text:
\begin{document}

\title{Donovan's conjecture and extensions by the centralizer of a defect group\footnote{This research was supported by the EPSRC (grant no. EP/T004606/1)}}

\author{Charles W. Eaton\footnote{Department of Mathematics, University of Manchester, Manchester M13 9PL. Email: charles.eaton@manchester.ac.uk} and Michael Livesey\footnote{Department of Mathematics, University of Manchester, Manchester M13 9PL. Email: michael.livesey@manchester.ac.uk}}

\date{19th June 2020}
\maketitle

\begin{abstract}
We consider Donovan's conjecture in the context of blocks of groups $G$ with defect group $D$ and normal subgroups $N \lhd G$ such that $G=C_D(D\cap N)N$, extending similar results for blocks with abelian defect groups. As an application we show that Donovan's conjecture holds for blocks with defect groups of the form $Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$ defined over a discrete valuation ring.
\end{abstract}

\section{Introduction}

Let $p$ be a prime and $k:=\overline{\mathbb{F}}_p$. Let $(K,\cO,k)$ be a $p$-modular system, so $\cO$ is a complete discrete valuation ring with residue field $k$. Donovan's conjecture states that for a given finite $p$-group $P$, there are only finitely many Morita equivalence classes amongst blocks of finite groups with defect groups isomorphic to $P$ (this may be stated over $k$ or $\mathcal{O}$). In reducing Donovan's conjecture to quasisimple groups, we must inevitably compare blocks $B$ of finite groups $G$ with those of normal subgroups $N$. The case where $N$ contains a defect group $D$ of $B$ was treated by K\"ulshammer in~\cite{ku95} for $k$-blocks, and by Eisele in~\cite{ei18} for $\mathcal{O}$-blocks. This paper concerns the problematic case of normal subgroups of index a power of $p$, where it suffices to assume $G=ND$. The subcase that $D$ is abelian was first considered, for $k$-blocks and with an additional splitting condition, in~\cite{kk96}. In~\cite{el19} the full $D$ abelian case was treated by introducing strong Frobenius numbers, related to the Morita Frobenius numbers introduced in~\cite{ke04}. The approach taken here also involves strong Frobenius numbers.

The purpose of this paper is to extend the reduction result above to the case that $G=C_D(D\cap N)N$. As an application, we show that Donovan's conjecture with respect to $\cO$ holds when $D \cong Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$ for some $n$. Blocks with defect group $Q_{2^m} \times C_{2^n}$ were studied by Sambale in~\cite{sa13} and the number of irreducible characters and Brauer characters computed. Donovan's conjecture for $\cO$-blocks with defect group $Q_8$ was proved by Eisele in~\cite{ei16}.

In order to make canonical choices of $k$ and $\mathcal{O}$, we choose $k$ to be the algebraic closure of the field $\mathbb{F}_p$ of $p$-elements and $\cO$ to be the ring of Witt vectors for $k$. There is a discussion of this in~\cite{eel19}. When we need to differentiate the versions of Donovan conjecture, the $R$-Donovan conjecture will relate to blocks defined with respect to the ring $R$, where $R$ may be $k$ or $\cO$.

We also consider the following, based on a question of Brauer, which is now often referred to as the weak Donovan conjecture:

\begin{conjecture}
\label{Brauerconj}
Let $P$ be a finite $p$-group. There is $c \in \NN$ such that for all blocks $B$ of finite groups $G$ with defect groups isomorphic to $P$, the entries of the Cartan matrix of $B$ are at most $c$.
\end{conjecture}

In~\cite{ke04} Kessar showed that the $k$-Donovan conjecture is equivalent to showing Conjecture \ref{Brauerconj} and that the Morita Frobenius number (defined in $\S$\ref{sec:strong_frobenius}) of a block is bounded in terms of the order of the defect groups. Variations on the Morita Frobenius number for blocks defined over $\cO$ were given in~\cite{el19}, including the \emph{strong $\cO$-Frobenius number} $\SOF(B)$, and in~\cite{eel19} the analogue of Kessar's result was shown for blocks defined over $\cO$. In~\cite{du04} D\"uvel reduced Conjecture \ref{Brauerconj} to quasisimple groups (although the result in~\cite{du04} is not quite strong enough for our purposes as stated). Our first reduction result concerns the other half of the problem, i.e., bounding the strong $\cO$-Frobenius numbers in terms of the defect groups:

\begin{theorem}
\label{sfreduce:theorem}
Let $G$ be a finite group and $B$ be a block of $\cO G$ with defect group $D$. In addition, let $N \lhd G$ such that $G=C_D(D\cap N)N$ and let $b$ be a block of $\cO N$ covered by $B$. Then $\SOF(B) \leq \SOF(b)$.
\end{theorem}

Combining with Lemma~\ref{Cartan_reduce:lemma}, we obtain the following:

\begin{theorem}
\label{Donovan reduce:theorem}
Let $P$ be a finite $p$-group. In order to verify Donovan's conjecture for $P$ for $\cO$-blocks, it suffices to check it for blocks of finite groups $G$ with defect group $D$ isomorphic to a subgroup of $P$ and no proper normal subgroup $N \lhd G$ such that $G=C_D(D\cap N)N$.
\end{theorem}

After analysing the blocks of quasisimple groups with defect groups $Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$ in Section \ref{sec:classification_application} using~\cite{an20} (in the latter case there are none), we show:

\begin{theorem}
\label{Q8Ctheorem}
There are only finitely many Morita equivalence classes of $\cO$-blocks of finite groups with defect groups isomorphic to $Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$, where $n \geq 0$.
\end{theorem}

The reader might ask what happens for some similar $p$-groups. We note that $Q_8 \times C_{2^n}$ and $Q_8 \times Q_8$ have particularly restricted subgroup structures, meaning that covered blocks of normal subgroups are either amenable to the application of Theorem \ref{sfreduce:theorem} or are easily dealt with by other methods. If, for example, the defect group were $Q_{16} \times C_{2^n}$, then we would have to consider the case of a normal subgroup with defect group $Q_8 \times C_{2^n}$, a case which does arise and to which Theorem \ref{sfreduce:theorem} would not apply. Similarly, $D_8 \times C_2$ is problematic as it contains a normal subgroup $(C_2)^3$. New methods will be needed for these cases.

The structure of the paper is as follows. In Section \ref{sec:strong_frobenius} we treat strong Frobenius numbers and prove Theorem \ref{sfreduce:theorem}. We prove Theorem \ref{Donovan reduce:theorem} in Section \ref{Donovan_reduce:sec}. In Section \ref{sec:classification_application} we show that there are no blocks of quasisimple groups with defect group $Q_8 \times Q_8$, and few with defect group $Q_8 \times C_{2^n}$ for $n \geq 1$. We give some preliminary reductions and results about blocks with the above defect groups in Section \ref{sec:background}, and the proof that Donovan's conjecture holds for these blocks in Section \ref{sec:Donovan_cases}.

\section{Strong Frobenius numbers and extensions by the centralizer of a defect group}\label{sec:strong_frobenius}

Throughout this section, let $G$ be a finite group and $B$ a block of $\cO G$ with defect group $D$. We denote by $\Irr(G)$ the set of irreducible characters of $G$ and $\Irr(B)$ the subset of $\Irr(G)$ of irreducible characters
lying in the block $B$. We write $kB$ for the block of $kG$ corresponding to $B$. We denote by $e_B\in \cO G$ the block idempotent for $B$ and by $e_\chi\in \QQ_c G$ the character idempotent for $\chi\in\Irr(G)$, where $\QQ_c$ is the universal cyclotomic extension of $\QQ$. If $A_1$ and $A_2$ are finitely generated $k$-algebras (respectively $\cO$-algebras), we write $A_1\sim_{\Mor}A_2$ if $A_1$ and $A_2$ are Morita equivalent as $k$-algebras (respectively $\cO$-algebras). We quote the following definition from~\cite[Definition 3.2]{el19}.

\begin{definition}
Let $q$ be a, possibly zero or negative, power of $p$. We denote by $-^{(q)}:k\to k$ the field automorphism given by $\lambda\to \lambda^{\frac{1}{q}}$. Let $A$ be a $k$-algebra. We define $A^{(q)}$ to be the $k$-algebra with the same underlying ring structure as $A$ but with a new action of the scalars given by $\lambda.a=\lambda^{(q)}a$, for all $\lambda\in k$ and $a\in A$. For $a\in A$ we define $a^{(q)}$ to be the element of $A$ associated to $a$ through the ring isomorphism between $A$ and $A^{(q)}$. Note that we have $kG\cong kG^{(q)}$ as we can identify $-^{(q)}: kG\to kG$ with the ring isomorphism:
\begin{align*}
-^{(q)}: kG&\to kG\\
\sum_{g\in G}\a_gg&\mapsto \sum_{g\in G}(\a_g)^qg.
\end{align*}

From now on, we identify $(kB)^{(q)}$ with the image of $kB$ under the above isomorphism. We define $B^{(q)}$ to be the unique block of $\cO G$ satisfying $k(B^{(q)})=(kB)^{(q)}$.

By an abuse of notation, we also use $-^{(q)}$ to denote the field automorphism of $\QQ_c$ defined by $\omega_p\omega_{p'}\mapsto \omega_p\omega_{p'}^{\frac{1}{q}}$, for all $p^{\nth}$-power roots of unity $\omega_p$ and $p'^{\nth}$ roots of unity $\omega_{p'}$ and also the ring automorphism
\begin{align*}
-^{(q)}: \QQ_c G&\to \QQ_c G\\
\sum_{g\in G}\a_gg&\mapsto \sum_{g\in G}(\a_g)^{(q^{-1})}g.
\end{align*}

If $\chi\in\Irr(G)$, then we define $\chi^{(q)}\in \Irr(G)$ to be given by $\chi^{(q)}(g) = \chi(g)^{(q^{-1})}$, for all $g\in G$. Note that if $\chi\in\Irr(B)$, then $(e_\chi)^{(q)}=e_{\chi^{(q)}}$ and $\chi^{(q)}\in\Irr(B^{(q)})$.
\end{definition}

For $R\in\{\ZZ,\QQ_c,\cO\}$, we define the $R$-linear map $\zeta_G:RG\to RG$, $g\mapsto g_pg_{p'}^p$, where $g_p$ and $g_{p'}$ are the $p$-part and $p'$-part of $g$ respectively. In general, the relevant ring $R$ should always be clear from the context. The following was proved in~\cite[Proposition 3.5]{el19}.

\begin{proposition}\label{prop:centre}
$\zeta_G$ restricted to $Z(\ZZ G)\subseteq \ZZ G$ induces a $\mathbb{Z}$-algebra automorphism of $Z(\ZZ G)$. Furthermore the algebra automorphism induced on $Z(\QQ_c G)\cong \QQ_c\otimes_{\ZZ} Z(\ZZ G)$ sends $e_\chi$ to $e_{\chi^{(p)}}$, for all $\chi\in \Irr(G)$.
\end{proposition}

We now generalise the above Proposition to deal with normal subgroups of index a power of $p$. In what follows, $A^H$ will denote the set of fixed points in $A$ under the action of $H$, where $A$ is an algebra with an action of a group $H$. In practice, $A$ will always be the group algebra of $G$ or one of its blocks with the natural conjugation action of $H\leq G$. For each $n\in\mathbb{N}$, we use $\omega_n\in \QQ_c$ to denote some fixed primitive $n^{\nth}$ root of unity.

