Let G be a group. For all a,b in G, Prove that 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 through that 's worth of 's until it's adjacent the on the left. Then all the 's will be together and all the '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 commutes with then commutes with for all , which would make the above somewhat cleaner.