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
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.