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Let $G$ be a group and $H<G$ such that $[G:H]<\infty$. There exists a subgroup $N\triangleleft G$ such that $[G:N]<\infty$.

I have to show this fact (that according to my book is due to Poincaré), but I think that the statement, written in this way, is trivial: for every group $G$, I can take $N=G$, in fact $G\triangleleft G$ and $[G:G]=1$.

Where am I wrong?

If I'm not wrong, do you know a similar statement?

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you might want to add group actions tag for this.... This could also be taken as a hint... –  Praphulla Koushik Feb 18 at 14:15
    
@PraphullaKoushik: hint to show what? is the statement not trivial as I think? –  user72870 Feb 18 at 14:17
    
It would be trivial if you consider trivial subgroup and would be non trivial if you want to look for non trivial subgroups.... –  Praphulla Koushik Feb 18 at 14:18
    
just tell me one thing.. are you familiar with group actions? –  Praphulla Koushik Feb 18 at 14:22

5 Answers 5

The non-trivial version of the statement is that $N$ can be taken to be a subgroup of $H$. In other words, the collection of finite index normal subgroups is cofinal in the collection of all finite index subgroups.

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Is this an answer (proof) :O –  Praphulla Koushik Feb 18 at 14:27
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@PraphullaKoushik, have you understood the question? It is not about a proof, but about the correct statement. –  Carsten Schultz Feb 18 at 15:23

The question is: If there exists a finite index proper subgroup, prove that there exists a finite index proper normal subgroup.

The group $G$ acts by left multiplication on the left cosets of $H$. This action corresponds to a homomorphism $G\rightarrow S_n$ from $G$ to the symmetric group on $n$ elements (why?). (Note that this map is not necessarily surjective or injective - it is just some homomorphism.) This map has a kernel...

Note: This is the solution that Praphulla Koushik is getting at. There is a solution which does not use group actions, but if my memory serves me well it is significantly longer.

Note II: mesel's comments, below, regard an earlier version of the answer where the action was by conjugation. Unpacking the "why?" is harder in this case - you need to apply the orbit-stabiliser theorem to get the map $G\rightarrow S_m\leq S_n$ where $m=|G:N_H(G)|$ is the index of the normaliser of $H$ in $G$. The kernel of this map is $G$ itself if and only if the subgroup $H$ is normal.

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I guess your solution is wrong.You said $G$ acts set of subgroups of index $n$.Then you will have homomorphism from $G$ to $S_k$ where $k$ is the number of subgroup of index $n$, which has no direct relation with $n$.But the idea is true, if I am wrong,correct me. –  mesel Feb 18 at 19:15
    
@mesel That is tied up in the "why?". However if I alter the action it makes it easier: rather than acting by conjugation act by left multiplication. I've edited the post. –  user1729 Feb 18 at 19:32
    
let $G=Z$ and $H=2Z$ then $|G:H|=2$ and $H$ is uniqe subgroup of index 2. So,your action is trivial $\implies$ $N=G$ which is not proper. –  mesel Feb 18 at 19:32
    
Now, it is ok now . I think the requiered condition is $N\leq H$. please see my answer also (now it is almost same). –  mesel Feb 18 at 19:35
    
@mesel If you act by conjugation then you are mapping to $S_m$ where $m$ is the index of the normaliser of $H$ in $G$. As $m\leq n$, this corresponds to a map to $S_n$. If the kernel is trivial then the subgroup itself is normal. –  user1729 Feb 18 at 19:55

Thanks everyone for the answers, even if I just asked for the correct statement and not to spoiler me the solution. I'm not familiar with group actions, so (with Nicky's hint) I tried to write a different solution.

Let $G$ be a group and $H<G$. If $[G:H]<\infty$, then there exists $N\triangleleft G$ such that $N\subseteq H$ and $[G:N]<\infty$.

Lemma 1. Let $H^g:=gHg^{-1}$ and $N:=\bigcap_{g\in G}H^g$. Then $H^g<G$, $N\triangleleft G$ and $N\subseteq H$.

Proof: Straightforward.

Lemma 2. Let $n=[G:H]$, then there are at most $n$ subgroups of $G$ of the form $H^g$.

Proof: Let $gH$ be a left coset of $H$ in $G$. Let $a, b\in gH$, then $H^a=H^g=H^b$. But the $n$ left cosets partition $G$.

Lemma 3. Let $H, K< G$, then $[G:H\cap K]\le [G:H][G:K]$.

Proof: Observe that for every $g\in G$, $g(H\cap K)=gH\cap gK$ and that we can choose $aH\cap bK$ in $[G:H][G:K]$ (not always different) ways.

Lemma 4. $[G:H]=[G:H^g]$.

Proof: Let $S:=\{xH:x\in G\}$ and $S^g:=\{xH^g:x\in G\}$. Define $f:S\rightarrow S^g$ such that $f(xH)=(gxg^{-1})H^g$. $f$ is well-defined and injective, in fact $xH=yH\Leftrightarrow xy^{-1}\in H\Leftrightarrow (gxg^{-1})(gyg^{-1})^{-1}=gxy^{-1}g^{-1}\in H^g \Leftrightarrow f(xH)=f(yH)$. Moreover, $f$ is surjective because $f(g^{-1}xgH)=xH^g$.

Now, putting things together, we have that $N$ defined in lemma 1 is the subgroup we wanted, because $$[G:N]\le [G:H]^n=n^n.$$

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If I would be your professor, I would have given you an $A^+$. Well done! –  Nicky Hekster Feb 18 at 19:58
    
To get an $A^{++}$, try to get the $n^n$ down to $n!$. –  user1729 Feb 18 at 20:36

The theorem follows from what is called Poincaré's Lemma in group theory.

Lemma Let $G$ be a group, $H$ and $K$ subgroups with finite index. Then the subgroup $H \cap K$ has finite index too.

Proof (sketch) for any $g \in G$, the coset $g(H \cap K) \subseteq gH \cap gK$.

The theorem now follows by observing that any conjugate $H^g$ of $H$ has the same (finite) index as $H$ in $G$ and the number of conjugates of $H$ is finite (it equals $[G:N_G(H)]$ which divides $[G:H]$).

Remark: see also this discussion!

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I guess correct statement sholud require $N \leq H$.

Let $G$ act on left coset of $H$ by left multiplication.Then we have homomorphism from $\phi$ from $G$ to $S_{[G:H]}$.Thus,$|G:Ker(\phi)|divides |G:H|!$ so $|G:Ker(\phi)|$ is finite. $Ker(\phi)$ is desired normal subgroup.

Since elements of $Ker(\phi)$ fixes every elements by left multiplacition, so it also fixes $H$ $\implies$ $Ker(\phi)| \leq H$.(Actually it is the largest normal subgroup contained in $H$)

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