1. ## 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)

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

Suppose $ab=ba$. We want to prove that $(ab)^n = a^nb^n$ for all positive integers $n$.

Certainly the statement is true for $n=1$. For the induction hypothesis, suppose it is true for $n=k$, where $k\geq 1$. This means that:

$(ab)^k=a^kb^k,$

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

From the induction hypothesis, we want to prove that $(ab)^{k+1}=a^{k+1}b^{k+1}$. Somewhat similar to what you do:

$(ab)^{k+1} = (ab)^k(ab)$

$= a^kb^kab,$

where I've used the induction hypothesis in the last equality. In $a^kb^kab$, we wish to interchange the two middle terms $b^k$ and $a$, i.e. we want to use that $b^ka=ab^k$. But why is this true? This in itself follows from induction over $k$, using the initial assumption that $ab=ba$.

So applying this, the calculation continues like:

$a^kb^kab = a^kab^kb$

$= a^{k+1}b^{k+1},$

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