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.
Best Answer
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.