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I'm having trouble with proving the following:

Let $H \subset G$ be a subgroup with finite index $n = [G:H]$
Prove:
$H$ is a normal subgroup of $G$ $\Rightarrow g^{n} \in H$ $\forall$ $g \in G$

So far, I've done this:

$H$ is a normal subgroup of $G$
$\Rightarrow ghg^{-1} \in H$ $(g \in G, h \in H)$
$\Rightarrow \exists$ $ h' \in H$ such that $ghg^{-1} = h'$

I think I have to use the index now, but I don't know how to complete this prove. Could you help me completing it? Thanks in advance!

Peter
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    You might consider the quotient group $G/H$ instead. Hint: What can you say about $(gH)^n$? – Josh Keneda Mar 09 '15 at 10:54
  • $(gH)^{n} = g^{n}H$. But how does that help me proving this exercise? – Peter Mar 09 '15 at 10:59
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    Alright, you're halfway there. What's the order of $G/H$? If I have a group of order $m$, do you know anything about $x^m$ for group elements $x$? I'll write up an answer in a bit, but see if you can finish from here. – Josh Keneda Mar 09 '15 at 11:08
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    Consider that $[G:H]$ is finite. There are therefore only a finite number of elements of the form $(gH)^n \in G/H$. – Arthur Mar 09 '15 at 11:08
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    @Arthur It's even better, since $H$ is normal we know that $G/H$ is a group with co-set multiplication, and it's order is $[G:H]=n$ – Snufsan Mar 09 '15 at 11:16
  • @Josh Keneda There is only one thing in Moya's answer I don't see immediately. Why is it true that if $gH \in G/H$ then $(gH)^{n}=H$? I understand that by Lagrange's theorem, the order of any element of G/H divides n. – Peter Mar 09 '15 at 12:27

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I'll just combine everything that's been said in the comments. $H\subset G$ is a normal subgroup, so $G/H$ is a group of order $n$. By Lagrange's theorem, the order of any element of $G/H$ divides $n$. In particular, if $gH\in G/H$, then $(gH)^n=H$.

From the comments, $(gH)^n=g^nH=H$, which implies that $g^n\in H$.

Moya
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  • Thank you for your answer. There is only one thing I don't see immediately. Why is it true that if $gH \in G/H$ then $(gH)^{n} = H?$ I understand that by Lagrange's theorem, the order of any element of G/H divides n. – Peter Mar 09 '15 at 12:11
  • So if $gH$ has order $m$, and $m$ divides $n$, so $mk=n$ for some $k\in \mathbb{Z}$. Then $(gH)^n=(gH)^{mk}=((gH)^m)^k=H^k=H$. – Moya Mar 09 '15 at 12:45
  • Thank you! its clear to me now. I was only wondering if this proof would also be correct if H was not a normal subgroup, but just a subgroup? – Peter Mar 09 '15 at 12:47
  • No: for a counterexample, take $S_3$ and the subgroup $H = {\rm{id},(12)}$. This has index $3$, but $(13)^3 = (13)\not\in H$. – Nicky Hekster Mar 09 '15 at 14:32
  • See also http://math.stackexchange.com/questions/545417/if-gh-n-is-it-true-that-xn-in-h-for-all-x-in-g/546031#546031 – Nicky Hekster Mar 09 '15 at 14:35