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 $mn$.

Let $r$ be the order of the element $ab$.
Since we have
\begin{align*}
(ab)^{mn}&=a^{mn}b^{mn} \quad \text{ (since } G \text{ is an abelian group)}\\
&=(a^m)^n(b^n)^m\\
&=1
\end{align*}

since $a^m=1$ and $b^n=1$.
This implies that the order $r$ of $ab$ divides $mn$, that is, we have
\[ r |mn. \tag{*}\]

Now, since $r$ is the order of $ab$ we have
\[1=(ab)^r=a^rb^r.\]
Then we have
\begin{align*}
1=1^n=a^{rn}b^{rn}=a^{rn}
\end{align*}

since $b^n=1$. This yields that the order $m$ of the element $a$ divides $rn$.

Since $m$ and $n$ are relatively prime, this implies that we have
\[m|r.\]

Similarly (switch the role of $n$ and $m$), we obtain
\[n|r.\]
Thus we have
\[mn|r \tag{**}\]
since $m$ and $n$ are relatively prime.

From (*) and (**), we have $r=mn$, and hence the order of the element $ab$ is $mn$.

Related Question.

As a generalization of this problem, try the following problem.

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 $n$.

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[…] Recall that if the orders $m, n$ of elements $a, b$ of an abelian group are relatively prime, then the order of the product $ab$ is $mn$. (For a proof, see the post “Order of the Product of Two Elements in an Abelian Group“.) […]

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[…] the post “Order of the product of two elements in an abelian group” for a similar problem about the order of elements in abelian […]

[…] Recall that if the orders $m, n$ of elements $a, b$ of an abelian group are relatively prime, then the order of the product $ab$ is $mn$. (For a proof, see the post “Order of the Product of Two Elements in an Abelian Group“.) […]