1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.