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...

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    $\begingroup$ Solved in ams.org/mathscinet-getitem?mr=1505882 $\endgroup$ – Jack Schmidt Jul 31 '13 at 21:53
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    $\begingroup$ math.stackexchange.com/questions/295740/… $\endgroup$ – Will Jagy Jul 31 '13 at 21:56
  • $\begingroup$ A quick glance at the article suggestst it indeed isn't as basic as was suggested in the question at the time. I'll look into it some more tomorrow. Thanks, also for the related MSE reference, that's basically about the failure of attempt 1! $\endgroup$ – HSN Jul 31 '13 at 22:06
  • $\begingroup$ I just put finite groups (maybe i put finite group) into the search window and asked for sort by date. You have a better chance of finding the question than I do, because when you see the correct question title it will jog your memory. Oh, once found, we can see previous versions under the edit history. People usually leave a question title alone unless it is really badly worded. $\endgroup$ – Will Jagy Jul 31 '13 at 22:11
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    $\begingroup$ mathoverflow.net/questions/118035/order-of-elements $\endgroup$ – Beginner Aug 1 '13 at 5:21

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 $S_o$).

  • $\begingroup$ I think you forgot to conclude... I am not sure what you were trying to say, but you have proven that there are infinitely many finite groups for each triple $a^{-1}+b^{-1}+c^{-1}\leq 1$. This is because if $N$ is the kernel of the map then $N$ contains a finite-index proper normal subgroup, $M$ say, and then $G/M$ has the required properties but has order strictly greater than $G/N$. I do not know if this was what you were trying to say, but it is rather nice all the same... $\endgroup$ – user1729 Aug 2 '13 at 9:44
  • $\begingroup$ @user1729: The question was "given a triple $(a,b,c)$ find a finite group with elements of such orders. The argument gives this group in the form of the group $F$. $\endgroup$ – Moishe Kohan Aug 2 '13 at 14:14
  • $\begingroup$ Ah, right, sorry, forgot precisely what the question was! $\endgroup$ – user1729 Aug 2 '13 at 14:18

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