Motivation + Problem: While doing some exercises in Aluffi's Algebra: Chapter 0, I came across a problem which asks the reader to prove that ${\mathbb Q}$ is not the product of two nontrivial groups. This is the standard product in the category of groups.
My Question: Since the chapter delves into categorical arguments, I tried it this way. We suppose not, that ${\mathbb Q} \cong G\times H$ and note that for any group $Z$ and any appropriate mappings $f,g$, we have the following diagram (sorry for the awkward TeXing, xypic doesn't seem to work here…),
$\displaystyle \begin{array}{ccccc} & & Z & & \\ & ^{f}\swarrow & \downarrow&_{\exists!\langle f,g\rangle} \searrow^{g} & \\ & G\longleftarrow_{\pi_{1}} & {\mathbb Q} & _{\pi_{2}}\longrightarrow H &\end{array}$
So, we find that there is always a unique mapping in the center if this is a product. The projections either inject an isomorphic copy of ${\mathbb Q}$ or are the zero mapping. We'd like to show that either $G$ or $H$ is trivial. I'm not sure how to show that at least one must be trivial.
My Attempt: Suppose $G \neq \{0\}$. We need to show $H = \{0\}$. Letting $Z = G$ and let $f = id$, we find that $G \cong {\mathbb Q}$. Moreover, the unique map in the center must be some multiplication mapping (which takes $x\mapsto qx$ for rational $q$; this is because $\pi_{1}$ must be a multiplication mapping if it is not the zero mapping). My guess here is that if $H$ is not trivial, it must also be an isomorphic copy of ${\mathbb Q}$ and we can allow $g$ to be some multiplication map like above such that the unique center mapping does not allow the right-hand triangle to commute.
Does this sort of argument work?
Best Answer
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.