A group of order $p^2$, where $p$ is prime, with exactly one proper subgroup is cyclic.

Question

I am not able to prove this, could any one help me?

$G$ be a finite group, $G$ has exactly one proper subgroup. We need to prove that $G$ is cyclic and $|G|=p^2$ where $p$ is prime.

Thank you.

See also the weaker: http://math.stackexchange.com/questions/429165/a-group-with-two-non-trivial-subgroups-is-cyclic/ — Jack Schmidt, Jul 19 '13 at 12:18

score 2 · Accepted Answer · answered Jul 19 '13 at 12:20

Let $G$ be your group and let $H$ be the unique proper subgroup. The Cauchy theorem implies that $G$ is a $p$-group for some prime $p$. Also $H$ has to be normal and neither $H$ nor $G/H$ have proper subgroups. It follows that $H \cong G/H\cong \mathbb{Z}/p\mathbb{Z}$. This proves $|G|=p^2$. Any $x\notin H$ must generate all of $G$, so that $G$ is cyclic.

score 1 · Answer 2 · answered Jul 19 '13 at 12:15

1

Hint: Let $C \leq G$ be a cyclic subgroup of maximal cardinality. Show that $C=G$.

answered Jul 19 '13 at 12:15

Seirios

33,157

A group of order $p^2$, where $p$ is prime, with exactly one proper subgroup is cyclic.

2 Answers2