Abstract Algebra question(s)?

As for your first question, unfortunately, no you didn't use the condition of being abelian there. Remember that:

(ab)^(mn)

means the product:

(ab)(ab)(ab)(ab)...(ab)

where there are mn terms. Then, of course, there are nm terms, since mn = nm. This is the commutativity of multiplication of integers, not the commutativity of the group operation. It still holds even if G was not abelian. Where you did use the commutativity of the group operation was in the step:

((ab)^m)^n = ((a^m)(b^m))^n

In a general group:

(ab)^m = abababab....ab

(a^m)(b^m) = aaaa...abbbb...b

You used the fact that G was abelian to rearrange the top product to obtain the bottom product. As for your other question, you know from the previous question that:

(ab)^(nm) = e

Now, this actually implies that the order of ab is a divisor of mn. This is typically proved as a more general result:

If a is an element of group G, and if a^n = e, then ord(a) divides n.

If you haven't yet covered this in lectures, we can prove it here. Let m = ord(a). We know, clearly, that m <= n. We also know, from the division algorithm that there exists a unique q and unique r such that 0 <= r < m and:

n = qm + r

So:

a^n = a^(qm + r)

e = (a^(qm))(a^r)

e = ((a^m)^q)(a^r)

e = (e^q)(a^r)

e = e(a^r)

e = a^r

But r < m = ord(a), so by the definition of ord(a), the only possible value for r must be 0. Therefore:

n = qm

i.e. m = ord(a) divides n.

Your first use of the property that G is abelian appears right here:

((ab)^m)^n = ((a^m)(b^m))^n

In general, what is (ab)^m? It is

(ab)(ab)...(ab) (m times).

If this is (aa...a)(bb....b) = a^mb^m, then it must be that a and b commute.

It doesn't matter if G is abelian or not when you conclude b^(nm) = b^(mn), that is just commutitivity of the integers m and n---or commutativity of b with itself which doesn't require an abelian group.

I'll look at the rest, but don't know if I can contribute or not. I'll come back if I can.

*** Update ***

I have a proof of the second part, but it is by contradiction. Perhaps you can mold it into a direct proof.

Let ord(ab) = k and suppose that k does not divide mn.

Then mn = αk + ß, where 0 < ß < k.

From the first part of the problem we have

e = (ab)^(mn)

= (ab)^(αk + ß)

= (ab)^(αk)(ab)^ß

=[(ab)^k]^α (ab)^ß

= e (ab)^ß

= (ab)^ß

But then (ab)^ß = e. As ß < k, this contradicts k = ord(ab). Hence k divides mn. ■