Let G be a group. For all a,b in G, Prove that$\displaystyle (ab)^n = a^nb^n$ for all n positive integers if and only if ab=ba.
Its been over two years since I've done induction in number theory.
Here is my write-up:
ab=ba => abab=aabb
Assume n=k holds such that (ab)^k=a^k * b^k
n=k+1 yields (ab)^(k+1) = a^(k+1) * b^(k+1)
Which implis ab(ab)^k = a(a)^k * b(b)^k
Which implies (ab)^n = a^n * b^n
You can progressively shuffle that rightmost $\displaystyle a$ through that $\displaystyle k$'s worth of $\displaystyle b$'s until it's adjacent the $\displaystyle a^k$ on the left. Then all the $\displaystyle a$'s will be together and all the $\displaystyle b$'s will be together, and you're left with what you're trying to prove.
There may be a quicker way of doing it, but when I first put this together for the ProofWiki site, I first proved by induction that if $\displaystyle a$ commutes with $\displaystyle b$ then $\displaystyle a$ commutes with $\displaystyle b^n$ for all $\displaystyle n$, which would make the above somewhat cleaner.