Generating pairs of projective special linear groups that fail to lift

TL;DR

This paper constructs explicit examples of pairs of groups, specifically free products of cyclic groups and finite projective special linear groups, that demonstrate the failure of a lifting property originally posed by Neumann.

Contribution

The authors introduce a new approach to produce infinitely many negative examples to Neumann's problem, including cases with infinite groups, using free products and finite linear groups.

Findings

Constructed explicit pairs of groups $(G,H)$ with $G$ as a free product of cyclic groups and $H$ as $ ext{PSL}(2,p)$.

Provided infinitely many negative examples for the case $n=2$.

Extended examples to include cases where $H$ is infinite.

Abstract

The following problem was originally posed by B.H. Neumann and H. Neumann. Suppose that a group $G$ can be generated by $n$ elements and that $H$ is a homomorphic image of $G$. Does there exist, for every generating $n$-tuple $(h_1,\ldots, h_n)$ of $H$, a homomorphism $\vartheta \colon G \to H$ and a generating $n$-tuple $(g_1,\ldots,g_n)$ of $G$ such that $(g_1^\vartheta,\ldots,g_n^\vartheta) = (h_1,\ldots,h_n)$? M.J. Dunwoody gave a negative answer to this question, by means of a carefully engineered construction of an explicit pair of soluble groups. Via a new approach we produce, for $n = 2$, infinitely many pairs of groups $(G,H)$ that are negative examples to the Neumanns' problem. These new examples are easily described: $G$ is a free product of two suitable finite cyclic groups, such as $C_2 \ast C_3$, and $H$ is a suitable finite projective special linear group, such as $\mathrm{PSL}(2,p)$ for a prime $p \ge 5$. A small modification yields the first negative examples $(G,H)$ with $H$ infinite.

Equations

$$p\mid m\vee \gcd (m,(q+1)/2)\ge 3\vee \gcd (m,(q-1)/2)\ge 3;$$

$$m=p\vee m\mid \frac{q+1}{\gcd (2,q-1)}\vee m\mid \frac{q-1}{\gcd (2,q-1)}.$$

$${\Gamma }_G=\{(g_1,g_2)\in G\times G\mid ⟨g_1,g_2⟩=G\}$$

$$\Phi :{\Gamma }_G$$

$$(g_1,g_2)$$

$${\alpha }_1:F\to F,$$

$${\alpha }_2:F\to F,$$

$${\alpha }_3:F\to F,$$

$$(g_1,g_2{)}^{{\alpha }_1}=(g_1^{-1},g_2),(g_1,g_2{)}^{{\alpha }_2}=(g_1{g}_2^{-1},g_2),(g_1,g_2{)}^{{\alpha }_3}=(g_2,g_1).$$

$$((g_1,g_2{)}^{\alpha }){\vartheta }^{\ast }=((g_1,g_2){\vartheta }^{\ast }{)}^{\alpha }.$$

$$c(\Omega )=\{[h_1,h_2]\mid (h_1,h_2)\in \Omega \}.$$

$$[x,y]=x^{-1}[x^{-1},y{]}^{-1}x=y^{-1}[x{y}^{-1},y]y=[y,x{]}^{-1}$$

$$c(\Omega )\subseteq \{[h_1,h_2{]}^g\mid g\in H\}\cup \{[h_2,h_1{]}^g\mid g\in H\}.$$

$$[[h_1,h_2]]=[\pm R(h_1),\pm R(h_2)]\in \mathrm{SL}(2,q)$$

$$\tau (h_1,h_2)=\mathrm{tr}([[h_1,h_2]])$$

$$\mathcal{T}(H)=\{\begin{array}{ll}{\mathbb{F}}_q\setminus \{2\}& \text{if q∈{2,4,8} or q≥13,}\\ {\mathbb{F}}_q\setminus \{1,2\}& \text{if q∈{3,9,11},}\\ {\mathbb{F}}_q\setminus \{0,2,4\}=\{1,3\}& \text{if q=5,}\\ {\mathbb{F}}_q\setminus \{0,1,2\}=\{3,4,5,6\}& \text{if q=7.}\end{array}$$

$$[A,B]=\left(\begin{array}{cc}-1& x\\ 0& -1\end{array}\right),\text{where x∈Fq.}$$

$$A=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)\text{for suitable a,b,c∈Fq.}$$

$$B^{-1}AB=A[A,B]=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)\left(\begin{array}{cc}-1& x\\ 0& -1\end{array}\right)=\left(\begin{array}{cc}-a& xa-b\\ -c& xc+a\end{array}\right)$$

$$A=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right).$$

$$B=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right),\text{where a,b,c,d∈Fp.}$$

