Finite group with elements of given order A few weeks ago, there was a queston on MSE that got edited, as soon as the question was answered well enough, according to OP. In the end, I think this question got reversed to its original question and all later questions were just forgotten. It is the last question I've seen, however, that interests me and I seem to be unable to find it, so I guess it hasn't been asked before or since.
The question was: Given $a,b,c\in\mathbb{N}$, find a finite group $G$ containing elements $x$ and $y$, such that $x$ has order $a$, $y$ has order $b$ and $xy$ has order $c$.
This question was supposed to be from a course on basic group theory, but I can't find an easy answer, nonetheless. The question therefore is what I'm asking as well. What I tried/noted so far: The group $G$ of course won't be abelian, in general. What I'd deem the two standard approaches didn't get me anywhere:
1) Try it via group presentations. Naïve attempt: $G=\langle x,y,z\mid x^a=1, y^b=1, z^c=1, xy=z\rangle$. This nicely satisfies the given requirements (it is made to), but it seems to fail to be finite. At least, I don't see how $x^2y=xz$ is going to be of finite order, unless $a=2$. This might be a problem, if $G$ has to be finite.
2) Some sort of twisted version of $C_a\times C_b\times C_c$, where $C_n$ is the cyclic group of order $n$ and where the product of generators of two cyclic groups is sent to the generator of the third. This again satisfies the requirements on order (where $x$ and $y$ are the generators of $C_a$ and $C_b$), but this time it fails to be well-defined. We already run into trouble for fairly small $a,b,c$: take $a=2, b=c=3$, call the identity elements $1$ and the generators of $C_a, C_b, C_c$ resp. $r,s,t$. Then $(1,s^2,1)(1,1,t^2) = (1,s,1)(r,1,1)(1,1,t)$ and if we work out the two left-hand terms first, this would be $(1,1,t^2)$, whereas it would be $(1,s^2,1)$ if the right-hand terms were multiplied first.
I still have the feeling there will be a group of order $abc$ that fulfills the requirements. Possibly even just a slightly altered version of one of the above. Or something really stupid I'm missing...
 A: As suggested above, start with the von Dyck group 
$$
G=V(a,b,c)=<X,Y,Z| XY=Z, X^a=1, Y^b=1, Z^c=1>.
$$ 
Such group is either finite (if $a^{-1}+ b^{-1}+ c^{-1}>1$) or embeds in the isometry group of Euclidean plane (if $a^{-1}+ b^{-1}+ c^{-1}=1$) or of the hyperbolic plane $SO(2,1)$ (if $a^{-1}+ b^{-1}+ c^{-1}<1$). In the first case you are done, so consider the remaining two. In both cases, the group $G$ is a finitely-generated matrix group. Therefore, it is residually finite by Malcev's theorem (1940), see e.g. arXiv:1306.2385 for a nice self-contain proof. Residual finiteness means that for every finite subset $S\subset G \setminus 1$, there exists a homomorphism $f: G\to F$, with $F$ finite and $f(S)$ disjoint from 1. You want to ensure that $x=f(X), y=f(Y), z=f(Z)$ have orders $a, b, c$ respectively. To get this, take the subset $S_o\subset G$ to consist of elements $X, X^2,..., X^{a-1}$, $Y, Y^2, ..., Y^{b-1}$, $Z, Z^2, ..., Z^{c-1}$. Now, apply Malcev's theorem. You obtain the finite group $F$ containing elements $x, y, z$ with $xy=z$ so that the orders of these elements are $a, b, c$ respectively. As observed in the comments, we also obtain infinitely many such groups provided that $a^{-1}+ b^{-1}+ c^{-1}\le 1$, as the group $G$ is infinite in this case (and, hence, we have infinitely many finite subsets $S$ to work with, provided that they all contain $S_o$).  
