Algebra Solutions: Powers distribute over products of commuting group elements

Let $G$ be a group and let $a,b \in G$ such that $ab = ba$ . Prove that $(ab)^n = a^n b^n$ for all $n \in \mathbb{Z}$ .

First we prove two technical lemmas.
Lemma 1. If $a,b \in G$ with $ab = ba$ , then $a b^n = b^n a$ for all $n \in \mathbb{Z}^+$ . We proceed by induction on $n$ . The statement is clear for the base case $n = 1$ . Now suppose the statement holds for some integer $n \geq 1$ ; then $ab^{n+1} = a b^n b = b^n a b = b^n b a = b^{n+1} a$ . Thus by induction the conclusion holds for all positive integers $n$ . $\square$
Lemma 2. If $a,b \in G$ with $ab = ba$ , then $a^m b^n = b^n a^m$ for all positive integers $n$ and $m$ . We proceed by induction on $n$ . The statement holds for the base case $n = 1$ by the previous lemma. Now suppose that for some $n \geq 1$ , we have $a^m b^n = b^n a^m$ for all $m$ . Then $a^m b^{n+1} = a^m b^n b = b^n a^m b = b^n b a^m = b^{n+1} a^m$ , using Lemma 1. Thus, by induction, the statement holds for all integers $n$ and $m$ . $\square$
We now move to the main result.
The statement is trivial if $n = 0$ .
We will prove the statement for $n > 0$ by induction. For the base case $n = 1$ , we have $(ab)^1 = ab = a^1 b^1$ . For the inductive step, suppose that for some positive integer $n$ , we have $(ab)^n = a^n b^n$ for all such $a$ and $b$ . Then $(ab)^{n+1} = ab(ab)^n = ab a^n b^n = a a^n b b^n = a^{n+1} b^{n+1}$ . Thus by induction, we have $(ab)^n = a^n b^n$ for all positive integers $n$ .
Suppose finally that $n < 0$ . Then $(ab)^n = ((ab)^{-n})^{-1} = (a^{-n} b^{-n})^{-1} = (b^{-n} a^{-n})^{-1} = a^n b^n$ . $\blacksquare$

Algebra Solutions

Powers distribute over products of commuting group elements

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