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 coset partition

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

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 $r$ $\mathrm{\square }$

Title	virtually abelian subgroup theorem
Canonical name	VirtuallyAbelianSubgroupTheorem
Date of creation	2013-03-22 18:58:42
Last modified on	2013-03-22 18:58:42
Owner	juanman (12619)
Last modified by	juanman (12619)
Numerical id	13
Author	juanman (12619)
Entry type	Theorem
Classification	msc 20F99
Classification	msc 20E99
Classification	msc 20E07
Synonym	subgroup theorem