\begin{proposition}\label{prop:zeta_auto}
Let $N\lhd G$ be of index a power of $p$. Then $\zeta_G$ induces a $\ZZ$-algebra automorphism of $(\ZZ G)^N$.
\end{proposition}

\begin{proof}
As noted in the proof of~\cite[Proposition 3.5]{el19}, $\zeta_G$ commutes with conjugation by any $g\in G$. Therefore, since $(\ZZ G)^N$ has a $\ZZ$-basis consisting of the $N$-conjugacy class sums of $G$, $\zeta_G$ maps $(\ZZ G)^N$ to itself. Hence, it is sufficient to prove that $\zeta_G$ induces a $\QQ_c$-algebra automorphism of $(\QQ_c G)^N\cong \QQ_c\otimes_{\ZZ}(\ZZ G)^N$. We do this by showing that $\zeta_G$ induces an isomorphism
\begin{align*}
(\QQ_c G)^Ne_\chi\to(\QQ_c G)^Ne_{\chi^{(p)}},
\end{align*}
for all $\chi\in\Irr(N)$. Note that $\Stab_G(\chi)=\Stab_G(\chi^{(p)})$ and if $g\notin\Stab_G(\chi)$, then
\begin{align*}
C_g^Ne_\chi=e_\chi C_g^Ne_\chi=C_g^Ne_{\chi^g}e_\chi=0,
\end{align*}
where $C_g^N\in\mathbb{Z}G$ is the sum of the elements in the $N$-conjugacy class containing $g$. We may, therefore, assume that $G=\Stab_G(\chi)$. For each $g\in G$ we define
\begin{align*}
e_{\chi,gN}:=\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^i e_{\chi'.\lambda^i},
\end{align*}
where $\chi'$ is an extension of $\chi$ to $\langle g\rangle N$ and $\lambda$ is the linear character of $\langle g\rangle N$ given by
\begin{align*}
\lambda:\langle g\rangle N&\to\QQ_c\\
g^iN&\mapsto\omega_{\ord(gN)}^i.
\end{align*}
If we define the $\QQ_c$-algebra automorphism
\begin{align*}
\tilde{\lambda}:\QQ_c (\langle g\rangle N)&\to \QQ_c (\langle g\rangle N)\\
h&\mapsto \lambda(h)h,
\end{align*}
for all $h\in \langle g\rangle N$, then
\begin{align}\label{algn:lambda_auto}
\begin{split}
\tilde{\lambda}(e_{\chi,gN})=&\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^i \tilde{\lambda}(e_{\chi'.\lambda^i})=\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^i e_{\chi'.\lambda^{i-1}}\\
=&\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^{i+1} e_{\chi'.\lambda^i}=\omega_{\ord(gN)}e_{\chi,gN}.
\end{split}
\end{align}
Therefore, $e_{\chi,gN}\in g(\QQ_cNe_\chi)$. Since we are making a choice of extension $\chi'$ of $\chi$, $e_{\chi,gN}$ is only defined uniquely up to multiplication by some power of $\omega_{\ord(gN)}$. With this in mind, we introduce the notation of $\alpha\approx\beta$, for $\alpha$, $\beta\in(\QQ_c G)^N$, if $\alpha=\mu\beta$ for some $p^{\nth}$-power root of unity $\mu\in\QQ_c$. Note that if when defining $e_{\chi,g^{-1}N}$ we choose the same extension $\chi'$ of $\chi$ as when defining $e_{\chi,gN}$, i.e.,
\begin{align*}
e_{\chi,g^{-1}N}:=\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^i e_{\chi'.\lambda^{-i}}=\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^{-i} e_{\chi'.\lambda^i},
\end{align*}
then
\begin{align*}
e_{\chi,gN}e_{\chi,g^{-1}N}=\sum_{i=1}^{\ord(gN)} e_{\chi'.\lambda^i}=e_\chi.
\end{align*}
Therefore, for any choice of $e_{\chi,g^{-1}N}$, $e_{\chi,gN}e_{\chi,g^{-1}N}\approx e_\chi$ and so we have shown that $(\QQ_c G)^Ne_\chi$ is a crossed product of $G/N$ with $Z(\QQ_c Ne_\chi)=\QQ_c e_\chi$ in the sense of K\"ulshammer~\cite{ku95}. In other words, $(\QQ_c G)^Ne_\chi=\bigoplus_{gN\in G/N}\QQ_c e_{\chi,gN}$. Now
\begin{align}\label{algn:idmpt}
\begin{split}
\zeta_G(e_{\chi,gN})=&\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^i \zeta_G(e_{\chi'.\lambda^i})=\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^i e_{(\chi'.\lambda^i)^{(p)}}\\
=&\sum_{i=1}^{\ord(gN)}\omega_{\ord(gN)}^i e_{\chi'^{(p)}.\lambda^i}\approx e_{\chi^{(p)},gN},
\end{split}
\end{align}
where the second equality follows from Proposition~\ref{prop:centre}. Therefore, $\zeta_G$ induces a $\QQ_c$-vector space isomorphism between $(\QQ_c G)^Ne_\chi$ and $(\QQ_c G)^Ne_{\chi^{(p)}}$.

Before proceeding we note that, for all $g,h\in G$, $(e_{\chi,gN})^{\ord(gN)}=e_\chi$. Also, as noted just after (\ref{algn:lambda_auto}), $e_{\chi,gN}=gxe_\chi$, for some $x\in (\QQ_c Ne_\chi)^\times$ and so $e_{\chi,gN}e_{\chi,hN}e_{\chi,gN}^{-1}=ge_{\chi,hN}g^{-1}\approx e_{\chi,{}^ghN}$. We use both these facts below in (\ref{algn:mult}).

It remains to show that for all $g,h\in G$,
\begin{align*}
\zeta_G(e_{\chi,gN}e_{\chi,hN})=\zeta_G(e_{\chi,gN})\zeta_G(e_{\chi,hN}).
\end{align*}
By (\ref{algn:idmpt}), $\zeta_G(e_{\chi,gN})=e_{\chi,gN}^{(p)}$ and $\zeta_G(e_{\chi,hN})=e_{\chi,hN}^{(p)}$ and so we need only show that $\zeta_G(e_{\chi,gN}e_{\chi,hN})=(e_{\chi,gN}e_{\chi,hN})^{(p)}$. Since $\zeta_G$ fixes coefficients that are $p^{\nth}$-power roots of unity, it suffices in turn to prove that $e_{\chi,gN}e_{\chi,hN}\approx e_{\chi,ghN}$. We prove this last statement via induction on the lowest layer of $gN$, $hN$ or $ghN$ in the upper central series of $G/N$. Note that
\begin{align*}
e_{\chi,gN}e_{\chi,hN}\approx e_{\chi,ghN}\Leftrightarrow e_{\chi,hN}e_{\chi,(gh)^{-1}N}\approx e_{\chi,g^{-1}N}\Leftrightarrow e_{\chi,(gh)^{-1}N}e_{\chi,gN}\approx e_{\chi,h^{-1}N}
\end{align*}
and so we may assume that $hN$ is in the lowest layer. If $h\in N$, then $e_{\chi,hN}=e_\chi$ and so $e_{\chi,gN}e_{\chi,hN}=e_{\chi,ghN}$. Now let $g,h$ be arbitrary elements of $G$ and set $p^n:=\max\{\ord(gN),\ord(hN),\ord(ghN)\}$. Then
\begin{align}\label{algn:mult}
\begin{split}
(e_{\chi,gN}e_{\chi,hN})^{p^n}&\approx (e_{\chi,gN})^{p^n}(e_{\chi,h^{g^{p^n-1}}N}\dots e_{\chi,h^gN} e_{\chi,hN})\approx (e_{\chi,h^{g^{p^n-1}}N}\dots e_{\chi,h^gN})(e_{\chi,h^{-1}N})^{p^n-1}\\
&\approx (e_{\chi,h^{g^{p^n-1}}N}e_{\chi,h^{-1}N})\dots(e_{\chi,h^{gh^{-(p^n-2)}}N}e_{\chi,h^{-1}N})\\
&\approx e_{\chi,[g^{-(p^n-1)},h]N}\dots e_{\chi,[h^{p^n-2}g^{-1},h]N}\approx e_{\chi,[g^{-(p^n-1)},h]\dots[h^{p^n-2}g^{-1},h]N}\\
&=e_{\chi,(gh)^{p^n}N}=e_\chi,
\end{split}
\end{align}
where the fourth and fifth relations follow from the inductive hypothesis. Now, since $e_{\chi,gN}e_{\chi,hN}\in gh(\QQ_c Ne_\chi)\cap(\QQ_c G)^N e_\chi=\QQ_c e_{\chi,ghN}$, we have $e_{\chi,gN}e_{\chi,hN}\approx e_{\chi,ghN}$.
\end{proof}

For the following definitions see~\cite{ke04} and~\cite{el19}.

\begin{definition}
The \textbf{Morita Frobenius number} $\mf(A)$ of a finite dimensional $k$-algebra $A$ is the smallest integer $n$ such that $A\sim_{\Mor} A^{(p^n)}$ as $k$-algebras. The \textbf{$\cO$-Morita Frobenius number} $\mfO(B)$ of $B$ is the smallest integer $n$ such that $B\sim_{\Mor} B^{(p^n)}$ as $\cO$-algebras. The \textbf{strong $\cO$-Frobenius number} $\SOF(B)$ of $B$ is the smallest integer $n$ such that there exists an $\cO$-algebra isomorphism $\phi:B\to B^{(p^n)}$ with the induced bijection of characters given by $\chi\mapsto \chi^{(p^n)}$, for all $\chi\in\Irr(B)$ or, equivalently, that $\phi$ restricted to $Z(B)$ coincides with $\zeta_G^{\circ n}$. Such an isomorphism $\phi$ is called a \textbf{strong Frobenius isomorphism of degree $n$}.
\end{definition}

Before the next lemma we need to give a brief overview of Picard groups of blocks. For a more detailed discussion see~\cite[$\S1$]{bkl18}.

Let $A$ be an $\cO$-algebra. The Picard group $\Pic(A)$ of $A$ consists of isomorphism classes of $A$-$A$-bimodules which induce $\cO$-linear Morita auto-equivalences of $A$. $\Pic(A)$ forms a group with the group multiplication given by tensoring over $A$. For each $\a\in\Aut(A)$ we define ${}_\a A$ to be the $A$-$A$-bimodule with the canonical right action of $A$ and the left action of $A$ given via $\a$. Moreover, ${}_\a A\cong A$ as bimodules if and only if $\a\in\Inn(A)$. In other words, we can view $\Out(A)=\Aut(A)/\Inn(A)$ as a subgroup of $\Pic(A)$.

Now set $A:=iBi$, where $i\in B^D$ is a source idempotent of $B$. We identify $D$ with its image in $A$ and denote by $\Aut_D(A)$ the group of $\cO$-algebra automorphisms of $A$ which fix $D$ pointwise. In addition, we set $\Out_D(A)$ to be the image of $\Aut_D(A)$ in $\Out(A)$, in other words $\Out_D(A)$ is the quotient of $\Aut_D(A)$ by the subgroup of inner automorphisms induced by conjugation with elements in $(A^D)^\times$. We set $\cT(B)$ to be the subgroup of bimodules in $\Pic(B)$ with trivial source, when considered as $\cO(G\times G)$-modules. Through the natural Morita equivalence between $B$ and $A$, we may identify $\Out_D(A)$ with the subgroup of $\cT(B)$ consisting of bimodules with vertex $\Delta D$.

The following Lemma and Theorem~\ref{sfreduce:theorem} were shown for $D$ abelian in~\cite[Theorems 3.15, 3.16]{el19}:

\begin{lemma}\label{lem:graded_units}
Let $N\lhd G$ such that $G=C_D(D\cap N)N$ and $b$ a block of $\cO N$ covered by a block $B$ of $\cO G$ with defect group $D$. Then $e_B=e_b$ and $b$ has defect group $D\cap N$. Viewing $b$ as a subalgebra of $B$, there is a choice of $a_{gN}\in (gb)^\times$ for each $gN\in G/N$ such that $B^N=\bigoplus_{gN\in G/N}a_{gN}Z(b)$.
\end{lemma}

\begin{proof}
By~\cite[Corollary 5.5.6]{nats88}, $B$ is the unique block of $\cO G$ covering $b$ and so $e_B$ is the sum of $e_b$ and its $G$-conjugates. However,~\cite[Theorem 5.5.10(v)]{nats88} gives that $b$ is $G$-stable, so that $e_B=e_b$ and $D \cap N$ is a defect group for $b$.

