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The problem: Let $a$ and $b$ be commuting elements of a group $G$. Let ord($a$) $=m$ and ord($b$) $= n$; let lcm($m,n$) denote the least common multiple of $m$ and $n$.

Prove that there is an element $c$ in $G$ whose order is lcm($m,n$). (HINT: Use the facts that (1) If $m$ and $n$ are relatively prime, then ord($ab$) $=mn$, and (2) If ord($a$) $=km$, then ord($a^k$) $=m$. Let $c=a^ib$ where $a^i$ is a certain power of $a$.)

My thoughts: $ab$ wouldn't always work, since $a$ and $b$ could just be inverses of each other. Aside from that observation, I don't think I have a clue on how to do this problem.

Asker
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  • But if b and a are inverses, their orders are the same $n=m$ and $lcm(m,n)=1$ – Javier Dec 01 '15 at 22:48
  • Good point about their orders being the same. I don't get the last part though - why would that mean $lcm(m,n)=1$? – Asker Dec 01 '15 at 22:52
  • @Javier Don't you mean $o(ab)=1$? If $n=m$,then $\operatorname{lcm}(m,n)=m$. – Tim Raczkowski Dec 01 '15 at 22:53
  • Oh, yes, my bad! – Javier Dec 01 '15 at 22:54
  • Have you tried using the hint? Can you do part (1) of the hint? – Eric Wofsey Dec 01 '15 at 22:57
  • But that is what you want with $c=a$ – Javier Dec 01 '15 at 22:57
  • @EricWofsey I've thought about separating things into two disjoint cases; one where $m$ and $n$ are relatively prime, and one where they are not. In the former case, $c=ab$. In the latter case, let the GCF of $n$ and $m$ be $z$. Let $m=gz, n=hz$. Then $lcm(m,n)=ghz$. Now maybe we can do some algebraic manipulations using (2) to get what we want. Since $ord(a)=zg$, $ord(a^z)=g$ by (2). And also, $ord(b)=hz$ by definition. Unfortunately, we can't guarantee that $g$ and $hz$ are coprime, so we can't reapply (1) to get $c=a^zb$. So, I am back at a dead end. – Asker Dec 01 '15 at 23:48

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It's not quite clear to me what proof the hint is trying to suggest, but the following minor modification of what the hint is saying should work. The idea is to reduce to the case where $m$ and $n$ are relatively prime. That is, you want to choose $i$ and $j$ such that $\operatorname{ord}(a^i)$ and $\operatorname{ord}(b^j)$ are coprime, but $\operatorname{ord}(a^i)\operatorname{ord}(b^j)=\operatorname{lcm}(m,n)$. Then by part (1) (the relatively prime case) applied to $a^i$ and $b^j$, $\operatorname{ord}(a^ib^j)=\operatorname{ord}(a^i)\operatorname{ord}(b^j)=\operatorname{lcm}(m,n)$. Can you find values of $i$ and $j$ which make this hold?

Eric Wofsey
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