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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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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0OK. Thanks a lot :) – 2012-11-15