Since $B^N$ is $G/N$-graded, it remains to find a unit $a_{gN}$ in the $gN$-graded component of $B^N$, for each $gN\in G/N$. Let $g\in C_D(D\cap N)$ and consider $c_g\in\Aut(B)$ given by conjugation by $g$. Now $c_g$ induces the element
\begin{align*}
M:=\cO_{\Delta c_g}\uparrow^{N\times N}e_b\in\Pic(b),
\end{align*}
where $\Delta c_g=\{(h,c_g(h))|h\in N\}\leq N\times N$ and $\cO_{\Delta c_g}$ is the trivial $\cO(\Delta c_g)$-module. In particular, $M\in\cT(b)$ and so, by~\cite[Theorem 1.1(i)]{bkl18}, $M$ has vertex $\{(h,c_g(h))|h\in D\cap N\}=\Delta(D\cap N)$. Therefore, by the comments preceding the lemma, $iMi\in\Out_{D\cap N}(ibi)$, where $i$ is a source idempotent for $b$. Now by~\cite[14.5, Proposition 14.9]{pu88}, $\Out_{D\cap N}(ibi)$ is a $p'$-subgroup of $\cT(b)$. In particular, $iMi$ has $p'$-order in $\Out(ibi)$ or, equivalently, $M$ has $p'$-order in $\Out(b)$. However, since $g\in D$, $c_g$ has order a power of $p$ meaning $M$ induces the trivial auto-equivalence and $c_g\in \Inn(b)$.

Let $gN$ be a left coset of $N$ in $G$, where we choose coset representative $g\in C_D(D\cap N)$. Set $\a_g\in b^\times$ such that $c_g$ is given by conjugation by $\a_g$. We set $a_{gN}:=g(\a_g)^{-1}$.
\end{proof}

Note that, if we assume the stronger condition $D=(D\cap N)Z(D)$, then $B^N$ can be replaced with $Z(B)$ in Lemma~\ref{lem:graded_units}. The result would then be proved via induction on $|G/N|$, since when $G/N$ is cyclic the $a_{gN}$'s constructed above are in $Z(B)$. We now prove Theorem~\ref{sfreduce:theorem}.
\newline

{\sc Proof of Theorem}~\ref{sfreduce:theorem}.
Set $n:=\SOF(b)$ and let $\phi:b\to b^{(p^n)}$ be a strong $\cO$-Frobenius isomorphism of degree $n$. Then, by Proposition~\ref{prop:zeta_auto}, we can extend $\phi$ to an isomorphism $\tilde{\phi}:B\to B^{(p^n)}$ by sending $a_{gN}$ to $\zeta_G^{\circ n}(a_{gN})$, for each left coset $gN$ of $N$ in $G$ and the $a_{gN}$'s are as in Lemma~\ref{lem:graded_units}. By construction, $\tilde{\phi}$ agrees with $\zeta_G^{\circ n}$ on $B^N$ so certainly it does on $Z(B)$ and we have $\SOF(B)\leq n$.

In the case $D=(D\cap N)Z(D)$, the proof of Theorem~\ref{sfreduce:theorem} can be greatly shortened as we may avoid reference to Proposition~\ref{prop:zeta_auto}. Indeed, given the comments following Lemma~\ref{lem:graded_units}, we only need that $\zeta_G$ induces an $\cO$-algebra automorphism between $Z(B)$ and $Z(B^{(p)})$. The proof then follows in very much the same vein as the abelian defect group case in~\cite[Theorems 3.16]{el19}.\hfill $\Box$

\section{A reduction theorem for Donovan's conjecture}
\label{Donovan_reduce:sec}

The following reduction result for Cartan invariants is presumably well-known, but we provide a proof. Write ${\rm c}(B)$ for the largest entry of the Cartan matrix of a block $B$.

\begin{lemma}
\label{Cartan_reduce:lemma}
Let $G$ be a finite group and $B$ a block of $kG$.
\begin{enumerate}[(i)]
    \item Let $N \lhd G$ have index $p^r$ and suppose $B$ covers a block $b$ of $kN$. Then ${\rm c}(B) \leq p^r {\rm c}(b)$.
    \item Let $Z\leq Z(G)$ be a $2$-group and $\overline{B}$ the corresponding block of $G/Z$. Then ${\rm c}(\overline{B}) \leq {\rm c}(B)$.
\end{enumerate}

\end{lemma}

\begin{proof}
\begin{enumerate}[(i)]
\item Let $B'$ be the unique block of $k\Stab_G(b)$ covering $b$. Then, by~\cite[Theorem C]{ku81}, $B'\sim_{\Mor}B$ and so we may assume that $b$ is $G$-stable and, as noted in the proof of Lemma~\ref{lem:graded_units}, that $e_B=e_b$. Therefore, by Green's Indecomposability Theorem, for every projective, indecomposable $b$-module $M$, the induced module $M\uparrow^G$ is a projective, indecomposable $B$-module. Furthermore, since every projective $B$-module $L$ is a summand of $e_B(L\downarrow_N\uparrow^G)=(e_b L\downarrow_N)\uparrow^G$, in fact every projective, indecomposable $B$-module is isomorphic to $M\uparrow^G$ for some projective, indecomposable $b$-module $M$.

Now fix some projective, indecomposable $b$-module $M$ and consider the composition factors of $M\uparrow^G$. Let $S$ be a simple $B$-module and $T$ a simple $b$-module appearing as a composition factor in $S\downarrow_N$. Certainly the multiplicity of $S$ among the composition factors of $M\uparrow^G$ is at most the multiplicity of $T$ among the composition factors of $(M\uparrow^G)\downarrow_N$. However, $(M\uparrow^G)\downarrow_N$ is the direct sum of $p^r$ projective, indecomposable $b$-modules, the $G$-conjugates of $M$. The claim follows.

\item For a $kG$-module $M$ we define the $k(G/Z)$-module ${}^ZM$ to be the fixed points of $M$ under left multiplication by $Z$. Note that ${}^Z(kG)\cong k(G/Z)$. Therefore, if $M$ is the projective cover of the simple $B$-module $S$, then ${}^ZM$ is the projective cover of the simple $\overline{B}$-module ${}^ZS=S$. In particular, ${\rm c}(\overline{B}) \leq {\rm c}(B)$.
\end{enumerate}
\end{proof}

We now prove Theorem~\ref{Donovan reduce:theorem}.
\newline
\newline
{\sc Proof of Theorem}~\ref{Donovan reduce:theorem}. Fix a finite $p$-group $P$.

Let $\cal{X}$ be the class of blocks $C$ of $\cO H$ for some finite group $H$ with defect group $Q$ isomorphic to a subgroup of $P$ and no normal subgroup $M \lhd H$ such that $H=C_Q(Q \cap M)M$. Suppose that there are only finitely many Morita equivalence classes amongst the members of $\cal{X}$. Then there is a largest Cartan invariant $c$ and a largest strong $\cO$-Frobenius number $s$ amongst blocks in $\cal{X}$.

Let $G$ be a finite group and $B$ a block of $\cO G$ with defect group $D$ isomorphic to a subgroup of $P$. We claim that ${\rm c}(B)\leq |D|c$ and $\SOF(B)\leq s$. Suppose that ${\rm c}(B)>|D|c$ and that $|G|$ is minimal with respect to these conditions. By the definition of the constant $c$, $B$ is not in $\cal{X}$ and so there is a proper subgroup $N \lhd G$ with $G=C_D(D \cap N)N$. Let $b$ be a block of $\cO N$ covered by $B$. Note that by Lemma~\ref{lem:graded_units} $b$ has defect group $D\cap N$. Then by Lemma~\ref{Cartan_reduce:lemma}(i)
\begin{align*}
{\rm c}(b) \geq [D:D \cap N]^{-1}{\rm c}(B) > [D\cap N]c,
\end{align*}
contradicting the minimality of $G$. A similar argument using Theorem~\ref{sfreduce:theorem} shows the bound on strong $\cO$-Frobenius numbers. We have shown that the Cartan invariants and the strong $\cO$-Frobenius numbers of all blocks with defect group isomorphic to $P$ are bounded, and so the result follows by~\cite[Corollary 3.11]{eel19}.\hfill $\Box$

\section{Blocks of quasisimple groups with defect groups $Q_8 \times C_{2^n}$ and $Q_8 \times Q_8$}
\label{sec:classification_application}

Let $G$ be a block of a finite group $G$ with defect group $D$ and maximal $B$-subpair $(D,b_D)$. Recall that $B$ is \emph{controlled} if for all $B$-subpairs $(Q,b_Q) \leq (D,b_D)$ and $g \in G$ with $(Q,b_Q)^g \leq (D,b_D)$, there are $c \in C_G(Q)$ and $n \in N_G(D,b_D)$ such that $g=cn$. We will use~\cite[Theorem 4.8]{st06} to observe that every block with defect group $Q_8^m \times A$ for $m \geq 0$ and $A$ an abelian $2$-group is controlled, and apply the classification of controlled blocks of quasisimple groups given in~\cite{an20}. To do so we first review some notation.

Let $\mathcal{F}$ be a saturated fusion system on a $p$-group $D$. A subgroup $Q \leq D$ is weakly $\cF$-closed if for any $\phi \in \Hom_\cF(Q,D)$ we have $\phi(Q)=Q$, and $Q$ is strongly $\cF$-closed if for any $P \leq Q$ and any $\phi \in \Hom_\cF(P,D)$ we have $\phi(P) \leq Q$.

The normalizer $N_\cF(D)$ is the fusion subsystem of $\cF$ on $D$ such that for all $P,Q \leq D$, the morphisms $\Hom_{N_\cF(D)}(P,Q)$ are those $\phi \in \Hom_\cF(P,Q)$ such that there is $\bar{\phi} \in \Hom_\cF(D,D)$ extending $\phi$.

A $p$-group $D$ is called \emph{resistant} if $\mathcal{F}=N_{\mathcal{F}}(D)$ whenever $\mathcal{F}$ is a saturated fusion system on $D$.

\begin{proposition}
Let $D=Q_8^m \times A$, where $m \geq 0$ and $A$ is an abelian $2$-group. Then $D$ is resistant.
\end{proposition}

\begin{proof}
Let $\mathcal{F}$ be a saturated fusion system on $D$. By~\cite[Theorem 4.8]{st06} $\mathcal{F}=N_{\mathcal{F}}(D)$ if and only if there is a central series $D=Q_n \geq Q_{n-1} \geq \cdots \geq Q_1 \geq 1$ with each $Q_i$ weakly $\mathcal{F}$-closed and $D$ strongly $\cF$-closed.

Now let $Q_1=\Omega_1(D)$, the (unique) largest elementary abelian subgroup of $D$, and $Q_2=D$, so $Q_2 \geq Q_1 \geq 1$ forms a central series. Since $Q_1$ is the unique elementary abelian subgroup of $D$ of maximal rank, it must be weakly $\mathcal{F}$-closed. Also $D$ is automatically strongly $\cF$-closed, so the result follows.
\end{proof}

Now let $\mathcal{F}$ be the fusion system on $D$ afforded by $B$ and $(D,b_D)$, sometimes written $\mathcal{F}_{(D,b_D)}(G)$. By Alperin's fusion theorem $B$ is controlled if and only if $N_G(D,b_D)$ controls strong fusion in $D$ (see~\cite[Proposition 4.24]{ab79}). This is equivalent to $\mathcal{F}=N_\mathcal{F}(D)$. We conclude that every block with defect group $D \cong Q_8^m \times A$, where $m \geq 0$ and $A$ is an abelian $2$-group, is controlled.

\begin{proposition}
\label{none_in_quasisimples:prop}
Let $G$ be a finite quasisimple group. Let $m \geq 1$ and $A$ be a finite abelian $2$-group. If $m > 1$, then there is no block of $G$ with defect groups isomorphic to $(Q_8)^m \times A$. If $B$ is a block of $G$ with defect group $Q_8 \times A$ where $A$ is nontrivial, then $G$ is a quotient of a classical group of Lie type not of type $\mathbf{A}$ or ${}^2\mathbf{A}$, defined over a field of order $q$ a power of an odd prime and $B$ corresponds to a non-quasi-isolated block of the corresponding group of Lie type. Furthermore, if $m=1$ and $A \cong C_4$, then $G \cong Sp_{2r}(q)$ for some $r$.
\end{proposition}

\begin{proof}
By the discussion above, all blocks with these defect groups are controlled. The result follows directly from~\cite[Theorem 1.1]{an20} and its proof.
\end{proof}

When we come to consider blocks of arbitrary groups with defect group $Q_8 \times Q_8$, we must deal with the case of blocks $b$ of $Sp_{2r}(q)$ with defect group $Q_8 \times C_4$. We show that there can be no overgroup of $Sp_{2r}(q)$ in $\Aut(Sp_{2r}(q))$ possessing a block covering $b$ with defect group $Q_8 \times Q_8$.

