${\mathbb Q}$ as a direct product.

If $\mathbb Q$ was isomorphic to $G\times H$ then $G$ and $H$ must be abelian and so $\mathbb Q$ would be a product $\mathbb Q \cong G\prod H$ in the category $Ab$, and as in $Ab$ finite products and coproducts agree, $\mathbb Q \cong G\coprod H$ holds as well. Let $i:G\to \mathbb Q$ and $j:H\to \mathbb Q$ be the canonical injections. In $Ab$ the canonical injections are monos and monos are injective functions. So, $G$ and $H$ are canonically subobjects of $\mathbb Q$. Now, the special property of $\mathbb Q$ is that the intersection of any two non-trivial subobjects in it have a non-trivial intersection (with $\mathbb Z$). However, in $Ab$ the canonical injections $G\to G\coprod H$ and $H\to G\coprod H$ intersect at the $0$ object, contradiction.

This is about as categorically as I could furnish the proof, I hope you like it.