What are necessary conditions for a cyclic group $G$ to be a direct sum of cyclic groups?

I saw somewhere that $G$ must be a non $p$-group. But I couldn't prove it.

Thank you for your hints/help


I think the only necessary condition is that the group $G$ is abelian. This condition is not sufficient. Every finitely-generated abelian group has a description as a direct sum of cyclic groups, but the direct product of infinitely many cyclic groups is not a direct sum.

Some $p$-groups are the direct sum of cyclic groups, for example $\mathbb Z/p\mathbb Z \oplus \mathbb Z/p\mathbb Z$.

After your edit: we can see that if a cyclic group is to be a direct sum of (at least two nontrivial) cyclic groups, our group must be finite (since $\mathbb Z$ is not), of order, say $N$, and that $N$ must be a composite number with no multiplicity in its prime factorization. The Chinese Remainder Theorem allows us to find decompositions in this case. What's more, this is a necessary condition: as an example, $\mathbb Z/p^2\mathbb Z$ would need to be isomorphic to $\mathbb Z/p \mathbb Z \oplus \mathbb Z /p \mathbb Z$, but the former has an element of order $p^2$ while the latter does not. This is the general obstruction to a finite cyclic group being the direct sum of at least two non-trivial cyclic groups.

  • $\begingroup$ Thank you @Rylee. Sorry I was actually missing an important condition in my question: G is cyclic: the counter-example won't work in this case $\endgroup$ – PerelMan Apr 14 at 21:42
  • 1
    $\begingroup$ Ah! I edited my answer. $\endgroup$ – Rylee Lyman Apr 14 at 21:50

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