Finite index subgroups of a virtually abelian group 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 ?
 A: 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 $G$.
