13

In my previous question i have got a comment that :

If a group is finitely generated, then there are finitely many subgroups of a given finite index.

I do not yet see the beauty of this problem but, i wanted to prove this atleast.

So,what i have tried is :

I fix $n\in \mathbb{N}$ and assume $H\leq G$ with $|G/H|=n$

As we have $H\leq G$ of finite index, we have the action of $G$ on left cosets of $H$ and

we get $\eta: G\rightarrow S_n$

I assume $\{g_1,g_2,\dots, g_r\}$ generate $G$

what i am thinking is once $H$ is fixed, the homomorphism is fixed $\eta$ (after all it is left multiplication of cosets of H). (otherway may not be true.. i dont see it immediately)

and any homomorphism of $G$ is fixed by all these $g_i :1\leq i\leq r$

each $g_i$ have $n!$ possibilities as its image.

So all $g_i : 1\leq i\leq r$ would have $n!n!\dots n!$ repeating $r$ times i.e., $(n!)^r$ times..

and as these homomorphism (i am not sure but somethings may not correspond to H) are finite.

Thus , I see that No.of $H$ should be atmost $(n!)^r$ (after excludind some useless cases)

So, I would like to say that

If a group is finitely generated, then there are finitely many subgroups of a given finite index.

I am very much sure that there are some gaps which i am unable to see..

please help e to fill this in detail (if this way is partially true)by giving some hints or please suggest me another approach (if this is a blunder).

Thank you

  • 2
    I don't see any problem with your reasoning. You prove that there are only finitely many actions of $G$ on a finite set, and a subgroup is equivalent to a transitive action with a chosen element of the set ($H$ is then the stabilizer of the element) – user8268 Sep 15 '13 at 14:19
  • @user8268 : i am not able to understand what you have meant by "You prove that there are only finitely many actions of G on a finite set, and a subgroup is equivalent to a transitive action with a chosen element of the set (H is then the stabilizer of the element)" .......finitely many actions on a finite set?? as my set is fixed, it will e unique.. i am calculating "how many sets".... Subgroup equivalent to transitive action???.... I would really like to know what exactly do you mean by that... AM i missing something?? –  Sep 16 '13 at 01:42
  • 4
    Yes, the claim which you are proving is a theorem by M. Hall and its proof from “Group theory” by A. G. Kurosh is the same as yours. Since the number of different homomorphisms from $G$ to $S_n$ is finite and any such homomorhism $\eta$ uniquely determines the respective subgroup $H$ (it seems because $g\in H$ iff $\eta(g)(1)=(1)$). So you can write your reasoning as an answer, accept it, and do not disturb MSE users. :-D – Alex Ravsky Sep 26 '13 at 05:29
  • @AlexRavsky I was looking for the proof in the book you recommended. Kindly can you please tell me which edition and if possible the page number of the proof in the book. Sorry for such a bad request. – user371231 Sep 23 '21 at 10:54
  • @user371231 The required Hall’s theorem is in the third extended edition (1967), the end of §38, p.250-251. – Alex Ravsky Sep 23 '21 at 18:17

1 Answers1

1

There is a missing step in your proof, at the line:

"what i am thinking is once $H$ is fixed, the homomorphism is fixed $\eta$ (after all it is left multiplication of cosets of $H$). (otherway may not be true.. i dont see it immediately)"

This question asks how to plug this gap.

user1729
  • 32,369
  • answer of the question you referred is not correct. – user371231 Sep 22 '21 at 08:01
  • @user371231 Thanks for pointing this out - I had meant to fix my answer there ages ago, but forgot (and now no longer have the time to think!). So I've deleted it, and edited this question. – user1729 Sep 22 '21 at 09:21