[Cross-posted from MSE]

Consider the Von Dyck group

$$ G = \langle x,y\mid x^a=y^b=(xy)^c=1\rangle $$ where $a,b,c\ge3$. Because $G$ is infinite and residually finite, it has an infinite family of finite quotients $\{\overline{G}_n\}$ such that $|\overline{G}_n|\rightarrow\infty$. I'm wondering if there is an expicit such family that can be constructed, giving a direct proof that $G$ is infinite.

Specifically, I'm trying to use Derek Holt's argument (that I copied in this MO answer), which can be used to show $G$ has quotients in $PSL(2,q)$ for infinitely many primes $q$. To summarize: Let $q$ be a prime such that $q-1$ is divisible by $2a$, $2b$, and $2c$. Let $\zeta$ be a primitive element of $\mathbb{F}_q^\ast$, and define $\zeta_t=\zeta^{(p-1)/t}$, so that $|\zeta_t|=t$. Then define the matrixes $X,Y\in SL(2,q)$ as \begin{equation*} \begin{aligned} X &= \begin{pmatrix}\zeta_{2a} & 1\\ 0 & \zeta_{2a}^{-1}\end{pmatrix} \end{aligned}\qquad \begin{aligned} Y = \begin{pmatrix}\zeta_{2b} & 0\\\lambda & \zeta_{2b}^{-1}\end{pmatrix} \end{aligned} \end{equation*} where $\lambda = (\zeta_{2c}-\zeta_{2a}\zeta_{2b})+(\zeta_{2c}^{-1}-\zeta_{2a}^{-1}\zeta_{2b}^{-1})$. Then the images $\overline{X},\overline{Y}\in PSL(2,q)$ generate a subgroup $\langle \overline{X},\overline{Y}\rangle$ that is a quotient of $G$.

It seems to me in this case we should actually have $\langle \overline{X},\overline{Y}\rangle=PSL(2,q)$. Is this true, and how hard is the proof? Equivalently (but perhaps easier to show), is it true that $\langle \overline{X},\overline{Y}\rangle$ contains a Sylow $q$-subgroup of $PSL(2,q)$?

Note: While I am mostly interested in the specific question of whether $\langle \overline{X},\overline{Y}\rangle=PSL(2,q)$, there are other ways to exhibit an infinite family $\{\overline{G}_n\}$. For example, by reducing to the (perhaps more tractable) case where $a,b,c$ are odd primes or $4$. Or if there are other explicit infinite families -- especially linear -- I'd like to hear about them!

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    $\begingroup$ If $\zeta_{2c} = \zeta_{2a}\zeta_{2b}$, then $\lambda = 0$ and $\langle X, Y \rangle$ is not irreducible, and so $\langle \overline{X},\overline{Y}\rangle$ is not $PSL(2,q)$, right? Say for example $a = b$ even and $c = a/2$. $\endgroup$ Commented Nov 10, 2023 at 3:38

1 Answer 1


Here is an old paper that gives one answer to your question about "other explicit infinite families":

G. A. Miller, Groups Defined by the Orders of Two Generators and the Order of their Product. Amer. J. Math. 24 (1902), no. 1, 96-100. JSTOR

Suppose that $2 \leq a \leq b \leq c$, and that $(a,b,c)$ is not one of $(2,2,c), (2,3,3), (2,3,4), (2,3,5)$.

By a direct calculation with permutations, Miller shows that for infinitely many $n$, there exists a transitive subgroup $G_n = \langle x,y \rangle$ of $S_n$ with $x^a = y^b = (xy)^c = 1$. Since $n \mid |G_n|$, we get $|G_n| \rightarrow \infty$ as $n \rightarrow \infty$.

Miller proceeds case-by-case, with many cases, giving explicit permutations in each case. For example, for $a = 2$, $b = 3$, $c = 6$, he defines

\begin{align*} M &= (1,2,3)(4,5,6)(7,8,9)(10,11,12) \cdots (6k-2,6k-1,6k) \\ L &= (3,4)(5,7)(6,8) (9,10)(11,13)(12,14) \cdots (6k-3,6k-2) \end{align*}

Then $ML = (1,2,4,7,6,3) \cdots$ is a product of $k$ disjoint $6$-cycles, and $\langle M,L \rangle \leq S_{6k}$ is transitive.

  • $\begingroup$ Given that one of the links in the question is to a post by the OP mentioning exactly this paper, I assume the basic premise of the question is that this is not explicit enough and/or not "direct" enough. $\endgroup$ Commented Nov 6, 2023 at 5:51
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    $\begingroup$ I interpreted the question as "does $G_{a,b,c}$ map onto ${\rm PSL}(2,q)$ for infinitely many $q$?". We know that the answer is yes unless $a$, $b$ and $c$ are all at most $5$ $\endgroup$
    – Derek Holt
    Commented Nov 6, 2023 at 9:02
  • $\begingroup$ @JoachimKönig: It might also be possible that link is referring to a different paper of Miller: "On the Product of Two Substitutions. Amer. J. Math. 22 (1900), no. 2, 185-190"; in which he proves that for $1 <a,b,c \leq n-2$ there exist $x,y \in S_n$ with $|x| = a$, $|y| = b$, $|xy| = c$. In the paper mentioned in my answer Miller constructs permutations which have $x^a = y^b = (xy)^c = 1$, but not necessarily e.g. $|x| = a$. $\endgroup$ Commented Nov 6, 2023 at 11:43
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    $\begingroup$ @MikkoKorhonen Yes, possibly. By the way, I was once naive enough to believe that I was the first to prove this theorem, before learning from your answer here mathoverflow.net/questions/118092/… that I was, at least for a short time, the last :) $\endgroup$ Commented Nov 6, 2023 at 12:33
  • $\begingroup$ The interpretation by Derek Holt is correct, but I did also mention I'm interested in other explicit families, and this is a new one to me, so +1. $\endgroup$
    – Steve D
    Commented Nov 6, 2023 at 13:34

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