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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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Can you say something about $K \cap H$? –  t.b. Aug 22 '11 at 20:10
    
@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 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. –  Zev Chonoles Aug 30 '11 at 21:33

1 Answer 1

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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Thank you Arturo. That was a wonderful explanation. –  user463498 Aug 22 '11 at 20:50
    
@Seth: In other words, if you have a class $\mathcal{X}$ of groups that is closed under subgroups, then the class of virtually-$\mathcal{X}$ groups is also closed under subgroups. –  Arturo Magidin Aug 22 '11 at 20:54
    
Arturo. Do you know a reference of the facts you've mentioned ? –  user463498 Aug 30 '11 at 21:13
    
@user463498: What facts? I proved everything I mentioned, in so far as I can see. The index inequality is standard. –  Arturo Magidin Aug 30 '11 at 21:30
    
This is true for quontients? If G is virtually abelian, any quotient of G is? –  Dennis Apr 6 '12 at 11:39

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