Prove that if $G$ is an Abelian group, then for all $a,b$ in $G$, $(ab)^{n} = a^{n}b^{n}$

Question

This question have already been asked on this site, but i could not understand the details so i ask it again. Also what i have done is that first for $n=1$ its trivial, for $n=2$ we have $a(ba)b=a(ab)b $. Since $ab=ba$ is given in problem. Does that make it true for $n=2$?

For $n=3$: $ababab=aaabbb$ then i get $baba=aabb$.

How do i prove it for $n=3$?

Answer 1

You ought to start earlier with $(ab)(ab)=a(b(ab))=a(a(bb)=(aa)(bb)$ as required.

Then by induction $(aa)(bb)(ab)=(aa)b(b(ab))=(aa)b(a(bb))=(aa)a(b(bb))=(aaa)(bbb)$ and so on.

Answer 2

Induction on $n$ should do it. The key to induction is $$(ab)^n=(ab)^{n-1}ab=a^{n-1}b^{n-1}ab=a^{n-1}b^na=aa^{n-1}b^n=a^nb^n$$