Let $r,s,t>1$ be positive integers. Must there exist a finite group $G$ with elements $x$ and $y$ such that $ord(x)=r$, $ord(y)=s$, and $ord(xy)=t$?


The answer is probably "yes." Is there a nice description of such a $G$?



Is $ord(xy)$ independent of $ord(x)$ and $ord(y)$ in a finite group?
Let $a$ and $b$ be elements of a group $G$. If $a$ has order $m$ and $b$ has
order $n$, what can we say about the order of $ab$? The next theorem shows
that we can say nothing at all.


THEOREM: For any integers $m,n,r>1$, there exists a finite group $G$ with
elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$
has order $r$.


PROOF: We shall show that, for a suitable prime power $q$, there exist elements $a$
and $b$ of $SL\_{2}(F\_{q})$ such that $a$, $b$, and $ab$ have
orders $2m$, $2n$, and $2r$ respectively. As $-I$ is the unique element of
order $2$ in $SL\_{2}(F\_{q})$, the images of $a$, $b$, $ab$ in
$SL\_{2}(F\_{q})/\{\pm I\}$ will then have orders $m$, $n$, and $r$
as required.


Let $p$ be a prime number not dividing $2mnr$. Then $p$ is a unit in the
finite ring $\mathbb{Z}/2mnr\mathbb{Z}$, and so some power of it, $q$ say,
is $1$ in the ring. This means that $2mnr$ divides $q-1$. As the group
$F\_{q}^{\times}$ has order $q-1$ and is cyclic,
there exist elements $u$, $v$, and $w$ of $F\_{q}^{\times}$ having
orders $2m$, $2n$, and $2r$ respectively.


Let
$$
a=\left(
\begin{array}{cc}
u & 1\\
0 & u^{-1}
\end{array}
\right)$$

and $$b=\left(
\begin{array}{cc}%
v & 0\\
t & v^{-1}%
\end{array}
\right)$$
(elements of $SL\_{2}(F\_{q})$), where $t$ has been chosen so that
$$
uv+t+u^{-1}v^{-1}=w+w^{-1}.
$$




The characteristic polynomial of $a$ is $(X-u)(X-u^{-1})$, and so $a$ is
similar to $diag(u,u^{-1})$. Therefore $a$ has order $2m$. Similarly $b$ has
order $2n$. The matrix
$$
ab=\left(
\begin{array}{cc}
uv+t & v^{-1}\\
u^{-1}t & u^{-1}v^{-1}%
\end{array}
\right) ,
$$
has characteristic polynomial
$$
X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,}
$$
and so $ab$ is similar to $diag(w,w^{-1})$. Therefore $ab$ has order
$2r$.



I don't know who found this beautiful proof. Apparently the
original proof of G.A. Miller is very complicated; see [MO24940](https://mathoverflow.net/questions/24913/quick-proofs-of-hard-theorems/24940#24940).


