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If $G$ is a virtually abelian group and $H$ is a finite index subgroup of $G$. Is it always true that $H$ is virtually abelian ?

Since $G$ is V.A, it has a finite index subgroup $K$ which is abelian.

If $H \subset K$, then $H$ is abelian and therefore V.A.

If $K \subset H$, then $H$ is V.A.

What about when neither subgroup is contained in the other ?

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    @user463498: I've merged your account with the account named "Seth" (unfortunately it was impossible to do the other way around). Because [you can only comment on your own questions, your own answers, and answers to your own questions](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757) when you have $\leq 50$ reputation points, you were unable to comment on Arturo's answer because this question was owned by the account "Seth". If you have trouble logging in in the future, comment or leave a "flag" for moderator help.2011-08-30

1 Answers 1

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Yes, $H$ is virtually abelian.

Lemma. Let $G$ be a group, $H$ and $K$ subgroups. Then $[H:H\cap K]\leq [G:K]$ (in the sense of cardinalities). In particular, if $K$ is of finite index in $G$, then $H\cap K$ is of finite index in $H$.

Proof. We define a function from the set of left cosets of $H\cap K$ in $H$, $\{h(H\cap K)\mid h\in H\}$, to the set of left cosets of $K$ in $G$, $\{gK\mid g\in G\}$ as follows: map $h(H\cap K)$ to $hK$.

We claim that this function is well-defined and one-to-one. Indeed, \begin{align*} h(H\cap K) = h'(H\cap K) &\Longleftrightarrow (h')^{-1}h\in H\cap K\\ &\Longleftrightarrow (h')^{-1}h\in K\\ &\Longleftrightarrow hK = h'K. \end{align*} (The second equivalence because (h')^{-1}h is always in $H$, hence it lies in $H\cap K$ if and only if it lies in $K$).

Thus, the cardinality of the set of left cosets of $H\cap K$ in $H$ (which is $[H: H\cap K]$) is less than or equal to the cardinality of the set of left cosets of $K$ in $G$ (which is $[G:K]$). $\Box$

Proposition. Let $\mathcal{X}$ be a class of groups that is closed under subgroups (that is, if $G\in\mathcal{X}$ and $K\lt G$, then $K\in\mathcal{X}$). If $G$ is virtually-$\mathcal{X}$ and $H$ is a subgroup of $G$, then $H$ is virtually-$\mathcal{X}$.

Proof. Let $K$ be of finite index in $G$ such that $K\in\mathcal{X}$. By the Lemma, $H\cap K$ is of finite index in $H$. Since $\mathcal{X}$ is closed under subgroups, $H\cap K\in\mathcal{X}$. Hence, $H$ is virtually-$\mathcal{X}$. $\Box$

Corollary. If $G$ is virtually abelian, and $H$ is a subgroup of $G$, then $H$ is virtually abelian.

Proof. The class of all abelian groups is closed under subgroups. $\Box$

So you don't even need to assume $H$ is of finite index in $G$.

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    Yes. But you're asking a related question rather than answering the question, so this isn't really the right place.2012-04-06