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?
Subgroup generated by n-th powers of elements
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$\begingroup$
group-theory
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0Question: Is G abelian ? – 2012-11-15
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1It doesn't have to be – 2012-11-15
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0Should the set of all nth powers be a group ? – 2012-11-15
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0No @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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0This is what I assumed. At the beginning, I thought that he wanted me to prove that its a subgroup – 2012-11-15
2 Answers
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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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0I 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 H – 2012-11-15
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0OK. Thanks a lot :) – 2012-11-15