Thread: Group, abelian

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)

Your idea seems quite right, but be sure to be clear about the details. I'll emphasize which details I'm thinking of:

Suppose . We want to prove that for all positive integers .

Certainly the statement is true for . For the induction hypothesis, suppose it is true for , where . This means that:



(notice that at this point, you (accidentally?) change the order of and in your calculation).

From the induction hypothesis, we want to prove that . Somewhat similar to what you do:





where I've used the induction hypothesis in the last equality. In , we wish to interchange the two middle terms and , i.e. we want to use that . But why is this true? This in itself follows from induction over , using the initial assumption that .

So applying this, the calculation continues like:





which proves the claim. Again, notice that for some reason, you are interchanging the 's and the 's during the calculation. If you want to do that, you need some sort of justification.