1

Is a direct sum of cyclic groups cyclic? I know every abelian group is a direct sum of cyclic groups of prime power orders, but I can't make use of this.

2 Answers2

1

Certainly not. Consider the Klein $4$-group $V \cong \Bbb Z/2\Bbb Z\oplus\Bbb Z/2\Bbb Z$. None of the four elements of $V$ generates $V$ since every nonidentity element is of order $2$.

Alex Ortiz
  • 24,844
1

Some direct sums of cyclic groups are cyclic. For example, if $\gcd(m,n)=1$ then $\mathbb Z/n\mathbb Z + \mathbb Z/m\mathbb Z$ is generated by $(1,1)$. But if $k=\gcd(m,n)>1$ then $k(1,1) =0$ in the direct sum, so $(1,1)$ fails to generate the whole group. And neither does any other element, since multiplying it by $k$ will yield the zero element.

  • Wanted to add a clarifying comment as I was confused for a minute: the fact that (1,1) fails to generate the whole group is irrelevant to the argument, many cyclic groups contain non-generating elements.I imagine you just mentioned (1,1) as an illustration. – JKEG Aug 19 '18 at 23:34
  • Should also add: all cyclic subgroups of $\mathbb{Z}_n + \mathbb{Z}_m$ have order at most gcd(m,n)<mn. So no cyclic subgroup is large enought to be the whole group. – JKEG Aug 19 '18 at 23:36