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

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

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

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

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

Conclusion:

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

The author said that:

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

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

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

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

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

So how did he come to the conclusion that $\displaystyle G$ 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 $\displaystyle G$ is any group and $\displaystyle a \in G$, it is easy to see that:
.......

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

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

So the answer to my problem in my last post:

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

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