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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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