Prove that $G$ is Abelian if and only if $(ab)^n =a^n b^n $ for all $a, b \in G$ and $ n \in \mathbb{Z} $.

I used proof by induction in the $ \rightarrow$ direction of the proof and I'm done with that.

Any help/hint on the $ \leftarrow$ direction.

  • $\begingroup$ Is this for all $n$ or just some particular $n$? $\endgroup$ – Jack Schmidt Jun 18 '13 at 16:21
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    $\begingroup$ That is for all n $\endgroup$ – Judiel Jun 18 '13 at 16:25
  • $\begingroup$ See Arturo's description math.stackexchange.com/questions/40996/… if you are interested in what happens for single $n$ (the group need not be abelian) $\endgroup$ – Jack Schmidt Jun 18 '13 at 17:57
  • $\begingroup$ Thanks @jack. This is what Im thinking. Can I say that suppose it is true for all n, then it is true for n+1 and n+2.. Then I will do what is written on the link to show its abelian. Am I leading to the right path? $\endgroup$ – Judiel Jun 18 '13 at 18:10
  • $\begingroup$ That is a correct way to do it. Peter's way is simpler. $\endgroup$ – Jack Schmidt Jun 18 '13 at 18:11

It should read "...for all $a,b\in G$ and $n\in\Bbb Z$."

Take $n=2$ and you get $abab= aabb$. Now use inverses.

  • $\begingroup$ So if n=2, then G is Abelian. How about if n is not 2? $\endgroup$ – Judiel Jun 18 '13 at 17:36
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    $\begingroup$ @Judiel You want the identity to hold for all $n\in\mathbb{Z}$, in particular for $n=2$. $\endgroup$ – egreg Jun 18 '13 at 17:54
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    $\begingroup$ @Judiel You seem to be misunderstanding the statement. It says "$G$ is abelian if and only if for each $n\in\Bbb Z$ and for each $a,b\in G$, $(ab)^n=a^nb^n$. We could have chosen any $n$ to prove the $\Leftarrow$ direction. The choice of $n=2$ was arbitrary but convenient. $\endgroup$ – Pedro Tamaroff Jun 18 '13 at 17:54

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