In what follows, for $r\in\mathbb{N}$ and $q$ a power of a prime, we set $I_r \in GL_r(q)$ to be the $r\times r$ identity matrix,
\begin{align*}
J_r:=
\begin{pmatrix}
0 & 0 & \cdots & 0 & 1 \\
0 & 0 & \cdots & 1 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 1 & \cdots & 0 & 0 \\
1 & 0 & \cdots & 0 & 0
\end{pmatrix}\in GL_r(q)
\end{align*}
and
\begin{align*}
\Omega_{2r}:=
\begin{pmatrix}
0 & J_r \\
-J_r & 0
\end{pmatrix}\in GL_{2r}(q).
\end{align*}
We define
\begin{align*}
Sp_{2r}(q)&:=\{x\in GL_{2r}(q)|x\Omega_{2r}x^T=\Omega_{2r}\},\\
CSp_{2r}(q)&:=\{x\in GL_{2r}(q)|x\Omega_{2r}x^T=\lambda \Omega_{2r} \text{ for some }\lambda\in\mathbb{F}_q^\times\}.
\end{align*}

\begin{lemma}
\label{noQ8xQ8ext:Lemma}
Let $H:=Sp_{2r}(q)$, where $4\mid(q-1)$ but $8\nmid(q-1)$ and $b$ a block of $\cO H$ with defect group $P \cong Q_8\times C_4$ labelled by some $s\in H^*=SO_{2r+1}(q)$ such that $m_{X-1}(s)=3$ and other $m_\Gamma(s)\leq 1$. Furthermore, let $H\lhd G$ such that $C_G(H)\leq H$ and $B$ a block of $\cO G$ lying above $b$ with defect group $D$. Then $D\ncong Q_8\times Q_8$.
\end{lemma}

\begin{proof}
We first describe $P\leq H$ (see~\cite{fs89} for a description of defect groups of finite classical groups). We denote by $i\in\mathbb{F}_q^\times$ a primitive $4^{\nth}$ root of unity. Then $P=P_1\times P_2\leq Sp_2(q)\times Sp_{2(r-1)}(q)\leq Sp_{2r}(q)$, where $P_1=\Syl_2(Sp_2(q))\cong Q_8$ is generated by
\begin{align*}
\begin{pmatrix}
i & 0 \\
0 & -i
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}
\end{align*}
and $P_2\cong C_4$.

Since $C_G(H)\leq H$, $G/H\hookrightarrow\Out(H)$ and as $8\nmid(q-1)$, $q$ is not a square and so $H$ has no field automorphisms of order $2$. It follows that $\Out(H)$ has a normal Sylow $2$-subgroup of order $2$ generated by the diagonal automorphism induced by
\begin{align*}
\begin{pmatrix}
iI_r & 0 \\
0 & I_r
\end{pmatrix}.
\end{align*}
Furthermore, this automorphism restricted to $Sp_2(q)\times Sp_{2(r-1)}(q)$ is induced by
\begin{align*}
(g_1,g_2):=\left(\begin{pmatrix}
i & 0 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
iI_{r-1} & 0 \\
0 & I_{r-1}
\end{pmatrix}\right)\in GL_2(q)\times GL_{2(r-1)}(q).
\end{align*}
Suppose $D \cong Q_8\times Q_8$. Then we may choose $D$ such that the unique block of $\cO HD$ covering $b$ has defect group $D$. Moreover, the image of $D$ in $\Out(H)$ is generated by the non-trivial diagonal automorphism.

We will have reached our desired contradiction once we have proved that no $h\in D\backslash P$ commutes with $P_1$. Since
\begin{align*}
C_{Sp_{2r}(q)}(Z(P_1))=C_{Sp_{2r}(q)}\left(\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix},
\begin{pmatrix}
I_{r-1} & 0 \\
0 & I_{r-1}
\end{pmatrix}\right)=Sp_2(q)\times Sp_{2(r-1)}(q),
\end{align*}
it is enough to show there exists no $g\in Sp_2(q)$ such that $g_1g^{-1}\in C_{CSp_2(q)}(P_1)$. This follows from the fact that $C_{CSp_2(q)}(P_1)=Z(CSp_2(q))$ and that conjugation by $g_1$ is not an inner automorphism of $Sp_2(q)$. These two facts can be readily checked.
\end{proof}

\section{Blocks with defect group $Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$}
\label{sec:background}

We begin by gathering together some information on subgroups and automorphism groups of these $2$-groups, easily verified by the reader. Blocks with defect group $Q_8 \times C_{2^n}$ were studied in~\cite{sa13}, where many of their numerical invariants were computed.

\begin{lemma}
\label{Dsubs:Lem}
\begin{enumerate}[(a)]
\item Let $P \cong Q_8 \times C_{2^n}$, where $n \geq 1$. Let $Q \leq P$ with $[P:Q]=2$.

\begin{enumerate}[(i)]
\item If $n=1$, then $Q \cong Q_8$ or $C_4 \times C_2$.

\item If $n \geq 2$, then $Q\cong C_4 \times C_{2^n}$, $Q_8 \times C_{2^{n-1}}$ or $C_4 \rtimes C_{2^n}$.
\end{enumerate}

In particular, every subgroup of $P$ has the form $1$, $C_{2^m}$, $C_2 \times C_{2^m}$, $C_4 \times C_{2^m}$, $Q_8$, $Q_8 \times C_{2^m}$ or $C_4 \rtimes C_{2^m}$ for some $m$, with $m\geq 2$ in the final case.

\item The proper subgroups of $Q_8 \times Q_8$ are isomorphic to the following: $1$, $C_2$, $C_4$, $C_2 \times C_2$, $C_4 \times C_2$, $C_4 \times C_4$, $Q_8$, $Q_8 \times C_2$, $Q_8 \times C_4$, $C_4 \rtimes C_4$ and $C_4 \rtimes Q_8$.
\end{enumerate}
\end{lemma}

\begin{lemma}
\label{Dautos:Lem}
\begin{enumerate}[(i)]
\item $\Aut(C_{2^n} \times C_{2^m})$ is a $2$-group if $m \neq n$, and is a $\{2,3\}$-group with Sylow $3$-subgroup of order $3$ if $m=n$.

\item $\Aut(Q_8 \times C_{2^n})$ is a $\{2,3\}$-group with Sylow $3$-subgroup of order $3$.

\item $\Aut(Q_8 \times Q_8)$ is a $\{2,3\}$-group with Sylow $3$-subgroup $C_3 \times C_3$.

\item $\Aut(C_4 \rtimes C_{2^n})$ is a $2$-group.

\item $\Aut(C_4 \rtimes Q_8)$ is a $2$-group.
\end{enumerate}
\end{lemma}

The key to our treatment of blocks with defect group $Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$ is that in most cases covered blocks of normal subgroups of index $2$ are nilpotent. This is covered in the following lemma.

\begin{lemma}
\label{cyclicnilpotent}
Let $B$ be a $2$-block of a finite group $G$ with defect groups isomorphic to (i) $C_{2^m} \times C_{2^n}$ for $m \neq n$, (ii) $C_4 \rtimes C_{2^n}$ for $n \geq 2$, or (iii) $C_4 \rtimes Q_8$. Then $B$ is nilpotent.
\end{lemma}

\begin{proof}
Let $D$ be a defect group for $B$.

(i) Since $D$ is abelian, it suffices to observe that $\Aut(D)$ is a $2$-group.

(ii) Since $C_4 \rtimes C_{2^n}$ is metacyclic, by~\cite[Theorem 3.7]{cg12} there is only one saturated fusion system on this $2$-group, and so $B$ must be nilpotent.

(iii) There is only one saturated fusion system on $C_4 \rtimes Q_8$ by~\cite[Table 13.1]{sa14}.
\end{proof}

We summarize the results of~\cite{sa13} that we need here:

\begin{proposition}[\cite{sa13}]
\label{numbersimples}
Let $B$ be a block with defect group $Q_8 \times C_{2^n}$ for some $n$. Then one of the following occurs:

\begin{enumerate}[(i)]
\item $k(B)=2^n\cdot 7$ and $l(B)=3$;
\item $B$ is nilpotent, $k(B)=2^n \cdot 5$ and $l(B)=1$.
\end{enumerate}
\end{proposition}

\begin{proof}
This follows from~\cite[Lemma 2.2, Theorem 2.7]{sa13}.
\end{proof}

The next result will be used frequently without reference throughout the remainder of the article.

\begin{proposition}\label{nilpotent_index_p:prop}
Let $G$ be a finite group and $N \lhd G$ with $G/N$ a $p$-group. Let $b$ be a block of $\cO N$ and $B$ the unique block of $\cO G$ covering $b$. Then $B$ is nilpotent if and only if $b$ is.
\end{proposition}

\begin{proof}
That $B$ nilpotent implies $b$ nilpotent is clear. The other direction is~\cite[Theorem 2]{ca87}.
\end{proof}

\begin{lemma}
\label{odd_index_Q_8xC2^n:lemma}
Let $G$ be a finite group and $N \lhd G$ such that $G/N$ is supersolvable of odd order. Let $B$ be a block of $\cO G$ with defect group $Q_8 \times C_{2^n}$ for some $n$ and let $b$ be a block of $\cO N$ covered by $B$. Suppose that $B$ covers no nilpotent block of any normal subgroup containing $N$. Then $B$ and $b$ are Morita equivalent.
\end{lemma}

\begin{proof}
Note that $B$ and $b$ must share a defect group, and that for any $M \lhd G$ with $N \leq M$ and any block $C$ of $\cO M$ covered by $B$ we have $l(C)=3$ by Proposition \ref{numbersimples}. By considering a chief series between $N$ and $G$ with prime factors it suffices to consider the case that $[G:N]$ is an odd prime $w$. If $b$ is not $G$-stable, then we are done. Suppose that $b$ is $G$-stable. By~\cite[Proposition 2.2]{kkl12} if $G$ acts as inner automorphisms on $b$, then $B$ and $b$ are Morita equivalent. By~\cite[Proposition 2.3]{kkl12} if $G$ does not act as inner automorphisms on $b$, then $B$ is the unique block of $\cO G$ covering $b$. Consider the action of $G$ on the three irreducible Brauer characters of $b$. If $w \geq 5$, then $G$ must fix every such Brauer character, and $l(B)=wl(b)=3w$, contradicting $l(B)=3$. If $w=3$, then either $G$ fixes each irreducible Brauer character, again a contradiction, or $G$ permutes the irreducible Brauer characters of $b$ transitively and $l(B)=1$, contradicting our assumption that $B$ is not nilpotent (using Proposition \ref{numbersimples}).
\end{proof}

\begin{lemma}
\label{coveringC4xC4:lem}
Let $G$ be a finite group and $N \lhd G$ with $[G:N]=2$. Let $b$ be a non-nilpotent block of $\cO N$ with defect groups isomorphic to $C_4 \times C_4$. Then the unique block $B$ of $\cO G$ covering $b$ cannot have defect group $Q_8 \times C_4$.
\end{lemma}

\begin{proof}
We assume the contrary. By Proposition \ref{numbersimples} we have $k(B)\geq 20$ and by~\cite[Theorem 1.1]{ekks14} $k(b)=8$. But by Clifford theory the number of irreducible characters of $G$ lying over irreducible characters of $b$ is at most $16$, and the result follows.
\end{proof}

A block of a finite group $G$ is described as quasiprimitive if every covered block of a normal subgroup of $G$ is $G$-stable. The following is proved in~\cite[Lemma 2.4]{ar19}.

\begin{lemma}
\label{solv_quotient:lemma}
Let $G$ be a finite group and $B$ be a quasiprimitive block of $\cO G$ with defect group $D$. Let $N \lhd G$ such that $G/N$ is solvable and $b$ the unique block of $\cO N$ covered by $B$. Then $ND/N \in \Syl_p(G/N)$.
\end{lemma}

Recall that a $p$-solvable group $G$ has $p$-length one if there are normal subgroups $N, M \lhd G$, with $N \leq M$, such that $N$ and $G/M$ are $p'$-groups and $M/N$ is a $p$-group. The abelian Sylow $2$-subgroup case of the following is well-known, and possibly also the $Q_8$ case, but since we do not know a reference we include a proof.

\begin{lemma}
\label{two_length:lemma}
Let $G$ be a solvable group with Sylow $2$-subgroups which are abelian or $Q_8$. Then $G$ has $2$-length one. If further $G$ has cyclic $2$-subgroups, then it is $2$-nilpotent.
\end{lemma}

\begin{proof}
We may assume that $O_{2'}(G)=1$,
so that $O_2(G) \neq 1$ and $C_G(O_2(G)) \leq O_2(G)$. Let $P \in \Syl_2(G)$. If $P$ is abelian, then $P \leq C_G(O_2(G)) \leq O_2(G)$ and we are done. Suppose that $P \cong Q_8$. Since $Z(P)$ is the unique subgroup of $P$ of order two, we have $Z(P) \leq O_2(G)$. If $O_2(G)=Z(P)$ or $O_2(G)=P$, then we are done. If $O_2(G) \cong C_4$, then $G/O_2(G)$ is a $2$-group and again we are done.

