Prove

**mandy123** · October 9th, 2008, 05:26 PM

Suppose that G is a group that has exactly one nontrivial proper subgroup. Prove that G is cyclic and |G|=p^2 where p is prime.

**ThePerfectHacker** · October 9th, 2008, 06:03 PM

Originally Posted by mandy123

Suppose that G is a group that has exactly one nontrivial proper subgroup. Prove that G is cyclic and |G|=p^2 where p is prime.

Let $H$ be the non-trivial proper subgroup.
Then $\{ e \} \subset H \subset G$ .

Pick $a \in G - H$ . And construct $\left< a \right>$ .

**jonnyfive** · October 21st, 2008, 08:19 PM

Could you please explain your proof a little more. Thanks.

**ThePerfectHacker** · October 22nd, 2008, 03:28 PM

Originally Posted by jonnyfive

Could you please explain your proof a little more. Thanks.

Note that once you construct $\left< a\right>$ it cannot be $H$ and cannot be $\{ e \}$ . Since that is a subgroup and it has no other proper non-trivial subgroups it follows that $\left< a \right> = G$ . Since the group is cyclic it is isomorphic to $\mathbb{Z}_n$ for some $n\geq 1$ . By the properties of cyclic groups we can show that only $n=p^2$ are the ones with this unique non-trivial proper subgroup property.

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