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Given integers $a, b, c$ all at least 2, I would like to identify the smallest group with an element $x$ of order $a$, element $y$ of order $b$ such that the product $yx$ has order $c$. For example, if $(a,b,c)=(2,3,7)$ the smallest group with elements $x, y, yx$ of orders $2,3,7$, respectively, is the simple group of order 168.

This problem is related to triangle groups but here I want to concentrate on finite groups, especially the smallest such.

This question arose in a discussion on permutation groups in a graduate algebra class. One can always create finite permutations $x, y$ with the desired properties but looking for small groups with these properties seemed to quickly become complicated. (This question may overlap with this MathOverflow question.)

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    $\begingroup$ As you no doubt know, $(2,3,4)$ is $S_{4}, (2,3,3)$ is $A_{4}$,$(2,3,5)$ is $A_{5}.$ Note that if $a,b,c$ are pairwise coprime, then no non-identity finite homomorphic image of $(a,b,c)$ is solvable. $\endgroup$ – Geoff Robinson Sep 6 '14 at 15:41
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    $\begingroup$ There is computational evidence that there always is a group with elements of order $a$ and $b$ whose product has order $c$ which embeds into the symmetric group of degree $\max(a,b,c)+2$, but nobody found a proof so far -- cf. mathoverflow.net/questions/118092. $\endgroup$ – Stefan Kohl Sep 6 '14 at 16:07
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    $\begingroup$ I think this question is very difficut - I don't believe that there have even been any reasonable upper bounds proved for general $a,b,c$. For specific $a,b,c$ that are not too big, it can probably be answered computationally. $\endgroup$ – Derek Holt Sep 6 '14 at 16:21
  • $\begingroup$ For simple quotients of Lie type, a criterion has been given by Claude Marion: degruyter.com/view/j/jgth.2010.13.issue-5/jgt.2010.014/… degruyter.com/view/j/jgth.2009.12.issue-5/jgt.2009.004/… $\endgroup$ – Ian Agol Sep 7 '14 at 4:37
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    $\begingroup$ Rather than using the SmallGroup library, you might want to use the lowindexnormalsubgroup command in magma on the triangle groups, you'll probably be able to get much further. $\endgroup$ – verret Sep 10 '14 at 5:48
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Here's an elaboration of my comment above in the pairwise coprime case, in an even more special case (the fact that in a finite solvable group we can't have three elements of pairwise coprime orders whose product is the identity is, I believe, due to P. Hall). If one insists, as in the question, on the minimality of the order, then if $a,b,c$ are distinct primes, any finite group $G$ of minimal order subject to containing elements $x,y,z$ of respective orders $a,b,c$ such that $xyz = 1_{G}$ (equivalently, $xy = z^{-1}$), is necessarily a finite simple group.

For let $N$ be a proper non-identity normal subgroup of $G.$ Then if none of $x,y,z$ is in $N,$ the minimality of $G$ is contradicted, as $xN yN zN = N$ in $G/N.$ If two of $x,y,z$ are in $N,$ then so is the third, and the minimality of $G$ is contradicted again.If (say) $x \in N,$ but neither $y$ nor $z$ lies in $N,$ we have a contradiction since $yNzN = N$ in $G/N$, but $yN$ and $zN$ have coprime orders. Hence there is no such normal subgroup $N$ and $G$ is simple.

Later edit: Here is a general argument which may be relevant, and admits various generalizations: Let $a>5$ be an odd prime, $b>1$ be an odd divisor of $a-1$ and $c>1$ be an odd divisor of $a+1$ (note that $a$ is then neither a Fermat nor a Mersenne prime). Then $G = {\rm PSL}(2,a)$ (which is isomorphic to a subgroup of $S_{a+1}) $ contains three elements $x,y,z$ of respective orders $a,b,c$ with $xyz = 1.$ This is because if $u$ is an element of order $a,v$ an element of order $b$ and $w$ an element of order $c$ in $G,$ then each non-trivial irreducible character $\chi$ of $G$ vanishes at one of $u,v,w.$ Hence $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi(u) \chi(v) \chi(w)}{\chi(1)}> 0.$ By a standard character-theoretic formula,this means that we may choose conjugates $x,y,z$ of $u,v,w$ respectively such that $xy = w^{-1},$ or $xyw = 1.$ It is also easy to check that $G = \langle x,y \rangle$ for any such $x,y$, since $\langle x,y \rangle$ must have more that one Sylow $a$-subgroup, so has $a+1$ Sylow $a$-subgroups and hence is all of $G.$

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  • $\begingroup$ It is worth noting that there are finite simple groups whose simplicity cannot be proved this way. Since the $(2,3,5)$ triangle group is isomorphic to $A_{5}$, $A_{6}$ and $PSU_{4}(2)$ cannot be proven simple this way. I am not sure if there are any other examples. $\endgroup$ – DavidLHarden Sep 6 '14 at 23:27
  • $\begingroup$ @DavidLHarden: Yes indeed. My point was just that in seeking the smallest such group, it might be useful to know that in several cases, the group in question is necessarily simple. $\endgroup$ – Geoff Robinson Sep 7 '14 at 7:34

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