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## Homework Statement

If ##a## and ##b## are commuting elements of ##G##, prove that ##(ab)^n = a^nb^n## for all ##n \in \mathbb{Z}##.

## Homework Equations

## The Attempt at a Solution

We prove two lemmas:

1) If ##a## and ##b## commute, then so do their inverses: ##ab=ba \implies (ab)^{-1} = (ba)^{-1} \implies b^{-1}a^{-1} = a^{-1}b^{-1}##.

2) If ##a## and ##b## commute, then ##b^n a = ab^n##: Base case is trivial. Suppose for some ##k## we have ##b^ka = ab^k##. Then ##b^{k+1}a = bb^ka = bab^k = abb^k = ab^{k+1}##.

Now to the actual result.

Clearly ##(ab)^0 = a^0b^0##. So first we prove the result for the positive integers, by induction. The base case is trivial. Suppose for some ##k \in \mathbb{Z}^+## we have ##(ab)^k = a^kb^k##. Then ##(ab)^{k+1} = (ab)^k(ab) = a^kb^kab = a^kab^kb = a^{k+1}b^{k+1}##.

Now we prove the result for negative integers. ##(ab)^{-n} = (b^{-1}a^{-1})^n = (a^{-1}b^{-1})^n = (a^{-1})^n(b^{-1})^n = a^{-n}b^{-n}##.

Does this argument work? Were the lemmas really necessary or could I have assumed they held since their proofs are trivial?