Could someone take a look at my solution to this exercise in Artin?
Let $a,b$ be elements of an abelian group of orders $m$ and $n$ respectively. What can you say about the order of their product, $ab$?
Solution. First, we consider a lemma that $(ab)^m = a^m b^m$ for $m \in \mathbb{N}$ for $a, b$ in some abelian group $G$. The base case holds since $$(ab)^1 = a^1 b^1 = ab.$$ Assuming the result for $n = k$: $$(ab)^k = a^k b^k.$$ Proving the result for $n = k + 1$: $$\begin{align} (ab)^{k+1} & = (ab)^k (ab) \\ &= a^k b^k (ab)\\ & = (a^k a)(b^k b)\\ & = a^{k+1} b^{k+1}, \end{align}$$ wherein the final step relies on associativity and commutativity in $G$. Hence, the result is proved for all natural numbers $n$.
By the definition of the order of elements, we have $a^m = e$ and $b^n = e$. Hence, $|ab| \leq mn$, as $$\begin{align} (ab)^{mn} &= a^{mn} b^{mn}\\ & = (a^m)^n (b^n)^m\\ & = e^n e^m\\ & = ee \\ &= e. \end{align}$$ Of course, for any integer multiple $k(mn)$ of $mn$, we also have $(ab)^{mn} = e$: $$\begin{align} (ab)^{kmn}& = a^{kmn} b^{kmn} \\ &= (a^m)^{kn} (b^n)^{km}\\ & = e^{kn} e^{km} \\ &= ee \\ &= e. \end{align}$$ However, there exist cases in which $|ab| < mn$. Take $a = (12)$ and $b = (1234)$ in the group $S_4$ under composition of permutations. We have $|a| = 2$, $|b| = 4$, but $|ab| = \text{lcm}(2,4) = 4 < 8$.
Hence, $|ab| \leq mn$.
Further, let $\ell = \text{lcm}(m,n)$. Hence, there exist positive integers $s$ and $t$ such that $\ell = sm = tn$. We have: $$\begin{align} (ab)^{\ell} &= a^{\ell} b^{\ell} \\ & = a^{sm} b^{tn}\\ & = (a^m)^s (b^n)^t\\ & = e^s e^t \\ &= ee\\ & = e. \end{align}$$