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 $G$?
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 $G$? 


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 $q1$. As the group $F_{q}^{\times}$ has order $q1$ 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 $(Xu)(Xu^{1})$, 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 $$ X^{2}(uv+t+u^{1}v^{1})X+1=(Xw)(Xw^{1})\text{,} $$ 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. 


Here's my comment as an answer: Take the $r,s,t$(ordinary) triangle group $T(r,s,t)=\langle x,y \  \ x^r=y^s=(xy)^t = 1 \rangle$, in which $x$, $y$, and $xy$ have the correct orders. See the section on ``von Dyck" groups here. As Anton mentions in his comment, $T(r,s,t)$ is infinite when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} \leq 1$. However, $T(r,s,t)$ is residually finite. The easiest way to see this is to use the facts that finitely generated linear groups are residually finite (due to Malcev, as Steve D mentions), and the fact that $T(r,s,t)$ is linear. To see that $T(r,s,t)$ is linear, note that when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = 1$, it is a discrete subgroup of the affine group $\mathbb{R}^2 \rtimes \mathrm{SL}_2(\mathbb{R})$, and when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} < 1$, it is a discrete subgroup of $\mathrm{Isom}^+(\mathbb{H}^2) \cong \mathrm{PSL}_2(\mathbb{R}) \cong \mathrm{SO}_0(2,1)$, where $\mathrm{SO}_0(2,1)$ is the identity component of $\mathrm{SO}(2,1)$. See this again. Now, since $T(r,s,t)$ is residually finite, there is a quotient $G$ in which $$x, x^2, \ldots, x^{r1}, y, y^2, \ldots, y^{s1}, (xy), (xy)^2, \ldots, (xy)^{t1}$$ are all nontrivial. This is the $G$ you seek. Also see Steve D's answer here. 

