1 votes]: $$4\sin40^{\circ}-\sqrt3=\frac{4\sin40^{\circ}\sin50^{\circ}-2\sin60^{\circ}\sin50^{\circ}}{\sin50^{\circ}}=$$
$$=\frac{2\cos10^{\circ}-2\cos90^{\circ}-\cos10^{\circ}+\cos110^{\circ}}{\sin50^{\circ}}=$$
$$=\frac{\cos10^{\circ}+\cos110^{\circ}}{\sin50^{\circ}}=\frac{2\cos60^{\circ}\cos50^{\circ}}{\sin50^{\circ}}=\cot50^{\circ}.$$


TITLE: Find the integer closest to $\ln(2013)$
QUESTION [18 upvotes]: I encounter such a problem, in a Maths contest, to find out the closest integer to $\ln(2013)$, without using a calculator. I really get stuck.
I tried to turn $\ln(2013)$ into $\ln(3)+\ln(11)+\ln(61)$, but nothing valuable obtained. I applied also Taylor series of natural log but it doesn't work. Any suggestions are welcomed.

REPLY [1 votes]: If you remember that $e^3 \approx 20$:
$$ln(2013) \approx ln(2000) = ln(20 \cdot 20 \cdot 5) = ln(20) + ln(20) + ln(5) \approx 3 + 3 + ln(5) $$
ln(5) is between 1 and 2 (because $e \approx 2.71$ and $e^2 \approx 7.4$), so all you need to known is if $ln(5)$ is greater or less than 1.5.
$e^{1.5} = \sqrt{e^3} \approx \sqrt{20} = \sqrt{4 \cdot 5} = 2 \cdot \sqrt{5} \approx 2 \cdot 2.2 = 4.4 < 5$, so $ln(5) > 1.5$.
=> $ln(2013) \approx 8$.


TITLE: References on the obstacle problem for the heat equation
QUESTION [5 upvotes]: Can you point out some references that deal with the obstacle problem for the heat equation?
$$(OP) \quad\begin{cases}
\max\{\Delta u -\partial_t u, \varphi - u \} = 0 & \text{ in } (0,T)\times \mathbb{R}^n \\
u(0,\cdot) = \varphi(0,\cdot) & \text{ in } \mathbb{R}^n
\end{cases}.$$

Works on elliptic obstacle problems appear to be much easier to find (see Wikipedia, for instance).

Since a bounty has been offered for this question, I'll write down what I feel is missing in the current (very nice) answer and that I'd like to see:

complete argument for the existence (with references too)
further details on the representation of solutions using the heat kernel
references about numerical analysis of the problem (and Matlab/Mathematica codes)
references on physical motivations for the problem.

REPLY [4 votes]: I will get you started, but there are lots of blanks to fill. We are
interested in the PDE
\begin{align*}
\min\left\{ -u_{t}+\Delta u,u-\varphi\right\}  & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\
u(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n}
\end{align*}
where $\varphi$ is a smooth function of polynomial growth.
Existence
Let $u^{0}$ be a classical solution of
\begin{align*}
-u_{t}^{0}+\Delta u^{0} & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\
u^{0}(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n}
\end{align*}
that is unique in an appropriately picked space of functions of polynomial
growth. Define inductively $u^{k}$ as a classical solution of
\begin{align*}
-u_{t}^{k}+\Delta u^{k}+k\min\left\{ u^{k-1}-\varphi,0\right\}  & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\
u^{k}(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n}
\end{align*}
that is unique in an appropriately picked space of functions of polynomial
growth. Now you would have to show that you can pass to limits ($k\rightarrow\infty$)
to obtain a solution of the original PDE.
The limiting solution is not, in general, twice differentiable in space. But, you should probably be able to establish sufficient conditions for it to be a once differentiable viscosity solution.
This technique is called a penalty method and is at least useful for establishing existence.
Uniqueness
Uniqueness can be handled in the space of viscosity solutions using a comparison principle argument.
I wrote about this in an expository post, but I'm sure this is available in different forms elsewhere. The setting of the post proves uniqueness in the space of bounded functions, but you should be able to generalize the arguments.
Heat Kernel
You can use smooth pasting to write the solution with the heat kernel, but you will probably not be able to find a "nice expression" for the free boundary, as this is believed to be hard.
References

Van Moerbeke, Pierre. "On optimal stopping and free boundary problems." Archive for Rational Mechanics and Analysis 60.2 (1976): 101-148.
Pham, Huyên. "Optimal stopping, free boundary, and American option in a jump-diffusion model." Applied mathematics & optimization 35.2 (1997): 145-164.


TITLE: If $\overline{k}$ is an algebraic closure of a field $k$, then $\overline{k}[x_{1}, \dots, x_{n}]$ is integral over $k[x_{1}, \dots, x_{n}]$.
QUESTION [2 upvotes]: I want to prove that if $\overline{k}$ is an algebraic closure of a field $k$, then $\overline{k}[x_{1}, \dots, x_{n}]$ is integral over $k[x_{1}, \dots, x_{n}]$.
It is used in exercise 11.3 of the book Introduction to commutative algebra by Atiyah-McDonald. I think the easiest way to prove is to show that, if $f\in\overline{k}[x_{1},\ldots,x_{n}]$, then $k[x_{1},\ldots,x_{n}][f]$ is a finitely generated $k[x_{1},\ldots,x_{n}]$-module, but I do not know how to prove it.

REPLY [4 votes]: More generally, show that if $R \to S$ is integral and $R \to T$ is arbitrary, then $T \to T \otimes_R S$ is integral. Prove this by using the fact $(*)$ that integral elements form a subring.
Apply this to $k \to \overline{k}$ and $k \to k[x_1,\dotsc,x_n]$. Of course, a direct proof without tensor products is also possible; but in any case I would just use the fact $(*)$.