# A group $G$ is Abelian iff $(ab)^n = a^n b^n$ for all $a,b \in G$ and $n \in \Bbb Z$

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.

• Is this for all $n$ or just some particular $n$? – Jack Schmidt Jun 18 '13 at 16:21
• That is for all n – Judiel Jun 18 '13 at 16:25
• 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) – Jack Schmidt Jun 18 '13 at 17:57
• 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? – Judiel Jun 18 '13 at 18:10
• That is a correct way to do it. Peter's way is simpler. – 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.
• @Judiel You want the identity to hold for all $n\in\mathbb{Z}$, in particular for $n=2$. – egreg Jun 18 '13 at 17:54
• @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. – Pedro Tamaroff Jun 18 '13 at 17:54