Let G be a group and let a, b 2 G. Show that (ab)n = anbn for all positive integers n

if and only if ab = ba.

---> Use a2b2 = (ab)2 = (ab)(ab) to show ab = ba. <---- Use induction.

I kinda get the forwards proof, but don't get the the back wards proof.

Let ab=ba.

We want to show (ab)^n=a^nb^n

I don't see how we use induction for this.

(ab)^1:

ab=ba

Assume true for k=n

(ab)^k

(ab)^k=b^ka^k

Show true for (ab)^(k+1)

(ab)^k+1=(ab)^kab

Since ab^k=b^ka^k

ab^k+1=b^ka^kab

=b^(k+1)a^(k+1)