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If $G$ has an element of order $p$ and an element of order $q$, where $p$ and $q$ are distinct primes, then the order of $G$ is a multiple of $pq$

Here is how I am working out my proof:

Suppose $x,y \in G$ and let $|x|=p$ and $|y|=q$ where $p$ and $q$ are distinct primes.

I am having trouble wording it and putting it together.

So the next thing, I wanna say is:

By Lagrange's theorem the order of $G$ is a multiple of $p$ and a multiple of $q$. Therefore, $G$ must be a multiple of $pq$.

However, it feels quite empty and missing something.

kero
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    You need to use that they are prime. For example if $|x|=6$ and $|y|=3$, then it may not be true that the order of $G$ is a multiple of $18$. – J126 Apr 02 '15 at 00:48
  • @JoeJohnson126, by the way this is just a random example that doesn't relate to the problem since 6 isn't prime. – kero Apr 02 '15 at 02:27
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    It's a counterexample in the case that $|x|$ and $|y|$ are not relatively prime. But, your argument, as stated, would apply to the situation where they aren't relatively prime. Thus your argument is lacking some detail. – J126 Apr 02 '15 at 10:08
  • @JoeJohnson126. Great thanks! – kero Apr 05 '15 at 04:22

1 Answers1

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You're correct, but you could add a little more detail, as below.

More generally, if $G$ has an element of order $m$ and an element of order $n$, then by Lagrange's theorem the order of $G$ is a multiple of both $m$ and $n$ and so is a multiple of $lcm(m,n)$. In particular, if $m$ and $n$ are coprime, then $lcm(m,n)=mn$, and the order of $G$ is a multiple of $mn$.

If $p$ and $q$ are distinct primes, then they are coprime and the result above holds.

lhf
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