I'm having trouble with proving the following:

Let $H \subset G$ be a subgroup with finite index $n = [G:H]$
$H$ 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!


marked as duplicate by Nicky Hekster, Najib Idrissi, Yiorgos S. Smyrlis, user149792, Jack D'Aurizio Mar 9 '15 at 21:31

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  • 3
    $\begingroup$ You might consider the quotient group $G/H$ instead. Hint: What can you say about $(gH)^n$? $\endgroup$ – Josh Keneda Mar 9 '15 at 10:54
  • $\begingroup$ $(gH)^{n} = g^{n}H$. But how does that help me proving this exercise? $\endgroup$ – Peter Mar 9 '15 at 10:59
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    $\begingroup$ 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. $\endgroup$ – Josh Keneda Mar 9 '15 at 11:08
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    $\begingroup$ Consider that $[G:H]$ is finite. There are therefore only a finite number of elements of the form $(gH)^n \in G/H$. $\endgroup$ – Arthur Mar 9 '15 at 11:08
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    $\begingroup$ @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 $[G:H]=n$ $\endgroup$ – Snufsan Mar 9 '15 at 11:16

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 $g^n\in H$.

  • $\begingroup$ 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. $\endgroup$ – Peter Mar 9 '15 at 12:11
  • $\begingroup$ So if $gH$ has order $m$, and $m$ divides $n$, so $mk=n$ for some $k\in \mathbb{Z}$. Then $(gH)^n=(gH)^{mk}=((gH)^m)^k=H^k=H$. $\endgroup$ – Moya Mar 9 '15 at 12:45
  • $\begingroup$ 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? $\endgroup$ – Peter Mar 9 '15 at 12:47
  • $\begingroup$ No: for a counterexample, take $S_3$ and the subgroup $H = \{\rm{id},(12)\}$. This has index $3$, but $(13)^3 = (13)\not\in H$. $\endgroup$ – Nicky Hekster Mar 9 '15 at 14:32
  • $\begingroup$ See also math.stackexchange.com/questions/545417/… $\endgroup$ – Nicky Hekster Mar 9 '15 at 14:35

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