# 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

$$K=\underset{1)}{\underset{\u23df}{(K\cap H)\bigsqcup (K\cap H){x}_{2}\bigsqcup \mathrm{\dots}\bigsqcup (K\cap H){x}_{s}}}\bigsqcup \underset{2)}{\underset{\u23df}{(K\cap H){y}_{s+1}\bigsqcup \mathrm{\dots}\bigsqcup (K\cap H){y}_{q}}}$$ 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 |