Dear group theorists,

Let $n \geq 1$ and $I$ be an infinite set (you may assume $I$ to be countable). Is the abelian group $(\mathbb{Z}/n)^{(I)}$ (direct sum of copies $\mathbb{Z}/n$) a direct summand of $(\mathbb{Z}/n)^I$ (direct product of copies $\mathbb{Z}/n$)? This question is motivated by that one.

If $n$ is prime, this follows from Linear Algebra. Of course, this is not constructive at all. Thus it's also true when $n$ is squarefree (use the Chinese Remainder Theorem). What happens otherwise? The smallest example is $n=4$. I don't know how to start ...