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If $G$ is an abelian group where $a,b \in G$ have orders $m$ and $n$ respectively, I'm trying to show that there exists an element with order equal to $lcm(a,b)$. This is a verification of my proof, but there are other good answers given here.

I'm wondering why I can't make a simpler proof. First, consider the case $a=b$. Then, $lcm(m,m) = m$, so $a$ satisfies the condition. Then, considering $a \neq b$ and $m \mid n$, $b$ satisfies the condition (or $a$ does, if $n \mid m$). Now, suppose that $a \neq b$, $m \nmid n$, $n \nmid m$.

Let $k = lcm(a,b)$. Since $G$ is an abelian group, $(ab)^k = a^kb^k$. Clearly, $a^kb^k = e$, since $n \mid k$ and $m \mid k$. Now it remains to be shown that $k$ is the smallest integer such that $(ab)^k = e$.

Consider all $l < k$ and $a^lb^l$. By construction, $k$ is the least integer such that $n \mid k$ and $m \mid k$. Therefore, at least one of $a^l \neq e$ or $b^l \neq e$ is true, so $(ab)^l \neq e$. Therefore $(ab)$ has order $k$, as desired.

user1447447
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  • It is not true that the order of $ab$ is the $\lcm$ of their orders. Say $b=a^{-1}$ for example. – lulu Jan 20 '19 at 17:56
  • That case is handled though, if $b = a^{-1}$, then $m = n$. – user1447447 Jan 20 '19 at 17:57
  • Then I don't understand your claim. I thought you were proving that the order of $ab$ was the lcm of their respective orders. What exact statement are you claiming? – lulu Jan 20 '19 at 17:58
  • I'm proving that there exists an element in an abelian group that has order equal to $lcm(m,n)$. In the case that $a \neq b$ and $n \nmid m$ and $m \nmid n$, this element is $ab$ (could be wrong!). The question is whether the proof given above is valid. – user1447447 Jan 20 '19 at 18:00
  • Got it. Maybe...but the argument seems incomplete. How do I know that the fact that, say, neither $a^l,b^l$ are the identity shows that $(ab)^l $ can't be the identity? I mean...you haven't (visibly) used your assumptions anywhere, I assume they come into play in that line, no? – lulu Jan 20 '19 at 18:08
  • Hmm, yes I might be wrong, if $n \nmid l $ and $m \nmid l$, then $(ab)^l = a^rb^q$, where $r \equiv l \mod m$ and $q \equiv l \mod n$, where at least one of $r, q \neq 0$. I would need to show that $a^{-r} \neq b^{q}$. – user1447447 Jan 20 '19 at 18:19
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    I think you'll end up reconstructing one of the standard proofs. No harm in trying though! – lulu Jan 20 '19 at 18:22

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Your proof is not right. There's no reason $a^lb^l\neq 1$ for $l<k$. Let $d=\mathrm{gcd} (m, n) $. Consider the elements $a^d$ and $b$. $a^d$ has order $m/d$, which is coprime to $n$. Thus if $(a^d)^lb^l=e$, then $a^{dl}$ and $b^l$ have the same order since they are inverses. This order must divide both $m/d$ and $n$, which are coprime, hence $a^{dl} =b^l =e$, so $l$ is at least $\mathrm{lcm} (m/d, n) =\mathrm{lcm} (m, n) $. Thus $a^db$ has the desired order.

Matt Samuel
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  • Hmm, but even if the order of $a^d$ and $b$ are the same, why is it the case that the order of $a^db$ isn't less than $l$? Since $a^lb^l \neq 1$ for all $a,b \in G, l < k$, it can't be the case that the order of the product of two elements must be divisible by the orders of those elements. – user1447447 Jan 20 '19 at 20:36
  • @user But the order of a power of an element divides the order of the element. That's the fact I use. – Matt Samuel Jan 20 '19 at 20:39
  • Ah I see, that makes sense, we know that there exists such an $l$ from my proof, the order of $a^{dl}$ must divide the order of $a^d$ and $b$, so it must divide $m/d, n$, which are coprime, so it must be $1$, which implies that $a^{dl} = b^l = e$, so $n,m/d \mid l$, so $l > lcm(m,n)$. Thanks! – user1447447 Jan 20 '19 at 20:58
  • @user Sure, no problem. – Matt Samuel Jan 20 '19 at 21:00