Let $G$ be a group. Let $a$ and $b$ be elements of $G$.
If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.

We claim that it is not true. As a counterexample, consider $G=S_3$, the symmetric group of three letters.
Let $a=(1\, 2), b=(1 \,3)$ be transposition elements in $S_3$.
The orders of $a$ and $b$ are both $2$.

Consider the product
\[ab=(1\, 2)(1 \,3)=(1 \, 3 \, 2).\]
Then it is straightforward to check that the order of $ab$ is $3$, which does not divide $4$ (the product of orders of $a$ and $b$).

Therefore, the group $G=S_3$ and elements $a=(1\, 2), b=(1 \,3)\in G$ serve as a counterexample.

Remark. (Abelian group case)

If we further assume that $G$ is an abelian group, then the statement is true.
Here is the proof if $G$ is abelian.

Let $e$ be the identity element of $G$.
\begin{align*}
(ab)^{mn} &=a^{mn}b^{mn} && \text{ since $G$ is abelian}\\
&=(a^m)^n(b^n)^m\\
&=e^n e^m && \text{since the order of $a, b$ are $m, n$ respectively}\\
&=e.
\end{align*}
Thus the order of $ab$ divides $mn$.

Related Question.

If the group is abelian, then the statement is true.

Problem. Let $G$ be an abelian group with the identity element $1$.
Let $a, b$ be elements of $G$ with order $m$ and $n$, respectively.
If $m$ and $n$ are relatively prime, then show that the order of the element $ab$ is $mn$.

Problem.Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively.
Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.

Also determine whether the statement is true if $G$ is a non-abelian group.

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