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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=\langle \{g_1,\cdots,g_m\}\rangle $, 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

Shaun
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Fardad Pouran
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    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
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    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 00:03
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    This is a duplicate of this question. – user1729 Sep 23 '20 at 14:45
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    This question (which was posted in 2013) had been closed as a duplicate of a question posted in 2020, so I voted to reopen it. – Derek Holt Mar 09 '21 at 07:31
  • @DerekHolt Can you kindly explain what you mean by "$H$ is the stabiliser of the point $1$"? – user371231 Sep 22 '21 at 08:47
  • @user371231 The action of $G$ defined by the homomorphism $\phi:G \to S_n$ is the action by multiplication on the $n$ left (or right if you prefer right actions) cosets of $H$ in $G$. So it would be more accurate to say that $H$ is the stabilizer of the coset $H$ in this action. But, we can number the cosets $g_1H,g_2H, \ldots, g_nH$ from $1$ to $n$, and it is standard practice to take $g_1 = 1_G$, so the coset $H$ is numbered $1$. That is why I referred to the stabilizer of point $1$. – Derek Holt Sep 22 '21 at 16:15

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