# 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 $1,\ldots,n$, with the coset $H$ labelled by $1$. This gives a homomorphism          $\phi:G\to S_{n}$. Now, $x\in H$ if and only if $Hx=H$, that is, $G$ fixes the coset $H$. Therefore, $H=\mathrm{Stab}_{G}(1)=\{g\in G\mid 1(g\phi)=1\}$, and this is completely determined by $\phi$. Now let $X$ be a finite generating set  for $G$. Then $\phi$ is determined by the images $x\phi$ 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 $H$.

## References

• 1
Title A finitely generated group has only finitely many subgroups of a given index AFinitelyGeneratedGroupHasOnlyFinitelyManySubgroupsOfAGivenIndex 2013-03-22 15:16:03 2013-03-22 15:16:03 avf (9497) avf (9497) 6 avf (9497) Theorem msc 20E07 Group FinitelyGenerated