Proposition. Let $\mathcal{G}=\{G_k\}_{k\in I}$ be a family of groups. Then $\mathcal{G}$ is full if and only if for any $i,j\in I$ such that $i\neq j$ we have that any homomorphism $f:G_i\to G_j$ is trivial.

Proof. ,,$\Rightarrow$ ' Assume that $f:G_i\to G_j$ is a nontrivial group homomorphism. Then define $$h:\bigoplus_{k\in I} G_k\to \bigoplus_{k\in I} G_k$$ as follows: if $t\in I$ is such that $t\neq i$ and $g\in\bigoplus_{k\in I}G_k$ is such that $g\in G_t$ then $h(g)=g$ If $g\in\bigoplus_{k\in I}G_k$ is such that $g\in G_i$ then $h(g)(j)=f(g(i))$ and $h(g)(k)=0$ for $k\neq j$ This values uniquely define $h$ and one can easily check that $h$ is not decomposable. $\square$ ,,$\Leftarrow$ ' Assume that for any $i,j\in I$ such that $i\neq j$ we have that any homomorphism $f:G_i\to G_j$ is trivial. Let $$h:\bigoplus_{k\in I} G_k\to \bigoplus_{k\in I} G_k$$ be any homomorphism. Moreover, let $i\in I$ and $g\in\bigoplus_{k\in I} G_k$ be such that $g\in G_i$ We wish to show that $h(g)\in G_i$

So assume that $h(g)\not\in G_i$ Then there exists $j\neq i$ such that $0\neq h(g)(j)\in G_j$ Let $$\pi:\bigoplus_{k\in I} G_k\to G_j$$ be the projection and let $$u:G_i\to \bigoplus_{k\in I} G_k$$ be the natural inclusion homomorphism. Then $\pi\circ u:G_i\to G_j$ is a nontrivial group homomorphism. Contradiction. $\square$

Corollary. Assume that $\{G_k\}_{k\in I}$ is a family of nontrivial groups such that $G_i$ is periodic for each $i\in I$ Moreover assume that for any $i,j\in I$ such that $i\neq j$ and any $g\in G_i$ $h\in G_j$ orders $|g|$ and $|h|$ are realitvely prime (which implies that $I$ is countable). Then $\{G_k\}_{k\in I}$ is full.

Proof. Assume that $i\neq j$ and $f:G_i\to G_j$ is a group homomorphism. Then $|f(g)|$ divides $|g|$ for any $g\in G_i$ But $f(g)\in G_j$ so $|g|$ and $|f(g)|$ are relatively prime. Thus $|f(g)|=1$ so $f(g)=0$ Therefore $f$ is trivial, which (due to proposition) completes the proof. $\square$