Problem: From Aluffi's Algebra: Chapter 0, Chapter II
3.5. Prove that $\mathbb Q$ is not the direct product of two nontrivial groups.
My attempt at a proof: Suppose $\mathbb Q\cong G\times H$ for some nontrivial groups $G$ and $H$. Let
$$\mathbb Q\ni 1\leftrightarrow(g, h)\in G\times H.$$
Choose a $(x, y)\in G\times H$ such that $x\ne e_G$ and $y\ne e_H$. Say
$$
(x, e_H)\leftrightarrow a/b\quad\text{and}\quad (e_G, y)\leftrightarrow c/d\text.
$$
Then
$$
(x, e_H)^b\leftrightarrow a\leftrightarrow(g, h)^a
$$
which means that
$
(x, e_H)^b = (g, h)^a
$
and hence
$h^a = e_H$. Similarly, $g^c = e_G$. Now if $a$ and $c$ are both nonzero, then $|(g, h)| < \infty$ in $G\times H$ which will mean that $|1| < \infty$ in $\mathbb Q$ which is false. Hence $a = 0$ or $c = 0$. This means that
$$(x, e_H)\leftrightarrow 0\quad\text{or} \quad (e_G, y)\leftrightarrow 0$$
which means that $x = e_G$ or $y = e_H$ since $0\leftrightarrow (e_G, e_H)$. This is the required contradiction. $\square$
Question: This seems like an overly complicated and artificial proof, as if there has to be a simpler and more elegant way here. Can you think of any?
Note: Aluffi covers only the very basic group theory until this point, hence the use of more advanced stuff is not allowed.
Best Answer
Sure. If $\mathbb{Q}\simeq G\times H$ then $\mathbb{Q}$ has two nontrivial subgroups $A,B\subseteq\mathbb{Q}$ such that $A\cap B=\{0\}$. But that cannot happen since any two non-zero rationals have a common multiple.
EDIT: According to comments you have not been introduced to subgroups yet. Which btw is weird to introduce homomorphisms before subgroups. But, hey, who am I to judge?
Anyway, if that's the case, then the argument above can be rewritten as follows: consider an isomorphism $f:G\times H\to\mathbb{Q}$ and consider nontrivial $g\in G$ and nontrivial $h\in H$. Then $f(g,e_H)$ and $f(e_G, h)$ are nonzero rationals and so they have a common nonzero multiple. The preimage of that common multiple has to belong to both $G\times\{e_H\}$ and $\{e_G\}\times H$ at the same time, and so it is $(e_G,e_H)$, which cannot happen, because $f(e_G,e_H)=0$.
Note that this is very similar to your reasoning though. I've just omitted few details to make the proof easier to read.