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Prove that if $H$ is a normal subgroup of a group $G$ and has index $n$, then for any $g\in G$ we have $g^n\in H$.

Give an example for if $H$ is not normal, the mentioned statement is not correct.
(Please give an example, except the symmetric group :D )

Martin Sleziak
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Antoni James
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    What are your thoughts on this? – Daniel Akech Thiong Feb 25 '16 at 08:45
  • You might think about $\mathbb{D}_n$ the dihedral group of order $2n$ with $n$ odd and $H:=\langle s\rangle$. Take $g=rs$. Justify that this is a counter example (remark that $g$ is of order $2$). – Clément Guérin Feb 25 '16 at 12:24
  • See also http://math.stackexchange.com/questions/618548/counterexample-that-a-in-g-an-notin-h-for-h-a-subgroup-of-finite-index and http://math.stackexchange.com/questions/545417/if-gh-n-is-it-true-that-xn-in-h-for-all-x-in-g for counterexamples – Martin Sleziak Feb 25 '16 at 15:18

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Hint: $x \in H \iff xH =H$. $xH$ is an element of the group $G/H$- a group because $H$ is normal. It follows that $H = e_{G/H} = (xH)^{[G:H]}=x^{[G:H]}H$. Look for a non-abelian group of small order for your counterexample.

Daniel Akech Thiong
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