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I was trying to prove that statement from Resident Dementor's answer, namely:

We can prove that if for an integer $n$ and every $a,b\in G$, $(ab)^n=a^nb^n$, then $$(aba^{-1}b^{-1})^{n(n-1)}=e$$ The proof is easy. In fact, $$(aba^{-1}b^{-1})^{n^2}=[(aba^{-1}b^{-1})^n]^n=[a^n(ba^{-1}b^{-1})^n]^n=...=a^nb^na^{-n}b^{-n}\\\ (aba^{-1}b^{-1})^{n}=(ab)^n(a^{-1}b^{-1})^n=a^nb^na^{-n}b^{-n}$$

I was able to understand only the first two equalities and i was trying to do different methods but no results. Can anyone demonstrate detailed proof, please?

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  • I don't see how $[a^n(ba^{-1}b^{-1})^n]^n=\dots =a^nb^na^{-n}b^{-n}$ either. – Shaun Dec 08 '17 at 14:20
  • @Shaun, It's strange, however 10 people upvote that question. Maybe it is true but I cannot derive the desired result. – RFZ Dec 08 '17 at 14:22

1 Answers1

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$$[a^n(ba^{-1}b^{-1})^n]^n$$ $$=[a^nba^{-n}b^{-1}]^n$$ $$=[a^nba^{-n}]^nb^{-n}$$ $$=a^nb^na^{-n}b^{-n}$$

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