In the case that $P$ is cyclic, the fact that $G$ is $2$-nilpotent follows from $\Aut(P)$ being a $2$-group.
\end{proof}

The following is by now a standard reduction when treating Donovan's conjecture, using Fong reductions and~\cite{kp90}.
\begin{proposition}
\label{reduce:prop}
Let $G$ be a finite group and let $B$ be a block of $\cO G$ with defect group $D$. There is a finite group $H$ with $[H:O_{p'}(Z(H))] \leq [G:O_{p'}(Z(G))]$ and a block $C$ of $\cO H$ with defect group $P \cong D$ such that $B$ is Morita equivalent to $C$ and the following are satisfied:
\begin{enumerate}
\item[(R1)] $C$ is quasiprimitive;
\item[(R2)] If $N \lhd H$ and $C$ covers a nilpotent block $c$ of $\cO N$, then $N \leq Z(H) O_p(H)$;
\item[(R3)] $O_{p'}(Z(H)) \leq [H,H]$.
\end{enumerate}
\end{proposition}

\begin{proof}
See the first part of the proof of~\cite[Proposition 4.3]{eel19}.
\end{proof}

For the purpose of this article, we call the pair $(H,C)$, where $C$ is a block of $\cO H$, \emph{reduced} if it satisfies conditions (R1), (R2) and (R3) of Proposition \ref{reduce:prop}. If the group is clear, then we just say $C$ is reduced. The property of being reduced is very restrictive in our situation:

\begin{proposition}
\label{structureQ8xC2n:prop}
Let $(G,B)$ be a reduced pair, where $B$ has defect group $D$.
\begin{enumerate}[(a)]
\item If $D \cong Q_8 \times C_{2^n}$, then either:
\begin{enumerate}[(i)]
\item there is $N \lhd G$ such that $G=ND$ and $D \cap N \cong Q_8$ (so $G/N$ is cyclic); or
\item there is $N \lhd G$, a quasisimple group that is a quotient of a group of classical Lie type other than $\mathbf{A}$ or ${}^2\mathbf{A}$, such that $G/N$ is $2$-nilpotent with cyclic Sylow $2$-subgroup $ND/N$ and $O_{2'}(G/N)$ is supersolvable. The unique block of $\cO N$ covered by $B$ is not quasi-isolated and $D \cap N$ has a subgroup isomorphic to $Q_8$.
\end{enumerate}
\item If $D \cong Q_8 \times Q_8$, then either:
  \begin{enumerate}[(i)]
  \item there are normal subgroups $N \lhd H \lhd G$ such that $H=ND$, $D \cap N \cong Q_8$ or $Q_8 \times C_2$, and $[G:H]$ is odd; or
  \item there are commuting, normal subgroups $N_1, N_2 \lhd G$ such that $N_1 \cap N_2 \leq Z(G)$, $D \cap N_1$, $D \cap N_2 \cong Q_8$ or $Q_8 \times C_2$, and $[G:N_1N_2]$ is odd.
  \end{enumerate}

\end{enumerate}
\end{proposition}

\begin{proof}
Let $G$ be a finite group and $B$ a block of $\cO G$ with defect group $D \cong Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$ for some $n \geq 1$, and suppose that $(G,B)$ is reduced. Write $E(G)$ for the layer of $G$, the central product of all of the components of $G$, and $F^*(G)$ for the generalized Fitting subgroup (see~\cite{asc00}). Then $F^*(G)=E(G)Z(G)O_2(G)$ by our assumption, and $C_G(F^*(G)) \leq F^*(G)$ because of general properties of $F^*(G)$. Write $E(G)$ for the central product $L_1 \circ \cdots \circ L_t$, where the $L_i$ are the components of $G$. Write $B_E$ for the unique block of $\cO E(G)$ covered by $B$. Then (using the fact that $B_E$ is $G$-stable) $D_E:=D \cap E(G)$ is a defect group for $B_E$ and $D_i:=D \cap L_i$ is a defect group for the unique block $b_i$ of $\cO L_i$ covered by $B_E$. By Lemma \ref{Dsubs:Lem} $O_2(G)$, $D_E$ and each $D_i$ is of the form $1$, $C_{2^m}$, $C_2 \times C_2$, $C_4 \times C_{2^m}$, $Q_8$, $Q_8 \times C_{2^m}$, $Q_8 \times Q_8$, $C_4 \rtimes C_{2^m}$ or $C_4 \rtimes Q_8$ for some $m$ (with $m \geq 2$ in the eighth case). Write $Z_E:=O_2(E(G))$. Note that $Z_E$ is central in $E(G)$. Now the unique block $\overline{B}_E$ of $\cO (E(G)/Z_E)$ corresponding to $B_E$ has defect group $\overline{D}_E:=D_E/Z_E \cong D_1Z_E/Z_E \times \cdots \times D_tZ_E/Z_E$ and has $2$-rank at most four (we are using that $\overline{B}_E$ corresponds to a block of a direct product of quasisimple groups of which $E(G)/Z_E$ is a quotient by a $2'$-group).

Suppose $t \geq 3$, or $t=2$ in case (a). Then at least one $D_iZ_E/Z_E$ is a cyclic $2$-group, so that the unique block $\overline{b}_i$ of $L_iZ_E/Z_E$ corresponding to $b_i$ is nilpotent. Let $M$ be the product of those $L_i$ such that $\overline{b}_i$ is nilpotent. Then $M$ must be a normal subgroup (since $G$ permutes the components). Let $B_M$ be the unique block of $\cO M$ covered by $B$ and $\overline{B}_M$ be the unique block of $MZ_E/Z_E$ corresponding to $B_M$. Now $\overline{B}_M$ is isomorphic to a product of nilpotent blocks and so is itself nilpotent ($\overline{B}_M$ corresponds to a block of a direct product of quasisimple groups of which $MZ_E/Z_E$ is a quotient by a $2'$-group). By~\cite{wa94} $B_M$ is then nilpotent, a contradiction to our assumption that $B$ is reduced. Hence $t \leq 2$, with equality only if $D \cong Q_8 \times Q_8$. It follows from Schreier's conjecture and Lemmas~\ref{Dsubs:Lem} and~\ref{Dautos:Lem} that $G/F^*(G)$ has a solvable subgroup of index at most two (namely the subgroup stabilizing the components if $t \geq 1$, and $G/F^*(G)$ itself if there are no components) and so $G/F^*(G)$ is itself solvable. Similarly $G/E(G)$ is solvable. By Lemma \ref{solv_quotient:lemma} $DF^*(G)/F^*(G)$ is a Sylow $2$-subgroup of $G/F^*(G)$ and $DE(G)/E(G)$ is a Sylow $2$-subgroup of $G/E(G)$, facts we will use frequently.

Suppose that $t=2$, in which case $D \cong Q_8 \times Q_8$. Note also that each component is normal in $G$ since $G/E(G)$ is of odd order (i.e., there is no involution permuting the two components). By considering all of the possible expressions of $Q_8 \times Q_8$ as a central product of two groups in Lemma \ref{Dsubs:Lem}, we must have (without loss of generality) $D_1$ and $D_2 \cong Q_8$ or $Q_8 \times C_2$, otherwise $B$ covers a block of a component with cyclic defect group, which forces a contradiction as in the previous paragraph. Note that $O_2(G) \leq Z(G)$ in this case. We are now in case (b)(ii) of the statement, with $N_i=L_i$.

Suppose that $t=0$. Then $F^*(G)=O_2(G)Z(G)$. Since $C_G(F^*(G)) \leq F^*(G)$, we have $G/F^*(G) \leq \Out(O_2(G))$. In particular $O_2(G)$ is a self-centralizing subgroup of $D$. Note that if $D = O_2(G)$, then $G\cong D\rtimes E$ for some $2'$-group $E$. It follows from Lemma~\ref{Dautos:Lem} that $G/C_E(D)$ is a subgroup of $SL_2(3) \times C_{2^n}$ or $SL_2(3)\times SL_2(3)$ and we must be in case (a)(i) or (b)(i) respectively. Suppose $O_2(G)$ is a proper subgroup of $D$. Note that, by Lemma \ref{Dautos:Lem}, all self-centralizing subgroups of $D$ have solvable automorphism group.

Suppose first that $D \cong Q_8 \times C_{2^n}$. By Lemma \ref{Dsubs:Lem} the only proper self-centralizing normal subgroup of $D$ is $C_4 \times C_{2^n}$ and we need only consider the case $O_2(G) \cong C_4 \times C_4$ as in the other cases the automorphism group is a $2$-group. In this case we must have that $G/O_2(G) \cong S_3$ and $B$ covers a non-nilpotent block of $O^2(G)$. However this cannot happen by Lemma \ref{coveringC4xC4:lem}. Now suppose that $D \cong Q_8 \times Q_8$. By Lemma \ref{Dsubs:Lem} the proper self-centralizing normal subgroups of $D$ are $C_4 \times C_4$, $Q_8 \times C_4$ and $C_4 \rtimes Q_8$, of which we need only consider $C_4 \times C_4$ and $Q_8 \times C_4$ since $C_4 \rtimes Q_8$ has automorphism group a $2$-group. If $O_2(G) \cong C_4 \times C_4$, then $G/F^*(G)$ has order $12$ and so has a non-trivial normal $2$-subgroup, a contradiction. If $O_2(G) \cong Q_8 \times C_4$, then $G/F^*(G) \cong S_3$. This forces $G \cong SL_2(3) \times Q_8$ and we are in case (a)(i).

Now suppose that $t=1$, so $F^*(G)=O_2(G)Z(G)L_1$.

Since, by Lemma \ref{cyclicnilpotent}, every block with defect group $C_4 \rtimes Q_8$, $C_4 \rtimes C_{2^m}$ or $C_4 \times C_{2^s}$ is nilpotent for $m \geq 2$ and $s \neq 2$, we may assume, by Lemma \ref{Dsubs:Lem}, that $D_1 \cong Q_8 \times Q_8$, $Q_8 \times C_{2^m}$, $Q_8$, $C_4 \times C_4$ or $C_2 \times C_2$ for some $m \geq 1$. We treat each of these cases in turn.

By Proposition \ref{none_in_quasisimples:prop} we cannot have $D_1 \cong Q_8 \times Q_8$.

