[dllthomas]

That seems too easy.

(ab)^2 = aabb a^2 b^2 = abab

ab = ba iff G is commutative, so for an Abelian group we can substitute for the middle bit

aabb = a(ab)b = a(ba)b = abab

Which won't hold if G is not commutative.

QED

I recognize that FizzBuzz is supposed to be easy, but it's supposed to recognize programmers with basic competence; I am not a mathematician. (But maybe I underestimate myself or overestimate some of those with advanced degrees in mathematics?)

[stephencanon]

In fairness, if you handed this in for homework in a sophomore algebra course, you likely wouldn't get credit (you've definitely proven the "only if"; the "if" is a bit murky). However, it's not too much of a stretch to clean it up into a proper proof.

[dllthomas]

In what way did the

aabb = a(ab)b = a(ba)b = abab

fail to prove the if?

edit: I did fail to re-state it, but figured it was obvious in the not-really-formal-proof setting. If that's all you meant by "a bit murky" then nevermind.

[stephencanon]

You have very clearly established the following:

    ab = ba --> (ab)^2 = a^2b^2

It is less obvious that you have proven that:

    (ab)^2 = a^2b^2 --> ab = ba

If I were grading a sophomore algebra class, I would expect to see something along the lines of:

    Suppose (ab)^2 = a^2b^2.
    Re-associating gives us a(ba)b = a(ab)b;
    multiplying on the left and right by the
    inverses of a and b gives the result.

In any domain outside of a sophomore algebra class, I happily accept much briefer and more hand-wavy proofs.

[dllthomas]

Ah, when I read your response I flipped the order of the problem around in recalling it, so had the if and only-if backwards. Yes, I was handwavy there but it seemed clear enough for the setting (which you seem to be granting anyway) - just wanted to be sure I wasn't misunderstanding something. Thanks :)