# Show order of product of two elements of a finite group is product of orders

Let $G$ be a finite abelian group and $x,y \in G$ so, such that the order of $x$ and $y$ is coprime. Then the order of the element $xy$ is the product of the orders of $x$ and $y$.

Thoughts. Let $p$ be the order of $x$ and $q$ be the order of $y$. Then all the elements of $\langle xy \rangle$ have the form $x^my^n$ for $m,n \in \mathbb{N}$. Obviously, $(xy)^{pq} = 1$. Thus the order of $xy$ divides $pq$. I hope this is right. Now I would like to use the coprimeness of $p$ and $q$ which means $\mathrm{lcm}(p,q) = pq$. Does from there the statement follow?

• In fact, all elements of $\langle xy \rangle$ have the form $(xy)^n = x^ny^n$ – Omnomnomnom Oct 10 '16 at 14:32

Put $o(x)=m$, $o(y)=n$ and $o(xy)=k$. Since $G$ is abelian, we have $(xy)^{mn}=x^{mn}y^{mn}=(x^m)^{n}(y^n)^m=1$. Hence $k \mid mn$. Since $gcd(m,n)=1$, the subgroups $\langle x \rangle$ and $\langle y \rangle$ have trivial intersection (if $g \in \langle x \rangle \cap \langle y \rangle$, then $o(g)$ divides both $m$ and $n$, so $g=1$). Now assume that $o(xy)=k$, then $x^ky^k=1$, whence $x^k=y^{-k} \in \langle x \rangle \cap \langle y \rangle$. So $x^k=1=y^{-k}$,which means that $m \mid k$ and also since $y^k=1$, $n \mid k$. But $m$ and $n$ are relatively prime, so also $mn=lcm(m,n) \mid k$, and you are done.
Note that since $p$ and $q$ are relatively prime, there exists a solution $n$ to the equation $$\begin{cases} n \equiv 0 \pmod p\\ n \equiv 1 \pmod q \end{cases}$$ so that $(xy)^n = y$. Similarly, there is an $n$ such that $(xy)^n = x$. Since $\langle xy \rangle$ contains elements of order $p$ and $q$, its order must be at least $\operatorname{lcm}(p,q)$.
In particular: if $ap + bq = 1$ for integers $a$ and $b$, then we may take $n = ap$, then $n = bq$.
• Um, pretty sure you need to justify that. How about $(xy)^n=e$ if and only if $x^n=y^{-n}$? Then say something about $\langle x \rangle \cap \langle y \rangle$. – Dustan Levenstein Oct 10 '16 at 14:37