2
$\begingroup$

[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!

$\endgroup$
1
  • 2
    $\begingroup$ 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

3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.