# Subgroups of given index of a finitely generated group

Let $G$ a finitely generated group. Prove that for a given natural $n$, there exists finitely many subgroups of index $n$ in $G$.

I tried to keep on this approach : Finitely generated group has only finitely many subgroups of given index

Here is my effort to edit or complete the above idea:

My main emphesize is on this theorem:

If $H$ is a subgroup of finite index $n$ in a group $G$, Then there exists homomorphism $\phi:G\rightarrow S_n$ such that $Ker(\phi)=Core(H)=\displaystyle\bigcap_{x\in G}{xHx^{-1}}$.

As $G=< \{g_1,\cdots,g_m\}>$, it's sufficient to find $\phi$'s effect over $g_i$s in order to determine whole $\phi$. Now $\phi(g_i)$ has $n!$ states to be chosen. Therefore, number of whole $\phi$s can be defined, are less than $(n!)^m$.

But if we suppose number of such $H$s is infinite, we must have this fact that infinite number of $Core(H)$s are equal.

Also if the left cosets of $H$ are $H,a_1H,\cdots,a_nH$, we can show $Core(H)=\displaystyle\bigcap_{i=1}^{n}{a_iHa^{-1}_i}$

Any idea is welcome for the rest. Thank You

• I'm not sure I understand why you dwell so much with the core of $\;H\;$ ...Isn't it enough to notice that each subgroup of index $\;n\;$ in the group determines a homomorphism to $\;S_n\;$ and the number of these homom's is finite as each of them is completely and uniquely determined by its action of the finite number of generators of $\;G\;$ ? – DonAntonio Nov 17 '13 at 19:29
• @DonAntonio That sounds like it just proves finitely many possible homomorphisms for a given subgroup (of given index), not finitely many subgroups (of given index). Is there a simple reason why distinct subgroups can't induce the same homomorphism? That seems to be what fardad is trying to figure out. – zibadawa timmy Nov 17 '13 at 19:33
• I see your (and perhaps also the OP's) point, @zibadawatimmy...Thanks. – DonAntonio Nov 17 '13 at 19:40
• The reason that distinct subgroups give distinct homomorphisms is that the subgroup $H$ is the stabilizer of the point $1$ in the image of the homomorphism to $S_n$ defined by $H$. So distinct subgroups have images with distinct stabilizers of $1$, so the homomorphisms cannot be equal. – Derek Holt Nov 17 '13 at 19:52
• thank you guys, i've never heard stabilizer before – Fardad Pouran Nov 18 '13 at 0:03