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

 $G=He\sqcup Hx_{2}\sqcup...\sqcup Hx_{q},$ *

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

$K$ is also virtually abelian

Proof: From $(*)$ above we have

 $K=K\cap G=K\cap(He\sqcup Hx_{2}\sqcup...\sqcup Hx_{q}),$
 $=(K\cap H)\sqcup(K\cap Hx_{2})\sqcup...\sqcup(K\cap Hx_{q}).$ **

Here we consider the two cases:
1) $x_{i}\in K$
2) $x_{j}\notin K$
In the first case $K=Kx_{i}$, and then $K\cap Hx_{i}=Kx_{i}\cap Hx_{i}=(K\cap H)x_{i}$. In the second, find $y_{j}\in K\cap Hx_{j}$ hence $K\cap Hx_{j}=Ky_{j}\cap Hy_{j}=(K\cap H)y_{j}$
So, in the equation $(**)$ above we can replace (reordering subindexation perhaps) to get

 $K=\underbrace{(K\cap H)\sqcup(K\cap H)x_{2}\sqcup...\sqcup(K\cap H)x_{s}}_{1)}% \sqcup\underbrace{(K\cap H)y_{s+1}\sqcup...\sqcup(K\cap H)y_{q}}_{2)}$

relation which shows that the index $[K:K\cap H]\leq[G:H]$.
It could be $<$ since it is posible that $K\cap Hx_{r}=\emptyset$ for some indexes $r$ $\Box$

Title virtually abelian subgroup theorem VirtuallyAbelianSubgroupTheorem 2013-03-22 18:58:42 2013-03-22 18:58:42 juanman (12619) juanman (12619) 13 juanman (12619) Theorem msc 20F99 msc 20E99 msc 20E07 subgroup theorem