A finitely generated group has only finitely many subgroups of a given index

Let $G$ be a finitely generated group and let $n$ be a positive integer. Let $H$ be a subgroup of $G$ of index $n$ and consider the action of $G$ on the coset space $(G:H)$ by right multiplication. Label the cosets $1,\mathrm{\dots },n$, with the coset $H$ labelled by $1$. This gives a homomorphism $\varphi :G\to S_n$. Now, $x\in H$ if and only if $Hx=H$, that is, $G$ fixes the coset $H$. Therefore, $H={\mathrm{Stab}}_G(1)=\{g\in G\mid 1(g\varphi )=1\}$, and this is completely determined by $\varphi$. Now let $X$ be a finite generating set for $G$. Then $\varphi$ is determined by the images $x\varphi$ of the generators $x\in X$. There are $|S_n|=n!$ choices for the image of each $x\in X$, so there are at most ${n!}^{|X|}$ homomorphisms $G\to S_n$. Hence, there are only finitely many possibilities for $H$.

Title	A finitely generated group has only finitely many subgroups of a given index
Canonical name	AFinitelyGeneratedGroupHasOnlyFinitelyManySubgroupsOfAGivenIndex
Date of creation	2013-03-22 15:16:03
Last modified on	2013-03-22 15:16:03
Owner	avf (9497)
Last modified by	avf (9497)
Numerical id	6
Author	avf (9497)
Entry type	Theorem
Classification	msc 20E07
Related topic	Group
Related topic	FinitelyGenerated