# Thread: If the order of a group is prime number then the group is cyclic?

1. ## If the order of a group is prime number then the group is cyclic?

Let $G$ be a group with a prime number $p$ of elements. If $a \in G$ where $a \neq e$,
then the order of $a$ is some integer $m \neq 1$.

But then the cyclic group $\langle a \rangle$ has $m$ elements.

By Lagrange's theorem, $m$ must be a factor of $p$.

But $p$ is a prime number, and therefore $m = p$.

It follows that $\langle a \rangle$ has $p$ elements, and is therefore all of $G$!

Conclusion:

If $G$ is a group with a prime number $p$ of elements, then $G$ is a cyclic group. Furthermore,
any element $a \neq e$ in $G$ is a generator of $G$.

The author said that:

"But then the cyclic group $\langle a \rangle$ has $m$ elements."

I understand that the cyclic group has $m$ elements. No problem there. I understand this statement is true for the generator $a$.

But why the group $G$ has to be definitely cyclic at the end?

Why the cyclic group $\langle a \rangle = G$?

Can't there be situations where $aaaa....a \notin G$ because we didn't say that $\langle a \rangle$ is a subgroup of $G$?

So how did he come to the conclusion that $G$ is cyclic?

2. ## Re: If the order of a group is prime number then the group is cyclic?

Don't worry I found the solution. The answer is on Pinter's Abstract Algebra book page-111:

If $G$ is any group and $a \in G$, it is easy to see that:
.......

c) the set of all the powers of $a$ is a subgroup of $G$

This subgroup is called the cyclic subgroup of $G$ generated by $a$.

So the answer to my problem in my last post:

$\text{For all element } a \in G, \langle a \rangle \text{ is a subgroup of } G$

That's it. That's all I wanted to know.