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I'm trying to solve this Putnam problem. The problem is "show that for a finite group with $n$ elements of order $p$, where $p$ is prime, either $n=0$ or $p\: \vert\: n+1$."

I'm trying to do this by showing we can create a subgroup of all elements such that $a^p=e$ (assuming $n\neq 0$, this group will have $n+1$ elements; all of our elements of order $p$ plus the identity). Then by Lagrange's Theorem, the order of an element divides the order of the group, so $p\:|\:n+1$.

I'm having trouble showing the closure of this subgroup though. I know that for a finite abelian group, $|ab|$ divides lcm$(|a|,|b|)$. If I can show this for a general group, then I can show that $|ab|=p$ because $p$ is prime. I'm having trouble doing this though. Anybody have any ideas? Or maybe I'm going about this problem completely wrong.

edit: never mind

Alex Mathers
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|ab| does not in general divide lcm(|a|,|b|). See the earlier post, with a link to a nice proof. Order of product of two elements in a group

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matt biesecker
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  • so there aren't any special circumstances when $|a|=|b|=p$ where $p$ is prime? – Alex Mathers Apr 23 '15 at 18:39
  • The statement of theorem is: "For any integers $m;n; r > 1,$ there exists a finite group G with elements $a$ and $b$ such that $|a|=m, |b|=n,$ and $|ab|=r.$" So i'ts possible to construct a non-abelian group with elements $a,b$ of order $p$, yet $|ab|$ not divisible by $p.$ – matt biesecker Apr 23 '15 at 19:33