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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 $p$. $\Box$

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    $\begingroup$ A group of order $1$ contains no element of prime order. $\endgroup$ – bof Nov 9 '15 at 9:13
  • $\begingroup$ Its OK. but his proof shows he want to consider $G\neq 1$. $\endgroup$ – Groups Nov 9 '15 at 9:16
  • $\begingroup$ 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.. $\endgroup$ – yoyostein Nov 9 '15 at 14:27
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    $\begingroup$ @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 $|G|$. $\endgroup$ – Morgan Rodgers Nov 9 '15 at 15:26
  • $\begingroup$ @MorganRodgers oh... Yes, you are right! Serious misunderstanding on my part.. $\endgroup$ – yoyostein Nov 9 '15 at 15:31
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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.

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  • $\begingroup$ But you can't let $g^{|g|\over p}$ equal $1$. Or am I misunderstanding? $\endgroup$ – a student Nov 9 '15 at 8:57
  • $\begingroup$ No on your second line. It starts with "Let ..." $\endgroup$ – a student Nov 9 '15 at 9:02
  • $\begingroup$ 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). $\endgroup$ – Morgan Rodgers Nov 9 '15 at 9:17
  • $\begingroup$ Now corrected with details. Thanks for noticing errors. $\endgroup$ – Groups Nov 9 '15 at 9:23
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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.

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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.

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    $\begingroup$ See whether he wants to prove Cauchy's theorem or something weaker. $\endgroup$ – Groups Nov 9 '15 at 9:13
  • $\begingroup$ @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 $\frac{|g|}{p}$. $\endgroup$ – Miz Nov 9 '15 at 9:14
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    $\begingroup$ 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 $\frac{|g|}{p} < |g|$). $\endgroup$ – Morgan Rodgers Nov 9 '15 at 9:15
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    $\begingroup$ 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 $G$). $\endgroup$ – Morgan Rodgers Nov 9 '15 at 9:52
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    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Nov 9 '15 at 10:28

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