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

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

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

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

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

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

$\displaystyle = a^kb^kab,$

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

So applying this, the calculation continues like:

$\displaystyle a^kb^kab = a^kab^kb$

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

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