characterization of full families of groups

Proposition. Let $𝒢={\{G_k\}}_{k\in I}$ be a family of groups. Then $𝒢$ is full if and only if for any $i,j\in I$ such that $i\ne j$ 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

$$h:\underset{k\in I}{\oplus }{G}_k\to \underset{k\in I}{\oplus }{G}_k$$

as follows: if $t\in I$ is such that $t\ne i$ and $g\in {\oplus }_{k\in I}{G}_k$ is such that $g\in G_t$, then $h(g)=g$. If $g\in {\oplus }_{k\in I}{G}_k$ is such that $g\in G_i$, then $h(g)(j)=f(g(i))$ and $h(g)(k)=0$ for $k\ne j$. This values uniquely define $h$ and one can easily check that $h$ is not decomposable. $\mathrm{\square }$

,,$\Leftarrow$” Assume that for any $i,j\in I$ such that $i\ne j$ we have that any homomorphism $f:G_i\to G_j$ is trivial. Let

$$h:\underset{k\in I}{\oplus }{G}_k\to \underset{k\in I}{\oplus }{G}_k$$

be any homomorphism. Moreover, let $i\in I$ and $g\in {\oplus }_{k\in I}{G}_k$ be such that $g\in G_i$. We wish to show that $h(g)\in G_i$.

So assume that $h(g)\notin G_i$. Then there exists $j\ne i$ such that $0\ne h(g)(j)\in G_j$. Let

$$\pi :\underset{k\in I}{\oplus }{G}_k\to G_j$$

be the projection and let

$$u:G_i\to \underset{k\in I}{\oplus }{G}_k$$

be the natural inclusion homomorphism. Then $\pi \circ u:G_i\to G_j$ is a nontrivial group homomorphism. Contradiction. $\mathrm{\square }$

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 $i\ne j$ 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 $i\ne j$ 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. $\mathrm{\square }$