# Finite group with element of prime order

• September 7th 2007, 05:38 PM
Finite group with element of prime order
Let G be a finite group with more than one element. Show that G has an element of prime order.
• September 7th 2007, 11:49 PM
red_dog
Let $a\in G, \ ord(a)=n$.
If $n$ is prime, we've done.
Else, let $p$ be a prime divisor of $n$.
Then $a^n=a^{pm}=(a^m)^p=e$.
Let $b=a^m\in G\Rightarrow b^p=e$.
Let $0. Then $b^q=(a^m)^q=a^{mq}$.
But $mq.
So $ord(b)=p$.
• September 8th 2007, 04:26 AM
topsquark
Quote:

Originally Posted by red_dog
Let $a\in G, \ ord(a)=n$.
If $n$ is prime, we've done.
Else, let $p$ be a prime divisor of $n$.
Then $a^n=a^{pm}=(a^m)^p=e$.
Let $b=a^m\in G\Rightarrow b^p=e$.
Let $0. Then $b^q=(a^m)^q=a^{mq}$.
But $mq.
So $ord(b)=p$.

If I am reading this right, then, the only case left to show is to prove that no group can exist where the order of all of its elements is 1. (I admit this is practically trivial, but it wasn't addressed in the above proof.)

-Dan
• September 8th 2007, 05:51 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
If I am reading this right, then, the only case left to show is to prove that no group can exist where the order of all of its elements is 1. (I admit this is practically trivial, but it wasn't addressed in the above proof.)

-Dan

The only element of order 1 is the identity. He said "with more than 1 element (identity)".