Thread:

1. 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.

2. 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 $0<q<p$. Then $b^q=(a^m)^q=a^{mq}$.
But $mq<mp=n\Rightarrow a^{mq}\neq e$.
So $ord(b)=p$.

3. 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 $0<q<p$. Then $b^q=(a^m)^q=a^{mq}$.
But $mq<mp=n\Rightarrow a^{mq}\neq e$.
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

4. 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)".
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