Suppose that $D_1 \cong Q_8 \times C_{2^m}$ for $m \geq 1$. Consider first the case $D \cong Q_8 \times C_{2^n}$. By Proposition \ref{none_in_quasisimples:prop} $L_1$ is of classical type other than $\mathbf{A}$ or ${}^2\mathbf{A}$ as in case (a)(ii) of the statement with $N=L_1$. That $G/N$ has cyclic Sylow $2$-subgroup $ND/N$ is immediate, and so by Lemma \ref{two_length:lemma} $G/N$ is $2$-nilpotent. Considering the outer automorphism groups of such quasisimple groups we have that $O_{2'}(G/L_1)$ is supersolvable as required. As a note to this last calculation, observe that unless $L_1$ has type $\mathbf{D}_4$ the only odd order elements of the outer automorphism group are field automorphisms, so that $O_{2'}(G/L_1)$ is cyclic. In the case of type $\mathbf{D}_4$ we may either analyse~\cite[Theorem 1.1]{an20} a little more deeply and observe that this case does not after all occur, or observe that a Hall $2'$-subgroup of $\Out(L_1)$ is a subgroup of $C_3 \times Y$ where $Y$ is cyclic, implying that $O_{2'}(G/L_1)$ is supersolvable.

Now suppose that $D_1 \cong Q_8 \times C_{2^m}$ for $m \geq 1$ and $D \cong Q_8 \times Q_8$. Then $1 \leq m \leq 2$. The case $m=2$ is ruled out by Proposition \ref{none_in_quasisimples:prop} and Lemma \ref{noQ8xQ8ext:Lemma}. If $m=1$, then we are in case (b)(i).

Suppose that $D_1 \cong Q_8$. Then since $G/L_1$ is a solvable group which has cyclic or $Q_8$ Sylow $2$-subgroup, by Lemma \ref{two_length:lemma}, it has $2$-length one. Let $N$ be the preimage in $G$ of $O_{2'}(G/L_1)$. Then the unique block $B_N$ of $\cO N$ covered by $B$ has defect group $Q_8$.  If $D/D_1$ is cyclic, then by Lemma \ref{two_length:lemma} $G/L_1$ is $2$-nilpotent and we are in case (a)(i) of the statement. If $D/D_1 \cong Q_8$, then let $H = O^{2'}(G)$ and we are in case (b)(i) of the statement.

Finally we claim that we cannot have $D_1 \cong C_4 \times C_4$ or $C_2 \times C_2$. Suppose that we do and so $D/D_1$ is an abelian. Let $M$ be the preimage in $G$ of $O_{2'}(G/L_1)$, and let $B_M$ be the unique block of $\cO M$ covered by $B$. Then $B_M$ has defect group $D_1$. Now there is a subgroup $M_1$ of $G$ containing $M$ with $[M_1:M]=2$. The unique block $B_{M_1}$ of $\cO M_1$ covering $B_M$ has defect group $Q_8 \times C_4$ or $C_4\rtimes Q_8$ (if $D_1 \cong C_4 \times C_4$) or $C_4 \times C_2$ (if $D_1 \cong C_2 \times C_2$), since it is a subgroup of $D$. In the $Q_8\times C_4$ case we obtain a contradiction by Lemma \ref{coveringC4xC4:lem}. In the $C_4\rtimes Q_8$ and $C_4\times C_2$ cases by Lemma \ref{cyclicnilpotent} $B_M$, and so by Proposition \ref{nilpotent_index_p:prop} $b_i$, must be nilpotent, a contradiction.

\end{proof}

\begin{corollary}
\label{sfQ8xC:cor}
Let $G$ be a finite group and $B$ a block of $\cO G$ with defect group $D$ isomorphic to $Q_8 \times C_{2^n}$ for $n \geq 0$. Then $\SOF(B) \leq (|D|^2)!$ and ${\rm c}(B) \leq 2^{n+3}$.
\end{corollary}

\begin{proof}
Morita equivalence preserves the Cartan invariants, and by~\cite[Proposition 3.12]{el19} Morita equivalence of $\cO$-blocks preserves $\SOF$, so by Proposition \ref{reduce:prop} it suffices to consider reduced blocks. Apply Proposition \ref{structureQ8xC2n:prop}.

First let $(G,B)$ be a reduced pair satisfying condition (a)(ii) of Proposition \ref{structureQ8xC2n:prop}. Let $B'$ be the unique block of $\cO O^2(G)$ covered by $B$. By Lemma~\ref{Cartan_reduce:lemma}(i) ${\rm c}(B) \leq [G:O^2(G)]{\rm c}(B')$ and by Theorem \ref{sfreduce:theorem} $\SOF(B) \leq \SOF(B')$. By considering the outer automorphism groups of the classical groups of Lie type (see for example~\cite[Table 5]{atlas}) we see that every normal subgroup of $O^2(G)$ is also normal in $G$, so $B'$ is quasiprimitive. Hence we see that $B'$ is also a reduced block satisfying (a)(ii). We may now assume that $O^2(G)=G$. Now let $B_N$ be the unique block of the quasisimple group $N$ covered by $B$. By Lemma \ref{odd_index_Q_8xC2^n:lemma} $B$ is Morita equivalent to $B_N$, and note that they share a defect group. Hence we may assume that $G=N$ since Morita equivalence preserves both of these invariants (see~\cite[Proposition 3.12]{el19} for the latter).

We make use of~\cite{bdr17}, to which we refer for notation. Assume for the moment that $G$ is a group of Lie type, i.e, the centre is largest possible. We note that $Z(G)$ is a $2$-group for the groups we are considering. Here the identity element is the only quasi-isolated element (see for example~\cite[Table 2]{bo05}) and so the principal block is the only quasi-isolated block.  However, for groups of these types the principal block of $G$ (or that of any quotient of $G$) cannot have the given defect groups, so we may assume our block is not quasi-isolated. We may now apply the Bonnaf\'e-Dat-Rouquier correspondence~\cite[Theorem 7.7]{bdr17}, so that $B$ is Morita equivalent to a block $C_1$ (with isomorphic defect group) of a proper subgroup $H_1$ of $G$. We note that the well-known error in~\cite{bdr17} does not apply in our situation, since we are working with $2$-blocks and the centre of $G$ is a $2$-group. If $G$ is not a group of Lie type (i.e., the centre is not largest possible), then we note that by, for example,~\cite[Proposition 4.1]{ekks14} the Bonnaf\'e-Dat-Rouquier correspondence induces a Morita equivalence modulo central $2$-subgroups and we may apply the same argument.

Applying Proposition \ref{reduce:prop} to $C_1$, it is Morita equivalent to a block $C_2$ in a reduced pair $(H_2,C_2)$ with isomorphic defect groups, where $[H_2:O_{2'}(Z(H_2))] \leq [H_1:O_{2'}(Z(H_1))]<[G:O_{2'}(Z(G))]$. We have ${\rm c}(B) = {\rm c}(C_1)$ and $\SOF(B)=\SOF(C_1)$. Now apply Proposition \ref{structureQ8xC2n:prop} to $(H_2,C_2)$. Either we are in case (a)(i) of Proposition \ref{structureQ8xC2n:prop} or we may repeat the above argument. Since the index of the $2'$-part of the centre strictly decreases each time we apply the Bonnaf\'e-Dat-Rouquier correspondence, repetition of this process must eventually end in case (a)(i) of Proposition \ref{structureQ8xC2n:prop}.

Now let $(G,B)$ be a reduced pair satisfying condition (a)(i), so there is $N \lhd G$ such that $G=ND$ and $G \cap N \cong Q_8$. Let $b$ be the unique block of $\cO N$ covered by $B$, noting that this has defect group $Q_8$. By~\cite{ei16} there is a unique Morita equivalence class of blocks with defect group $Q_8$ and a given Cartan matrix, so $\mfO(b)=1$. By Theorem \ref{sfreduce:theorem} and~\cite[Corollary 3.11]{el19} $\SOF(B) \leq \SOF(b) \leq |Q_8|^2!$. Considering Cartan invariants, ${\rm c}(b) \leq 8$ and so, by Lemma~\ref{Cartan_reduce:lemma}(i), ${\rm c}(B) \leq 2^{n+3}$.
\end{proof}

We remark that we cannot at present so easily obtain a similar bound on the strong $\cO$-Frobenius number for blocks with defect group $Q_8 \times Q_8$, since we do not know how this invariant behaves with respect to normal subgroups of $p'$-index.

We further remark that in order to bound only the strong Frobenius number of a quasisimple group we could have used~\cite{fk18}.

\section{Donovan's conjecture for blocks with defect group $Q_8 \times C_{2^n}$ or $Q_8 \times Q_8$}
\label{sec:Donovan_cases}

We are now in a position to verify Donovan's conjecture for $Q_8 \times C_{2^n}$ and $Q_8 \times Q_8$.
\newline
\newline
{\sc Proof of Theorem}~\ref{Q8Ctheorem}.

By Proposition~\ref{reduce:prop}, in verifying Donovan's conjecture it suffices to consider reduced blocks.

Consider first $Q_8\times C_{2^n}$. From~\cite[Corollary 3.11]{eel19} we need only bound strong $\cO$-Frobenius numbers and Cartan invariants for reduced blocks with defect groups isomorphic to $Q_8\times C_{2^n}$, hence the result follows in this case by Corollary \ref{sfQ8xC:cor}.

Now consider blocks with defect groups $Q_8 \times Q_8$. By Proposition~\ref{structureQ8xC2n:prop} either: (i) there are normal subgroups $N \lhd H \lhd G$ with $H=ND$ and $D \cap N \cong Q_8$ or $Q_8 \times C_2$, and $[G:H]$ odd; or (ii) there are commuting $N_1, N_2 \lhd G$ with $N_1 \cap N_2 \leq Z(G)$ such that $D \cap N_1$, $D \cap N_2 \cong Q_8$ or $Q_8 \times C_2$, and $[G:N_1N_2]$ is odd.

By~\cite[Corollary 4.18]{ei18} it suffices to show that there are only finitely many possibilities for the Morita equivalence class of the unique block of $\cO H$ or $\cO N_1 N_2$ covered by $B$. Hence we may assume that $G=H$ in case (i) and $G=N_1N_2$ in case (ii).

As above, from~\cite[Corollary 3.11]{eel19} we need only bound strong $\cO$-Frobenius numbers and Cartan invariants for  such blocks.

In case (i) we have $\SOF(B) \leq 16^2!$ and ${\rm c}(B) \leq 2^6$ using arguments as in Corollary \ref{sfQ8xC:cor}, noting that $D\cong Q_8 \times Q_8$ with $D\cap N\cong Q_8$ or $Q_8 \times C_2$ satisfies the conditions of Theorem \ref{sfreduce:theorem}. Suppose that we are in case (ii). Note that $G \cong (N_1 \times N_2)/W$ for some group $W \leq Z(N_1 \times N_2)$. Now $B$ corresponds to a block $A$ of $N_1 \times N_2$ with $O_{2'}(W)$ in its kernel and defect group $Q_8 \times Q_8$, $Q_8 \times Q_8 \times C_2$ or $Q_8\times Q_8\times C_2\times C_2$. Write $A_i$ for the block of $N_i$ covered by $A$, with defect group $Q_8$ or $Q_8 \times C_2$. As above we have $\SOF(A_i)\leq 16^2!$, for $i=1,2$. Hence $\SOF(A) \leq (16^2!)^2$. By~\cite[Proposition 3.17]{el19} we have $\SOF(B) \leq \SOF(A)$. Finally we have ${\rm c}(A_i) \leq 16$, for $i=1,2$, so by Lemma~\ref{Cartan_reduce:lemma}(ii) ${\rm c}(B) \leq {\rm c}(A) \leq 16^2$.
\hfill $\Box$

\medskip

The authors are aware of only three Morita equivalence classes of blocks with defect group $Q_8 \times C_{2^n}$ for each $n \geq 0$, namely those with representatives the principal blocks of $\cO (Q_8 \times C_{2^n})$, $\cO (SL_2(3) \times C_{2^n})$ and $\cO(SL_2(5) \times C_{2^n})$. Similarly the known Morita equivalence classes of blocks with defect group $Q_8 \times Q_8$ have representatives the principal blocks of $\cO(Q_8 \times Q_8)$, $\cO(Q_8 \times SL_2(3))$, $\cO((Q_8 \times Q_8)\rtimes C_3)$ (SmallGroup(192,1022), where $C_3$ acts on $(Q_8 \times Q_8)/Z(Q_8 \times Q_8)$ with only one fixed point), $\cO(Q_8 \times SL_2(5))$, $\cO(SL_2(3) \times SL_2(3))$, $\cO(SL_2(3) \times SL_2(5))$ and $\cO(SL_2(5) \times SL_2(5))$, and a non-principal block of $(Q_8 \times Q_8) \rtimes 3_+^{1+2}$, where the centre of $3_+^{1+2}$ acts trivially.