$$[A,B]=\left[\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\right]=\left(\begin{array}{cc}{a}^2+c^2& ab+cd\\ ab+cd& b^2+d^2\end{array}\right).$$

$$\mathrm{tr}([A,B])=a^2+b^2+c^2+d^2=(a+d{)}^2+(b-c{)}^2-2=s^2+2.$$

$$[(123),(145)]=(321)(541)(123)(145)=(142).\text{qed}$$

$$t:x\mapsto -\frac{1}{x},$$

$$u_0:x\mapsto 2x,$$

$$v_1=u^{w_1},\text{ where }{w}_1:x\mapsto \frac{3x+4}{x+4},$$

$$v_2=u^{w_2},\text{ where }{w}_2:x\mapsto \frac{6x+3}{x+3},$$

$$v_3=u^{w_3},\text{ where }{w}_3:x\mapsto \frac{5x+2}{x+2},$$

$$[u,v_1]$$

$$[u,v_2]$$

$$[u,v_3]$$

$$A=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),B=\left(\begin{array}{cc}0& 3\\ 3& 0\end{array}\right),C=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right),D=C^2=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right).$$

$$[A,C]=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right),[A,D]=\left(\begin{array}{cc}1& 2\\ 2& 0\end{array}\right),[B,D]=\left(\begin{array}{cc}1& 2\\ 3& 2\end{array}\right)$$

Full text

M. J. Dunwoody gave a negative answer to this question, by means of a carefully engineered construction of an explicit pair of soluble groups. Via a new approach we produce, for $n=2$, infinitely many pairs of groups $(G,H)$ that are negative examples to the Neumanns’ problem. These new examples are easily described: $G$ is a free product of two suitable finite cyclic groups, such as $C_{2}\ast C_{3}$, and $H$ is a suitable finite projective special linear group, such as $\operatorname{PSL}(2,p)$ for a prime $p\geq 5$. A small modification yields the first negative examples $(G,H)$ with $H$ infinite.

The third author acknowledges the support from the Alexander von Humboldt Foundation and from the Forscher-Alumni-Programm of the Heinrich-Heine-Universität Düsseldorf (HHU); she thanks HHU for its hospitality.

1. Introduction

The following question about lifting finite generating tuples along homomorphisms of groups was originally raised by B. H. Neumann and H. Neumann [14].

Neumanns’ Problem**.**

Suppose that $G$ is a group which can be generated by $n$ elements and let $H$ be a homomorphic image of $G$. Does there exist, for every generating $n$-tuple $(h_{1},\ldots,h_{n})$ of $H$, a homomorphism $\vartheta\colon G\to H$ and a generating $n$-tuple $(g_{1},\ldots,g_{n})$ of $G$ such that $(g_{1}^{\,\vartheta},\ldots,g_{n}^{\,\vartheta})=(h_{1},\ldots,h_{n})$?

By means of an ingenious counting trick, W. Gaschütz showed in [4] that under an additional finiteness condition the answer is yes; indeed, a significantly stronger result holds.

Gaschütz’ Lemma**.**

Let $G$ be a group which can be generated by $n$ elements and let $\vartheta\colon G\to H$ be a surjective homomorphism with finite kernel. Then, for every generating $n$-tuple $(h_{1},\ldots,h_{n})$ of $H$, there exists a generating $n$-tuple $(g_{1},\ldots,g_{n})$ of $G$ such that $(g_{1}^{\,\vartheta},\ldots,g_{n}^{\,\vartheta})=(h_{1},\ldots,h_{n})$.

A profinite version of Gaschütz’ Lemma has manifold applications in the theory of profinite groups; compare [9, §2]. For a recent generalisation to metrisable compact groups see [2].

However, in general one cannot expect that arbitrary generating tuples will lift along a single, fixed epimorphism. For instance, take $\vartheta\colon C_{\infty}\to C_{5}$, a homomorphism from an infinite cyclic group onto a group of order $5$. Then, of course, only two of the four generators of $C_{5}$ lift along $\vartheta$ to generators of $C_{\infty}$. As Gaschütz pointed out, this simple example illustrates the inherent limitation in his theorem, but falls short of settling the Neumanns’ Problem (in the negative).

Eventually M. J. Dunwoody [3, §3] gave a negative answer to the Neumanns’ Problem (and indeed negative answers to several related questions), by means of a carefully engineered construction of an explicit pair of $2$-generated groups $(G,H)$, where $H$ is a homomorphic image of $G$, but $H$ admits a generating pair that does not lift back to a generating pair of $G$ along any homomorphism. In this example, $H$ is a (nilpotent of class $2$)-by-(nilpotent of class $2$) group of order $5^{3}\,11^{10}$; the group $G$ is an extension of an (infinite) abelian group by $H$.

