2
$\begingroup$

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?

  • 0
    This is what I assumed. At the beginning, I thought that he wanted me to prove that its a subgroup2012-11-15

2 Answers 2

4

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

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

0

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

  • 0
    OK. Thanks a lot :)2012-11-15