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This is Exercise 1.12.1 of Goodman's "Algebra: Abstract and Concrete".

Let $G\cong H\times K$ be a direct product of finite groups. Show that every element of $G$ has order dividing $\operatorname{lcm}(\vert H\vert, \vert K\vert)$.

Here $H, K$ are finite groups (of course).

My Attempt:

Let $g=(h, k)\in G$. Then $g^{\vert G\vert}=e_G=(e_H, e_K)$. Let $\gamma=\operatorname{ord}_G(g)$. We want to show that $$\gamma\mid\operatorname{lcm}(\vert H\vert, \vert K\vert).$$

I don't know where to go from here.

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Hint : By definition of $\operatorname{ord}_G(g)$ it suffices to show that $$g^{\operatorname{lcm}(|H|,|K|)}=e_G.$$

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    Thank you. I've written my own answer based on yours. Does that work? – 2017-02-25
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    @Shaun Yes that's right. – 2017-02-25
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Let $\mu=\operatorname{lcm}(\vert H\vert, \vert K\vert)$. Then $\mu=\vert H\vert u$ and $\mu=\vert K\vert v$ for some $u, v\in\mathbb{Z}$ by definition of $\mu$.

Also, for $g=(h, k)\in G$, $$\begin{align} g^\mu&=(h, k)^\mu \\ &=(h^\mu, k^\mu) \\ &=((h^{\vert H\vert})^u, (k^{\vert K\vert})^v) \\ &=(e_H^u, e_K^v) \\ &=(e_H, e_K) \\ &=e_G. \end{align}$$

Thus $\operatorname{ord}_G(g)\mid \mu$ by definition of $\operatorname{ord}_G(g)$.