1 Attachment(s)

Finite group with element of prime order

I have this problem due this Thursday, and I used a solution on here to help me figure it out. The problem is this: "Let G be a non-trivial finite group. Prove that G contains an element of prime order." I have solved the proof, but have a little trouble understanding the explanation from the solution I found, which my professor noticed. His comment towards my solution was this: "In the sentence beginning "Then, suppose (there exists) q in Z ...", what are you trying to say? Of course if q<p then mq<mp, but what does that matter? What would that contradict?" I understand that this new finding contradicts the definition of the order of the element, but I don't know how to further explain this in my proof. Any help would be appreciated as soon as possible. Thanks!

Re: Finite group with element of prime order

I'm not sure what you meant to do. You coulda have just used the fundamental theorem of arithmetic.

If n is a prime, it has a unique prime decomopsition n = $\displaystyle p_1..p_n$ $\displaystyle a^{n} = a^{{p_i...p_n}^{p_1}} = e$, since each $\displaystyle p_i..p_n \leq n$, $\displaystyle a^{p_i...p_n}$ is indeed an element of prime order $\displaystyle p_1$