# What are smallest finite images of triangle groups?

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.)

• 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. Sep 6 '14 at 15:41
• 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. Sep 6 '14 at 16:07
• 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. Sep 6 '14 at 16:21
• 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/… Sep 7 '14 at 4:37
• 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. Sep 10 '14 at 5:48

## 1 Answer

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 $G.$

• 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. Sep 6 '14 at 23:27
• @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. Sep 7 '14 at 7:34