# Cyclic group with exactly 3 subgroups: itself $\{e\}$ and one of order $7$. Isn't this impossible?

Suppose a cyclic group has exactly three subgroups: $G$ itself, $\{e\}$, and a subgroup of order $7$. What is $|G|$? What can you say if $7$ is replaced with $p$ where $p$ is a prime?

Well, I see a contradiction: the order should be $7$, but that is only possible if there are only two subgroups. Isn't it impossible to have three subgroups that fit this description?

If G is cyclic of order n, then $\frac{n}{k} = 7$. But if there are only three subgroups, and one is of order 1, then 7 is the only factor of n, and $n = 7$. But then there are only two subgroups.

Is this like a trick question?

edit: nevermind. The order is $7^2$ right?

-
The cyclic group of order $7^2$ has exactly three subgroups. – Mariano Suárez-Alvarez Oct 13 '11 at 22:35
Hint $\frac{n}{k}=7$, what is your $k$? What can it be? – N. S. Oct 13 '11 at 22:35
Hint: Think about the cyclic group of order $49$. – André Nicolas Oct 13 '11 at 22:36
@MarianoSuárez-Alvarez ha yeah, I realized this as I was editing my question :) – iDontKnowBetter Oct 13 '11 at 22:37

Hint: how can a number other than $7$ not have other factors?