# Proof that finite group contains an element of prime order

I tried to prove the following claim but it seemed a bit too easy:

Let $G$ be a finite group. Then $G$ contains an element of prime order.

Please could someone tell me if my proof is correct or if I'm missing something?

Let $g$ be any non-identity element of $G$. Let $p$ be any prime factor of $|g|$. Then $g^{|g|\over p}$ has prime order $p$. $\Box$

• A group of order $1$ contains no element of prime order. – bof Nov 9 '15 at 9:13
• Its OK. but his proof shows he want to consider $G\neq 1$. – Groups Nov 9 '15 at 9:16
• Consider $G=\mathbb{Z}_6$, the additive cyclic group. Choose $p=3$, $g=3$, the $g^{(6/3)}=3+3=0$, with order 1! I believe this is a counter example to the proof. The theorem you are trying to proof is correct, but the proof is not.. – yoyostein Nov 9 '15 at 14:27
• @yoyostein You don't get to choose $p$ at the beginning. You take an element, for example $g = 3$. Then you choose $p$ that divides $|g| = 2$, so your only choice is $p=2$. Now $3^{2\over2} = 3$, which has prime order $2$. Notice that the prime is chosen to divide $|g|$, not $|G|$. – Morgan Rodgers Nov 9 '15 at 15:26
• @MorganRodgers oh... Yes, you are right! Serious misunderstanding on my part.. – yoyostein Nov 9 '15 at 15:31

Yes. This is right. To ensure it:

If $|g|$ is a prime, then we are done. Let $p$ be a prime factor of $|g|$, so that $|g|>p$.

Let $h=g^{\frac{|g|}{p}}$. Then $h^p=g^{|g|}=1$, which implies that order of $h$ divides $p$, so it is $1$ or $p$.

If order of $h$ is $1$ then this means, $h=1$ i.e. $g^{|g|/p}=1$, i.e. $|g|\leq |g|/p$, this is contradiction.

• But you can't let $g^{|g|\over p}$ equal $1$. Or am I misunderstanding? – a student Nov 9 '15 at 8:57
• No on your second line. It starts with "Let ..." – a student Nov 9 '15 at 9:02
• What is the purpose of your last sentence? You cannot have the order of $h$ equal to $1$ with your setup, since $g$ is chosen to not be the identity (that's the reason for my downvote). – Morgan Rodgers Nov 9 '15 at 9:17
• Now corrected with details. Thanks for noticing errors. – Groups Nov 9 '15 at 9:23

Your answer is correct, with one small adjustment. You need the assumption that $|G| > 1$, so that there is in fact a nonidentity element $g$ to select. Other than that, your proof is good; it is in fact an easy one-line proof once you see the key.

No the proof is not correct. It is possible that $g^{\frac{|g|}{p}}=e$. The theorem you are stating is commonly called Cauchy's theorem. There is an elementary proof here using the class equation.

• See whether he wants to prove Cauchy's theorem or something weaker. – Groups Nov 9 '15 at 9:13
• @Groups Your answer above is flawed. $h^p=1$ does not mean that the prder of $h$ is $p$ or $1$ it can be any divisor of $\frac{|g|}{p}$. – Miz Nov 9 '15 at 9:14
• If $g$ has order $|g|$, then $|g|$ is the smallest power of $g$ that is equal to $e$. Therefore $g^{\frac{|g|}{p}} \neq e$ (because $\frac{|g|}{p} < |g|$). – Morgan Rodgers Nov 9 '15 at 9:15
• The linked proof shows it for every $p$ dividing $|G|$, which is Cauchy's theorem. The OP only needs to show there is a single element having prime order (any prime will do; in their proof, $p$ is an arbitrary prime dividing $|g|$, where $g$ is an arbitrary element of $G$). – Morgan Rodgers Nov 9 '15 at 9:52
• The proof in the question is correct (well, with a small adjustment in case $g$ has prime order itself), and the exercise is a way weaker statement than Cauchy's theorem. – Tobias Kildetoft Nov 9 '15 at 10:28