Is $(\mathbb{Q},+)$ the direct product of two non-trivial subgroups?

Is this statement true or false? I am really not having any idea how to prove or a counterexample, please help.

Is $(\mathbb{Q},+)$ a direct product of two non-trivial subgroups?

Solution 1:

HINT: If $\Bbb Q=H\times G$, then there is an obvious homomorphism $h:\Bbb Q\to G$ with kernel $H$. Now consider this earlier question.

Solution 2:

If $G$ is an abelian group and $G\simeq G_1\times G_2$, then there exist $H_1,H_2\le G$ two subgroups such that $G=H_1+H_2$ and $H_1\cap H_2=\{0\}$ (and viceversa!). Furthermore, $H_i$ can be chosen isomorphic to $G_i$, $i=1,2$.

If $H_i\le\mathbb Q$, $H_i\neq \{0\}$, then $H_i\cap\mathbb Z\neq \{0\}$ (why?), and therefore $(H_1\cap\mathbb Z)\cap(H_2\cap\mathbb Z)\neq \{0\}$. In particular, $H_1\cap H_2\neq\{0\}$.

Solution 3:

Let $H,K \leq \mathbb{Q}$ be two subgroups such that $\mathbb{Q} \simeq H \times K$.

Notice that $H$ is divisible: If $h \in H$ and $n \in \mathbb{N}^*$, there exist $(h_0,k_0) \in H \times K$ such that $\displaystyle \frac{h}{n}=h_0+k_0$; so $nk_0 = h-nh_0 \in H \cap K=\{0\}$, hence $k_0=0$ and $h/n \in H$.

If there exists $\displaystyle h:=\frac{p}{q} \in H \backslash \{0\}$, then $\displaystyle 1=q \cdot \frac{h}{p} \in H$ ($H$ is divisible). So $\mathbb{Z} \subset H$, and because $H$ is divisible, $\mathbb{Q} \subset H$.

Therefore, $H= \mathbb{Q}$ and $K= \{0\}$.

Solution 4:

Direct product decompositions $A \cong B \times C$ of an abelian group are in natural bijection with idempotent endomorphisms $e : A \to A$ of $A$, where the idempotent corresponding to $A \cong B \times C$ is the projection onto $B$, so that $B = \text{im}(e)$ and $C = \text{ker}(e)$. So if you understand the ring of endomorphisms of $A$ you can compute all idempotents in it, and if there aren't any nontrivial ones then $A$ doesn't admit any nontrivial decompositions.

The endomorphism ring of $\mathbb{Q}$ is $\mathbb{Q}$ (exercise), which has no nontrivial idempotents.