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$\begingroup$

How can I prove that, if we have a group G, then subgroup of G generated by all n-th powers of elements from G is normal subgroup of G?

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    Question: Is G abelian ?2012-11-15
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    It doesn't have to be2012-11-15
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    Should the set of all nth powers be a group ?2012-11-15
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    No @Amr, it shouldn't, but that's why Martin wrote about the subgroup *generated* by the n-th powers.2012-11-15
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    This is what I assumed. At the beginning, I thought that he wanted me to prove that its a subgroup2012-11-15

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Not only a normal subgroup but in fact a fully invariant subgroup , since for any endomorphism $\,\phi:G\to G\,$ ,we have:

$$\forall\,x\in G\,\,\,,\,\,\phi (x^n)=(\phi x)^n\Longrightarrow \phi(G^n)\subset G^n$$

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Hint: $yx^ny^{-1}=(yxy^{-1})^n$

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    I assumeed that the set $H=${$x^n|x$ is in G} is a group. Now it suffices to show that for all y in G $yHy^{-1}$ is a subset of H2012-11-15
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    OK. Thanks a lot :)2012-11-15