\bigskip

\begin{center} ACKNOWLEDGEMENTS \end{center}

We thank Mandi Schaeffer-Fry, Lucas Ruhstorfer and Jay Taylor for some useful conversations. This research was supported by EPSRC grant no. EP/T004606/1.

config: arxiv


TITLE: Matroids relaxations of a given matroid
QUESTION [5 upvotes]: Let $\mathcal{M}$ be a rank-$d$ matroid on $[n]$. Say a matroid $\mathcal{N}$ is a relaxation of $\mathcal{M}$ if $\mathrm{rank}(\mathcal{N})=d$, $\mathrm{groundset}(\mathcal{N})=[n]$, and every independent set of $\mathcal{M}$ is an independent set of $\mathcal{N}$ (observe that this notion of relaxation is in the labeled sense).

Has anyone seen the set of all relaxations of $\mathcal{M}$ naturally show up in some context?

I'm particularly interested in the algorithmic task of listing all relaxations of a given matroid $\mathcal{M}$, and in any heuristic to make an implementation more efficient than performing the enumeration by brute force; it would be even better if such an implementation already exists.
Actually, I only care about those relaxations of $\mathcal{M}$ that are realizable over a fixed field, say $\mathbb{C}$, which brings me to the next question:

Does anyone know of an implementation of a (finite) algorithm that decides whether a matroid is representable and provides a representation in case it is? (that such a finite algorithm exists can be proved using Groebner bases)

Here I'm explicitly disregarding any complexity issues; I just want to be able to do compute a realization for some small matroids.
Note: the first question is different from the "opposite" one of listing all matroids that specialize a fixed matroid, which has been asked here. I wonder how different the task of listing all relaxations is.

Update: I implemented the following brute-force algorithm in sage to enumerate the relaxations of a rank-$d$ matroid on $[n]$:
For every subset $\mathcal{F}$ of the nonbases of $\mathcal{M}$, check whether $bases(\mathcal{M})\cup \mathcal{F}$is the collection of bases of a matroid.
According to sage's method .is_valid() for matroids, the outcome for all the matroids I tested was that all $2^{\#\{nonbases(\mathcal{M})\}}$ subsets of $\binom{[n]}{d}$ obtained form the bases of a matroid, which wasn't what I expected.

Can it be that there's a bug in sage's .is_valid() method, or is it actually to be expected that most naive relaxations of a matroid (as enumerated above) are actually matroids?

(the reward is independent from the update in the question)

REPLY [4 votes]: You are describing what are usually called "weak maps" between matroids.
So a bijection $\varphi: E(\mathcal{N}) \to E(\mathcal{M})$ is called a weak map if for every independent set $I$ of $\mathcal{M}$ the inverse image $\varphi^{-1}(I)$ is independent in $\mathcal{N}$. In your case, you are asking when the identity function is a weak map, which just requires every independent set of $\mathcal{M}$ to be independent in $\mathcal{N}$.
There is lots of stuff known about weak maps (and strong maps) and other things. Given a bunch of matroids on the same ground set, the weak-map relation is a partial order. The best place to start is around page 280 of the second edition of Oxley's book Matroid Theory, which refers to the original papers.
The "minimal" relaxation that can be done is to take a single non-basis $N$ of $\mathcal{M}$ and form a matroid whose bases are ${\mathrm{bases}} (\mathcal{M}) \cup \{N\}$. This will work if (and, I think, only if) $N$ is a circuit-hyperplane of $\mathcal{M}$.
So you certainly do not expect all of your relaxations to pass Sage's is_valid routine.
In fact, I wanted to do a test to convince myself that it was working ok, so I ran the following code, just testing all the relaxations of a single randomly chosen matroid (in this case the matroid called J). All this does is create an array of matroids, each of which are specified by their bases, for which I am using the bases of J together with one of the non-bases. Then I ask is_valid() which of those matroids are valid.
m = matroids.named_matroids.J()
bases = [x for x in m.bases()]
nonbases = [x for x in m.nonbases()]
circuithyps = [x for x in m.circuits() if x in m.hyperplanes()]
tstsets = [bases + [nonbases[i]] for i in [0..len(nonbases)-1]]
tstmats = [Matroid(x) for x in tstsets]
tstcheck = [m.is_valid() for m in tstmats]
tstcheck

and I obtained the following result:
[True, False, False, False, False, False, False, False, False, False, False, False, False, True, False, False, False, False, False, False]

which when we check the number of circuit-hyperplanes, is exactly what we expected.
[frozenset(['c', 'b', 'e', 'd']), frozenset(['h', 'e', 'g', 'f'])]

You should let us see the code that you used that produced the incorrect answers, so we can tell whether it is a Sage bug, a sage.matroids bug, or user-error!
Finally, Stefan van Zwam has written a Mathematica program using Groebner bases to determine whether a matroid is representable, but the problem is hard unless you only want to know about matroids with less than 10 elements. I am not sure if he intends to add it to the Sage code-base or whether he even has it still.  Ask him: https://www.math.lsu.edu/~svanzwam/programming.html


TITLE: The best way to factorize?
QUESTION [0 upvotes]: $f(t) = t^3 -t^2 +t + 7$.
Just made it up, but looking through previous tests, they come up a lot when trying to find eigenvalues. How would I easily factorize this or make it=0?
Wow, thanks for the quick and thorough answers.
The one I'm looking at is t^3 -t^2 -t +1
What is the fastest way to mathematically prove 1 and -1 as eigenvalues? (Hopefully without the cubic equation :)

REPLY [1 votes]: The example you have given only has $1$ real root and that is
$$t = \frac{1}{3}\left[1-\frac{2^\frac{2}{3}}{(3 \sqrt{267}-49)^\frac{1}{3}}+(2 (3 \sqrt{267}-49))^\frac{1}{3}\right]$$
as computed by Wolfram|Alpha. In general you can find the roots of a cubic using the cubic equation.
Doesn't the above look awful? If you were to pick a random cubic polynomial then there would likely be no resort but to apply the cubic formula - but the cubics you are given aren't random. Someone has to write those questions and they don't want you have have to solve really difficult equations. In every case I've seen for schoolwork (unless someone has made a mistake), at least one of the eigenvaules has been rational - for which you can apply the rational root theorem - but more often a root is simply guessable. In most cases checking the function at small integer values yields a root. Once you have a single root $r$ (and assuming you haven't been able to guess the others) you can divide your polynomial by $x-r$ and then apply the quadratic formula to find the remaining roots.


TITLE: Prove that any tree with $n$ nodes has at least $\frac {n}{2}$ nodes with degree $1$ or $2$. $n \ge 2$
QUESTION [1 upvotes]: Here is the question:

A tree is a connected simple graph with no cycle.
Prove that any tree with $n$ nodes has at least $\frac {n}{2}$ nodes with degree $1$ or $2$. $n \ge 2$

I try to use mathematical induction to prove this with the base case of $2$. Every tree will have $n-1$ edges if there are $n$ vertices. So the Base Case will have at least one node with degree of $1$. However, it seems like it is very hard to prove the $k+1th$ case. Any method to approach this question?
Thanks

REPLY [3 votes]: Suppose that there were fewer than $n/2$ nodes with degree $1$ or $2$. If $n$ is even then this means there are at least $n/2+1$ nodes with degree $3$ or more, and if $n$ is odd then there are at least $(n+1)/2$ nodes with degree $3$ or more. Of course, since trees are connected, the rest of the nodes have degree at least $1$.
Now take the total sum of the degrees. In the even case it is at least
$3n/2+3 + n/2-1 = 2n + 2$, so by the handshaking lemma there are at least
$n+1$ edges. In the odd case the sum is at least $3(n+1)/2 + (n+1)/2 = 2(n+1) = 2n+2$ as well, so again there are at least $n+1$ edges.
But this is absurd, since we know there are only $n-1$ edges in a tree on $n$ nodes.


TITLE: Eigenvalues of a Symmetric Positive Semi-Definite (PSD) matrix after rank one update
QUESTION [5 upvotes]: I have a Symmetric Positive Semi-Definite matrix $A$ which i know its eigenvalue and eigenvectors. let $v$ and $u$ be a random column vector. i want to know if it is possible to have eigenvalues of matrix $A+uv^T$.
I don't need its eigenvectors, but it is required to have the most precise eigenvalues. We know that $uv^T$ is also a rank one PSD matrix.
Is there a close form to this problem?
${\bf PS:}$ As you see i found the solution, but after implementation of this i can see a very small error in result, it would be appreciated if anyone know why this is happening, because we didn't used any approximation to get the result and it should be exact.

REPLY [3 votes]: Complementing Denis's answer, you can look at the paper Some Modified Matrix Eigenvalue Problems by Golub (http://www.jstor.org/stable/2028604) where this issue is treated at some length.


TITLE: directional derivative with increasing path
QUESTION [0 upvotes]: $f(x,y,z) = xe^{y^2-z^2}$
$P(1,2,-2)$
$r(t)=(t)i+(2cos(t-1))j-(2e^{t-1})k$
The directional derivative goes along the path r(t) (I assume this means parallel and r(t) is increasing)
For this problem, I have calculated the gradient but how do I perform the dot product if $r$ has the variable $t$? Should I leave it as just the multiplication of the terms in the gradient and path?

REPLY [0 votes]: You can express the gradient of f in terms of t, using the fact that the directional derivative is computed along points in the path r. Substitute x(t), y(t), z(t) in $\nabla f$.


TITLE: Properties of a derivative of a function $f(x)$ expressed as a function of $f$
QUESTION [1 upvotes]: Consider a differentiable function $f(x)$ and let $u(f(x)) = f'(x)$.
Normally we solve for $u$ as a function of $x$, but we can also express it as a function of $f$.
Some examples:
If $f(x) = e^{-x}$, then $f'(x) = -e^{-x} = - f(x)$, hence $u(f) = -f$.
If $f(x) = x^{-1}$, then $f'(x) = -x^{-2} = -(x^{-1})^2 = -f(x)^2$, so $u(f) = -f^2$.
If $f(x) = \tan(x)$, then $f'(x) = \sec^2(x) = 1 + \tan^2(x) = 1 + f(x)^2$, thus $u(f) = 1 + f^2$.
These show up in elementary differential equations like $\frac{dy}{dx} = -y$.
My first question is if it is always possible to express $u$ solely in terms of $f$ and not $x$.
Intuitively, it seems to be so, for while the mapping from $\ln(x)$ to $x^{-1}$ or $\sin(x)$ to $\cos(x)$
may not be easily expressible in terms of elementary functions,
it is certainly a well-defined function wherever $f'(x)$ is defined,
although the existence of everywhere continuous, nowhere differentiable functions
means $u$ does not always exist.
Naturally, I welcome any counter-example which forces $u$ to depend on $x$ and not $f$ only.
Secondly, when, if ever, does $u$ have the same form as $f$? That is, for what function $f(x)$ does $u(t) = f(t)$?
(Using $t$ instead of $x$ to avoid confusion.)
Is this function unique, up to a constant?
I feel that this ought to have come up in an elementary analysis or differential equations course,
but I don't recall it doing so. I'd especially be glad to see a reference to an in-depth discussion in a textbook or a paper.
My own cursory searches have been fruitless.

REPLY [0 votes]: Trivially, almost yes. If $\exists f^{-1}$, you can write $u(f(x))=f′(x)$ equivalently as $u(y)=f′(f^{-1}(y))$. Obviously, the global existence of $f^{-1}$ is false in general, but you have local existence near the points $c$ s.t. $f'(c)\ne 0$ (Inverse Function Theorem).
Now, think yourself in the second question.


TITLE: Which statement is true or are they all false?
QUESTION [1 upvotes]: Which of the following statements are true?

If the covariance of two random variables is zero, the random variables are independent.
If X is a continuous random variable, the continuity correction is used to approximate probabilities pertaining to X with a discrete distribution.
If E and F are mutually exclusive events which occur with nonzero probability, E and F are independent.
If X and Y are independent random variables, then given that their moments exist and E[XY] exists, E[XY]=E[X]E[Y].

