Let $(G, \ast)$ be a group such that it has only two subgroups. Show that $(G, \ast)$ is cyclic.

Question

Let $(G, \ast)$ be a group such that it has only two subgroups. Show that $(G, \ast)$ is cyclic.

I know this:

In group $G$, we always have two subgroups. One is $\{e\}$ and other one is $G$ itself.

Where to go from this? How to prove this?

Answer 1

Take an element $x$ distinct of $e$ it generates a subgroup which is $G$.

Answer 2

An element $a\in G$ generates a subgroup. If $a\neq e$ then what subgroup does it generate?

Answer 3

If $G=1$ then $G$ has only one subgroup which is $G$ itself.
Let $G$ be a group with only two subgroups. Then $G\neq 1$.
There exists $g\in G$ where $g\neq 1$.
Note that $\langle g \rangle \leq G$ and the only two subgroups of $G$ are $1$ and $G$.
Since $\langle g \rangle \neq 1$, we have $\langle g\rangle=G$.
By definition, $G$ is cyclic.

Answer 4

Among the subgroups of $G$ there are $\{1\}$ and $G$. If the group has only two subgroups, these are all. This is your starting point and it's good.

Now observe that $G\ne\{1\}$. Take $g\in G$, $g\ne1$. What can you say about $\langle g\rangle$?

Further exercise: show that $|G|$ is prime.