virtually abelian subgroup theorem

Let us suppose that $G$ is virtually abelian and $H$ is an abelian subgroup of $G$ with a the finite right cosetpartition

$$G=He\bigsqcup H{x}_2\bigsqcup \mathrm{\dots }\bigsqcup H{x}_q,$$*

so if $K$ is any other subgroup in $G$ we are going to prove:

$K$ is also virtually abelian

Proof: From $(*)$ above we have

$$K=K\cap G=K\cap (He\bigsqcup H{x}_2\bigsqcup \mathrm{\dots }\bigsqcup H{x}_q),$$

$$=(K\cap H)\bigsqcup (K\cap H{x}_2)\bigsqcup \mathrm{\dots }\bigsqcup (K\cap H{x}_q).$$**

Here we consider the two cases:
1) $x_i\in K$
2) $x_j\notin K$
In the first case $K=K{x}_i$, and then $K\cap H{x}_i=K{x}_i\cap H{x}_i=(K\cap H)x_i$.In the second, find $y_j\in K\cap H{x}_j$ hence $K\cap H{x}_j=K{y}_j\cap H{y}_j=(K\cap H)y_j$
So, in the equation $(**)$ above we can replace (reordering subindexation perhaps) to get

$$K=\underset{1)}{\underbrace{(K\cap H)\bigsqcup (K\cap H)x_2\bigsqcup \mathrm{\dots }\bigsqcup (K\cap H)x_s}}\bigsqcup \underset{2)}{\underbrace{(K\cap H)y_{s+1}\bigsqcup \mathrm{\dots }\bigsqcup (K\cap H)y_q}}$$

relation which shows that the index $[K:K\cap H]\le [G:H]$.
It could be since it is posible that $K\cap H{x}_r=\mathrm{\varnothing }$ for some indexes $r$$\mathrm{\square }$