We say that an abelian group $G$ is mixed if it has elements $ \neq 0$, that are of finite order (torsion elements), as well as elements of infinite order (torsion-free elements). We denote the torsion subgroup of $G$ as $T(G)$. We call a mixed group "splitting" if it can be written as a direct sum $G = T(G) \oplus H$, where $H$ is a torsion-free subgroup of $G$.
In the given exercise, we have a group $G=\underset{p\in\mathbb{P}}{\prod}\mathbb{Z}/p\mathbb{Z}$ the product of $\mathbb{Z}/p\mathbb{Z}$ over all prime numbers $p$. We can demonstrate without difficulty that the torsion subgroup $T(G)$ is equal to the direct sum of $\mathbb{Z}/p\mathbb{Z}$ for all prime numbers $p$ : $T(G)=\underset{p\in\mathbb{P}}{\bigoplus}\mathbb{Z}/p\mathbb{Z}$
Assume that $G$ is splitting, and we have $G=\bigoplus_p\mathbb{Z}/p\mathbb{Z}\oplus H$, where $H$ is a torsion-free subgroup of $G$.
I tried to find a contradiction, but I failed. I would like your help with this problem.