# $G$ is an infinite abelian group such that $G \cong H$ for every non trivial subgroup $H$ of $G$ , then is $G$ cyclic?

If $G$ is an infinite abelian group such that $G \cong H$ for every non trivial subgroup $H$ of $G$ , then is $G$ cyclic , or equivalently asking , then is $[G:H]$ finite for every non trivial subgroup $H$ of $G$ ?

I know that the assumption holds for $G=\mathbb Z$ , so I am asking is this the only group upto isomorphism for which the stated assumption holds . Also I know that if for an infinite group $G$ , $[G:H]$ is finite for every non trivial subgroup of $H$ then $G$ is cyclic .

Please help . Thanks in advance

• – Watson Jun 14 '16 at 11:59

## 1 Answer

Let $x\in G$, $x\ne0$; then $\langle x\rangle$ is a (non trivial) cyclic subgroup of $G$. Since $\langle x\rangle\cong G$, we have that $G$ is cyclic.

• off-course .. and I missed that :) thanks – user228169 Apr 30 '16 at 8:11
• @user228169 Now a quiz: does there exist a non cyclic abelian group such that its proper (non trivial) subgroups are all cyclic? Can you classify such groups, if they exist? – egreg Apr 30 '16 at 8:23
• For prime $p$ , all proper subgroups of the abelian non cyclic group $\mathbb Z_p \times \mathbb Z_p$ is cyclic . Let me think about the classification and an infinite example .. – user228169 Apr 30 '16 at 8:27
• Any finite abelian non cyclic group with the stated property is of the form $\mathbb Z_p \times Z_p$ for some prime $p$ – user228169 Apr 30 '16 at 8:51
• @user228169 Correct. Infinite groups are a bit harder. 😀 – egreg Apr 30 '16 at 9:36