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

**x3bnm** · February 3rd 2013, 11:21 AM

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?

**x3bnm** · February 3rd 2013, 12:39 PM

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.

**Deveno** · February 4th 2013, 05:07 AM

to answer your earlier question as asked:

no, for any element a of G, any power of a is ALSO in G, by closure.

**x3bnm** · February 4th 2013, 06:43 AM

Originally Posted by Deveno

to answer your earlier question as asked:

no, for any element a of G, any power of a is ALSO in G, by closure.

Thanks Deveno.

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