Since $\mathbb{Z}/4\mathbb{Z}$ is an abelian group of order $4$, $4=2^2$, and $2$ is a prime, we know $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ is isomorphic to $\mathbb{Z}/4\mathbb{Z}$. So $\mathbb{Z}/4\mathbb{Z}$ is semisimple as a $\mathbb{Z}$-module, right?
Thanks for your attention.
I notice you posted another question regarding the socle of $\mathbb{Z}_{16} \times \mathbb{Z}_{25}$. In my answer to that question there is a characterization of the socle of $\mathbb{Z}_m$ for general $m \in \mathbb{N}^*$. In this particular case, the socle of $\mathbb{Z}_4$ is isomorphic to $\mathbb{Z}_2$ (this can be seen very easily, even without resorting to the systematized theory I presented a bit of in my answer to the above mentioned question, since $\mathbb{Z}_4$ has only one nontrivial proper submodule, which happens to be simple); as the module in question contains its socle as a proper submodule, it cannot be semisimple.