I know that 1 is false and I am pretty sure that 4 is false, but I am not sure about 2 and three. I do not know what they are talking about in number 3 when they say continuity correction. Is 3 false because even though they are mutually exclusive the event A would occur if event B did not occur?

REPLY [0 votes]: False, e.g. $X=\cos2\pi t$, $Y=\sin2\pi t$, with $t$ uniform on $[0,1]$. In other words the coords of a circular distribution.
Not sure what 'the continuity correction' means. It's sometimes used to approximate a discrete distribution by a continuous one, to 'smooth out the gaps'.
False as $0=P(EF)\ne P(E)P(F)$.
True. You only need moments of order $\le2$.


TITLE: Confusion regarding proof of L'Hopital rule.
QUESTION [5 upvotes]: I was reading Spivak's Calculus. I have a query with regards to the proof for L'Hopital's rule for the 0/0 indeterminate form.
Theorem Statement :-
"Suppose that
$
\lim_{x\to a} f(x) = 0 $ and
$\lim_{x\to a} g(x) = 0
$
and suppose that $ \lim_{x\to a} \frac {f'(x)}{g'(x)} $ exists .
Then $ \lim_{x\to a} \frac {f(x)}{g(x)} $ exists and
$$
\lim_{x\to a} \frac {f(x)}{g(x)} = \lim_{x\to a} \frac {f'(x)}{g'(x)}
$$
"
I have a confusion with regards to the last part of the proof.
Using the Cauchy Mean Value Theorem it is shown that there exists a number $\alpha_x$ in $(a,x)$ such that $$
\ \frac {f(x)}{g(x)} = \ \frac {f'(\alpha_x)}{g'(\alpha_x)}
$$
Now $\alpha_x$ approaches $a$ as $x$ approaches $a$ because $\alpha_x$ is in $(a,x)$ , it follows that
$$
\lim_{x\to a} \frac {f(x)}{g(x)} = \lim_{x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{\alpha_x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{y\to a} \frac {f'(y)}{g'(y)}
$$
My query is regarding the last two of the above equations. I understand that as $x$ approaches $a$ so does $\alpha_x$ , but then how are we treating $\alpha_x$ as a 'dummy variable' and replacing it with $y$. Is not $\alpha_x$ a dependent variable on x.
(I am also having some trouble in seeing how the step
$$\lim_{x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)} = \lim_{\alpha_x\to a} \frac {f'(\alpha_x)}{g'(\alpha_x)}$$ is justified )
Thanks in advance. I am new to these forums and looking forward to the discussion.

REPLY [3 votes]: These equalities hold because you assumed that the limit $\displaystyle\lim_{x\rightarrow a} \dfrac{f'(x)}{g'(x)}$ exists.  That means it doesn't really matter how you approach $a$ (i.e. either take $x\rightarrow a$ or $\alpha_x\rightarrow a$), you have to arrive at the same value for the limit.  This also explains why your last equality is okay, since you have some variable approaching $a$, it doesn't matter how you get there; essentially if the limit exists, then the variable respect to which you are taking the limit is automatically a dummy variable.
If you want to be more precise, you can use the $\varepsilon$-$\delta$ definition of the limit.  Say $L$ is your limit, which we assume exists.  Then for all $\varepsilon>0$ there exists some $\delta>0$ such that if $|a-x|<\delta$ then $\left|L-\dfrac{f'(x)}{g'(x)}\right|<\varepsilon$. But $|a-\alpha_x|<|a-x|<\delta$, so we know that for each $\varepsilon$, we can use the same $\delta$ to see that if $|x-a|<\delta$, then $\left|L-\dfrac{f'(\alpha_x)}{g'(\alpha_x)}\right|<\varepsilon$.  This immediately implies $\displaystyle \lim_{x\rightarrow a}\dfrac{f'(\alpha_x)}{g'(\alpha_x)} = L$.


TITLE: Is infinite linear combination of concave function still concave
QUESTION [0 upvotes]: Let $f_n(x)$ be  concave functions.
I know that
\begin{align}
g(x)=\sum_{n=1}^N a_nf_n(x)
\end{align}
where $a_n  \ge 0$ and $N<\infty$ is concave.
My question is if the sum becomes infinite sume is $g(x)$ still concave?
To me it seems yes, since in the proof we do not assume anyting about finitness of the sum:
\begin{align}
g((1-t)x+ty)&=\sum a_nf_n((1-t)x+ty) \ge  \sum a_n ( (1-t)f_n(x)+tf_n(y))\\
&=(1-t) g(x)+tg(y)
\end{align}

REPLY [0 votes]: As you may or may not have guessed from the comments, if you know that the sum converges (pointwise) for every argument you want to check it for (that is, in this case, for $(1-t)x+ty, (1-t)x$, and $ty$ then the conclusion is valid.
The proof of the inequality would usually be done by reducing it to a finite sum and some $\varepsilon, \delta $ logic, but it's rather straightforward.


TITLE: Finding every element of order $2$ in $D_6$
QUESTION [1 upvotes]: Find every element of order $2$ in
$$
D_6 = \{e, \sigma, \sigma^2, \sigma^3, \sigma^4, \sigma^5, \rho, \sigma\rho, \sigma^2\rho, \sigma^3\rho, \sigma^4\rho, \sigma^5\rho\}
$$
The answers say that the elements of order $2$ are: $\sigma^3$, $\rho$, $\sigma\rho$, $\sigma^2\rho$, $\sigma^3\rho$, $\sigma^4\rho$, $\sigma^5\rho$.
I can see why this is the case for $\sigma^3$ as $\sigma^3\sigma^3 = \sigma^6 = 1$. Similarly $\rho\rho = \rho^2 = 1$. However I am having problems conceptually grasping why $\sigma\rho$, $\sigma^2\rho$, $\sigma^3\rho$, $\sigma^4\rho$, $\sigma^5\rho$ are of order $2$.

REPLY [0 votes]: Label the vertices of the hexagon $0,\dots,5$. A rotation maps vertex $k$ to $k+1\pmod 6$, and a reflection maps vertex $k$ to $6-k\pmod 6$.
Then $\sigma^k\rho$ maps vertex $j$ to $j+k \pmod 6$ and then to $6-(j+k)=6-j-k \pmod 6$. Repeating the operations maps $6-j-k \pmod 6$ to $6-j-k+k=6-j \pmod 6$, and then back to $j$, i.e. where it started.


TITLE: Defining the addition of systems of linear equations
QUESTION [0 upvotes]: In the book Matrices and Linear Algebra by Hans Schneider and George Philip Barker, I encounter the exercise of adding two systems of linear equations.
Let the first system be:
$$\begin{cases} a_{11}x_1 +a_{12}x_2 = y_1\\a_{21}x_1 + a_{22}x_2 = y_2 \end{cases}$$
and the second system be:
$$\begin{cases} b_{11}x_1 + b_{12}x_2 = z_1\\b_{21}x_1 + b_{22}x_2 = z_2 \end{cases}.$$
The sum of these two systems is defined as the system whose first equation is the sum of each system's first equations, and the second equation is the sum of each system's second equations:
sum = $\begin{cases} (a_{11} + b_{11})x_1 + (a_{12} + b_{12})x_2 = y_1 + z_1\\(a_{21} + b_{21})x_1 + (a_{22} + b_{22})x_2 = y_2 + z_2 \end{cases}$
Defining the equation this way would mean that the order in which the equations appear in each system matters, so why do we define the addition as such? What practical use are we trying to get out of this definition?

REPLY [1 votes]: This is just addition as matrix equations. So the first equation is $Ax=y$, and the second is $Bx=z$. Addition yields
$$
(A+B)x=y+z,
$$
as you wrote. Like adding equations in a field $K$, we can also add equations in $K^n$.


TITLE: Riesz Representation Theorem, true for pre-Hilbert spaces and any functional?
QUESTION [1 upvotes]: I have some doubts about the Riesz theorem.
Firstly can you check my proof?
Fa = for all
Fa (H,<,>) a pre Hilbert space
Fa x in H different from the zero vector
Fa F a functional on H
Fa T: (H->R) -> H : T(F) = F(x) x/||x||^2 a map from a functional on H to H
Then F(x) = F(x) < x, x > / ||x||^2 = < F(x) x/||x||^2, x > = < T(F), x >
End of proof, where F(x) = < T(F), x > for any vector x in a pre Hilbert space.
If my proof is correct, it implies that H doesn’t have to be an Hilbert space but only a pre Hilbert one and that the functional doesn’t have to be neither linear and continuous, but only a functional.
Is it true or I’m missing something?

REPLY [0 votes]: The Riesz Representation theorem is:
$$\forall F \in H^*, \exists y \in H, \forall x \in H, F(x) = \langle y, x\rangle$$
What you have shown is
$$\forall F \in H^*, \forall x \in H, \exists y \in H, F(x) = \langle y, x\rangle$$
$y$ or $T(F)$ in the Riesz Representation theorem is picked knowing only $F$, not $x$. This single value of $y$ must work for every $x$. In your result, you pick $x$ first, and then define $T(F)$ in terms of it. Thus your $y$ changes from one value of $x$ to another.


TITLE: Extract vector from vector-matrix equation
QUESTION [0 upvotes]: We have an equation of the following form:
$$\frac{(u \times w)^T}{(u \times w) \cdot v} \cdot \frac{v \times w}{(v \times w) \cdot u} = \omega$$
where $u,v$ known unit vectors and $\omega$ are known scalar.
Is it possible to solve this equation with respect to an unknown vector $w$?

REPLY [2 votes]: To avoid confusing $\omega$ with $w$, I will write $c$ for $\omega$.
The denominators in the equations are, up to a sign, $(u\times v)\cdot w$. Therefore you need $\{u,v\}$ to be linearly independent in order that the problem is well posed. Suppose this is indeed the case. Let $p=u\times v$. Then the equation can be rewritten as
$$
(u\cdot v)\|w\|^2 - (u\cdot w)(v\cdot w) + c(p\cdot w)^2 = 0,\tag{1}
$$
subject to the constraint that $p\cdot w\ne0$. Then $(1)$ can be rewritten as
$$
w^T\,\underbrace{\left[(u\cdot v)I_3-\frac12(uv^T+vu^T)+cpp^T\right]}_A\,w=0.\tag{2}
$$
Since $A$ is symmetric, it can be orthogonally diagonalised as $A=QDQ^T$. Let $x=Q^Tw$ and $q=Q^Tp$. The constraint $p\cdot w\ne0$ thus becomes $q^Tx\ne0$. Hence the problem boils down to solving
\begin{align}
d_1x_1^2+d_2x_2^2+d_3x_3^2&=0,\tag{3}\\
q_1x_1+q_2x_2+q_3x_3&\ne0.\tag{4}
\end{align}
Clearly, $(3)$ is solvable only if $d_i=0$ for some $i$ or $d_id_j<0$ for some $i,j$, i.e. only if $A$ is singular or indefinite. And when $(3)$ is solvable, it is not hard to see if it is still solvable with the constraint $(4)$. Once a solution $x$ is found, we may set $w=Qx$.


text:
\section{Conclusion}
\label{conclu}
In this paper we proposed a class of novel high-order positivity-preserving
finite difference schemes for the 1D and multi-D ideal MHD systems.
In the 1D case, a positivity-preserving limiting technique was applied to modify
high-order WENO-HCL flux with the first-order Lax-Fridrichs flux to produce positive density and pressure.
In multi-D cases, the limiting technique was also applied to the
hyperbolic solver, followed by a constrained transport technique evolving the magnetic
potential to control the divergence errors.
The main advantage of the proposed schemes is, the high order of accuracy, a discrete divergence-free condition and positivity of solutions can be attained  at the same time.
The overall scheme shares the same CFL constraint as the low-order Lax-Fridrichs scheme,
without extra restriction resulting from the limiting process.
We demonstrated the effectiveness and efficiency of the positivity-preserving schemes by 1D, 2D and 3D numerical examples.
A strict proof for high-order accuracy of the proposed limiting technique will be part of our future work.