TITLE: $H$ has finite index $\implies aH$ has finite index
QUESTION [4 upvotes]: Just out of curiosity,

If $H$ is a subgroup of finite index in $G$ and $a\in G$, is it always true that $aH$ and $Ha$ are of finite index in $G$?

I think I can prove it as follows: If $a_1H,\ldots,a_kH$ be all the distinct left cosets of $H$, then $(a_1a^{-1})aH,\ldots,(a_ka^{-1})aH$ are distinct left cosets of $H$, and thus $aH$ is of finite index. Similarly, $Ha$ is of finite index.
Is there something wrong in the above proof? I find this result, if correct, extremely natural, but I don't find it in my abstract algebra textbook.

REPLY [7 votes]: If you are trying to show that $aHa^{-1}$ has the same index in $G$ as $H$, you can define a bijection from $G/H$ to $G/(aHa^{-1})$ by sending $gH$ to $(ga^{-1})aHa^{-1}$. This is a bijection because it has an inverse: right multiplication by $a$. Namely,
$(ga^-1)aHa^{-1}(a) = H$


TITLE: If the intersection of all non-zero ideals of an integral domain R is non-zero, show that R is a field.
QUESTION [2 upvotes]: Let $R$ be a commutative ring with unity.  Call $R$ a *-ring if the intersection of all non-zero ideals of $R$ is non-zero.  Note that we consider $R$ itself an ideal.  If $R$ is an integral domain that is a *-ring, show that $R$ is a field.
My idea was to show that a maximal ideal $I$ is the zero ideal, but I'm not really getting anywhere.  Doesn't seem like too hard of a problem, so any hints would be amazing.

REPLY [1 votes]: May I present a few clarifications on what I assume is the direction of reasoning that Lord Shark the Unknown intended: as introduced in the above answer $I$ is indeed the minimum in the ordered set of non-zero ideals; let us fix $a \in I\setminus \{0_A\}$
(for linguistic reasons, I prefer to denote my rings with $A, B, C...$).
Remark first of all that by minimality $I$ will be generated by any of its nonzero elements. Consider now an arbitrary $x \in A^{\times}=A \setminus \{0_A\}$; since $A$ is a domain $ax \neq 0_A$ and equally $ax \in Aa=I$; thus it follows that $I=A(ax)=Aa$ and from this, since $a \in Aax$, we infer the existence of $y \in A$ such that $a=axy$; since non-zero elements are cancellable in a domain, this last equation leads to $xy=1_A$, by which we can conclude that every non-zero element is invertible and hence $A$ is actually a field.

P.S. Actually an even stronger claim can be shown to hold true: if $A$ is a nonzero (commutative) integral domain such that the intersection $\displaystyle\bigcap_{x \in A^{\times}}(x) \neq \left(0_A\right)$ of all nonzero principal ideals be nonzero, then $A$ is a field. Indeed, let us denote the intersection in question by $I$ and consider a fixed $a \in I^{\times}\colon=I \setminus \{0_A\}$. Since in particular $I \subseteq \left(a^2\right)$, it follows that $a \in Aa^2$ and thus there exists $b \in A$ such that $a=ba^2$. Simplification by the nonzero element $a$ entails $ab=1_A$, signifying that $a \in \mathrm{U}(A)$ is a unit and subsequently $I=A$. This has the consequence that for any $x \in A^{\times}$ we have $A=(x)$, in other words $A^{\times} \subseteq \mathrm{U}(A)$ -- any nonzero element is invertible -- and hence $A$ is indeed a field.


TITLE: Equivalence of associative Lie bracket
QUESTION [0 upvotes]: Let $L$ be a Lie-algebra. I have to show that the Lie-bracket $[\,,]$ is
associative if and only if $[x,y]\in C(L)$ for all $x,y\in L$, where $C(L)$ denotes the center of $L$.
I know that $[x,y]\in C(L)\Leftrightarrow [[x,y],z]=0\,\,\,\forall z\in L$.
But from here I do not know how to continue.
Can anyone give me hint?

REPLY [0 votes]: A Lie algebra $L$ is associative if and only if it is two-step nilpotent (including abelian), i.e., if and only if $[L,L]\subseteq Z(L)$, see this duplicate:
Does there exist a non-trivial, associative Lie algebra?


TITLE: proof verification $\frac{3+2\sqrt{6}}{1-\sqrt{6}}$ is an algebraic integer
QUESTION [1 upvotes]: Is $$\frac{3+2\sqrt{6}}{1-\sqrt{6}}$$
  an algebraic integer?

An algebraic integer means an algebraic number in some algebraic number field $K\supset \Bbb Q$ that is the root of a monic polynomial $f\in \Bbb Z[x]$. Here I guess we are in $\Bbb Q({\sqrt{6}})=\Bbb{Q}(\alpha)/(\alpha^2-6)$, and with that:
$$\beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies \beta= \frac{(3+2\sqrt{6})(1+\sqrt6)}{-5}\implies -5\beta =15+5\sqrt{6}\implies (-5\beta-15)^2=25(6)\implies 5^2\beta^2+2(15)(5)\beta+15^2-5^2(6)=0,$$
and thus $5^2x^2+6(5)^2x+5^2(3)\in\Bbb{Z}[x]$ is the desired polynomial.
Is this correct?

REPLY [3 votes]: Your approach - finding the minimal polynomial and observing that it is a monic polynomial over $\Bbb Z$ - is correct, but the following is probably easier:
$$
\frac{3 + \sqrt6}{1 - \sqrt6} =
\frac{3 + \sqrt6}{1 - \sqrt6}
\frac{1 + \sqrt6}{1 + \sqrt6} =
\frac{15 + 5 \sqrt6}{-5} =
-3 + \sqrt6.
$$
Since $\sqrt 6$ is an algebraic integerer, so is $-3 + \sqrt6$ and its minimal polynomial is $(x + 3)^2 - 6 = x^2 + 6x + 3$.

REPLY [0 votes]: $$ \beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies $$
$$\beta= \frac{(3+2\sqrt{6})(1+\sqrt 6)}{-5}\implies $$
$$-5\beta =15+5\sqrt{6}\implies $$
$$ \beta =-3-\sqrt 6$$
$$ \beta ^2 + 6\beta +3=0 $$