Proposition. Let $\{ G_i\}_{i\in I}$ be a full family of groups. Then each $G_i$ is Hopfian (co-Hopfian) if and only if $\bigoplus_{i\in I}G_i$ is Hopfian (co-Hopfian).

Proof. ,,$\Rightarrow$ '' Let $$f:\bigoplus_{i\in I}G_i\to \bigoplus_{i\in I}G_i$$ be a surjective (injective) homomorphism. Since $\{ G_i\}_{i\in I}$ is full, then there exists family of homomorphisms $\{ f_i:G_i\to G_i\}_{i\in I}$ such that $$f=\bigoplus_{i\in I} f_i.$$ Of course since $f$ is surjective (injective), then each $f_i$ is surjective (injective). Thus each $f_i$ is an isomorphism, because each $G_i$ is Hopfian (co-Hopfian). Therefore $f$ is an isomorphism, because $$f^{-1}=\bigoplus_{i\in I} f_{i}^{-1}.\ \ \square$$

,,$\Leftarrow$ '' Fix $j\in I$ and assume that $f_j:G_j\to G_j$ is a surjective (injective) homomorphism. For $i\in I$ such that $i\neq j$ define $f_i:G_i\to G_i$ to be any automorphism of $G_i$ . Then $$\bigoplus_{i\in I} f_i:\bigoplus_{i\in I}G_i \to \bigoplus_{i\in I} G_i$$ is a surjective (injective) group homomorphism. Since $\bigoplus_{i\in I}G_i$ is Hopfian (co-Hopfian) then $\bigoplus_{i\in I}f_i$ is an isomorphism. Thus each $f_i$ is an isomorphism. In particular $f_j$ is an isomorphism, which completes the proof. $\square$

Example. Let $\mathcal{P}=\{p\in\mathbb{N}\ |\ p\mbox{ is prime}\}$ and $\mathcal{P}_0$ be any subset of $\mathcal{P}$ . Then $$\bigoplus_{p\in P_0}\mathbb{Z}_{p}$$ is both Hopfian and co-Hopfian.

Proof. It is easy to see that $\{\mathbb{Z}_{p}\}_{p\in\mathcal{P}}$ is full, so $\{\mathbb{Z}_{p}\}_{p\in\mathcal{P}_0}$ is also full. Moreover for any $p\in P_0$ the group $\mathbb{Z}_{p}$ is finite, so both Hopfian and co-Hopfian. Therefore (due to proposition) $$\bigoplus_{p\in P_0}\mathbb{Z}_{p}$$ is both Hopfian and co-Hopfian. $\square$