How to prove that $\mathbb Z/p$ is not a direct summand of any direct sum of copies of $\mathbb Z/n$?

Question

How can I prove that $\mathbb Z/p$ ($p$ is a prime) cannot be a direct summand of any arbitrary direct sum of copies of $\mathbb Z/n$, where $p^2$ divides $n$?

Answer 1

Assume that $\mathbb{Z}/p$ is a direct summand of $\bigoplus_{i \in I} \mathbb{Z}/n$. We may assume $n=p^2$ after dividing by $p^2$. The image is generated by some element of order $p$, say $b=(b_i \bmod p^2)_{i \in I}$. Since $p b_i \equiv 0 \bmod p^2$, we may write $b_i = p a_i$. If $b_i=0$, we choose $a_i=0$. It follows that $a:=(a_i)_{i \in I}$ is an element of the direct sum such that $b=pa$. But then every homomorphism to $\mathbb{Z}/p$ kills $b$, a contradiction.