It is largely unknown how ‘frequently’ negative examples to the Neumanns’ Problem should be expected to occur, when one restricts $G$ or $H$ to more special classes of groups. The main purpose of the present paper is to provide, for $n=2$, an infinite family of negative examples $(G,H)$ to the Neumanns’ Problem whose nature is rather different from the groups arising from Dunwoody’s construction. Indeed, our main examples arise from free products of cyclic groups $G\cong C_{2}\ast C_{m}$ mapping onto finite projective special linear groups $H\cong\operatorname{PSL}(2,q)$.

Theorem 1.1**.**

For the following pairs of $2$-generated groups, the group $H$ is a homomorphic image of $G$, but there exist generating pairs $(h_{1},h_{2})$ of $H$ such that for all homomorphisms $\vartheta\colon G\to H$ and all generating pairs $(g_{1},g_{2})$ of $G$, we have $(g_{1}^{\,\vartheta},g_{2}^{\,\vartheta})\neq(h_{1},h_{2})$.

1. (i)

$G=C_{2}\ast C_{3}$* and $H=\operatorname{PSL}(2,q)$, where $q$ is a prime power such that $q\geq 4$, but $q\neq 9$;* 2. (ii)

$G=C_{2}\ast C_{p}$* and $H=\operatorname{PSL}(2,q)$, where $q=p^{k}$ is a power of a prime $p\geq 3$ such that $q\geq 7$, but $q\neq 9$;* 3. (iii)

$G=C_{2}\ast C_{m}$* and $H=\operatorname{PSL}(2,q)$, where $m\in\mathbb{N}$ and $q=p^{k}$ is a power of a prime $p$ such that $q\equiv_{4}3$, but $q\neq 3$, and*

[TABLE] 4. (iv)

$G=C_{3}\ast C_{3}$* and $H=\operatorname{PSL}(2,q)$, where $q$ is a prime power such that $q\geq 5$.*

The groups of the form $C_{2}\ast C_{m}$, where $m\geq 3$, arise in number theory as Hecke groups, acting on the upper half of the complex plane; e.g., see [7]. The most prominent member of this family is the modular group $\operatorname{PSL}(2,\mathbb{Z})\cong C_{2}\ast C_{3}$, which clearly maps onto $\operatorname{PSL}(2,p)$ for any prime $p$. As expected, the infinite dihedral group $G\cong C_{2}\ast C_{2}$ does not yield negative examples to the Neumanns’ Problem; see Proposition 4.2.

As a first step toward Theorem 1.1, one needs to verify that the groups $H$ are actually homomorphic images of the relevant free product $G$. Fortunately, the subgroups of finite projective special linear groups of degree $2$ were completely described by L. E. Dickson; compare [6, Thm. II.8.27]. Based on this classification, A. M. Macbeath [10] studied triples $(A,B,C)$ of $\operatorname{SL}(2,q)$ such that $ABC=1$ with a view toward determining $\langle\overline{A},\overline{B}\rangle\leq\operatorname{PSL}(2,q)$ according to the traces of $A,B,C$. In particular, he showed that $\operatorname{PSL}(2,q)$ is $(2,3)$-generated unless $q=9$; we refer to [15] for a complete overview about $(2,3)$-generation of arbitrary (projective) special linear groups over finite fields and references to related results.

In [8], U. Langer and G. Rosenberger use Macbeath’s results to determine necessary and sufficient conditions for $\operatorname{PSL}(2,q)$ to be the quotient of a given triangle group, with torsion-free kernel. Generalising a result of J. L. Brenner and J. Wiegold, they establish (ibid., Satz 5.3) the following theorem. For every prime power $q\not\in\{2,4,5,9\}$ and all integers $m\geq 2$, $n\geq 3$, the group $\operatorname{PSL}(2,q)$ is $(m,n)$-generated provided that $m,n$ are possible element orders in $\operatorname{PSL}(2,q)$.

We recall that $m$ is a possible element order in $\operatorname{PSL}(2,q)$, where $q=p^{k}$ is a power of a prime $p$, if and only if

[TABLE]

This explains, for instance, the assumptions in Theorem 1.1(iii), following the condition $q\equiv_{4}3$. Generation properties of $\operatorname{PSL}(2,q)$ for $q\in\{2,4,5,9\}$ can, of course, be dealt with by direct computation: $\operatorname{PSL}(2,9)\cong\operatorname{Alt}(6)$ is not $(2,3)$-generated, but it is $(3,3)$-generated; $\operatorname{PSL}(2,5)\cong\operatorname{Alt}(5)$ is $(2,5)$-generated, but generating pairs lift, as we state in Proposition 4.1.

In Dunwoody’s example and all our examples thus far, the group $G$ is $2$-generated and infinite, whereas the homomorphic image $H$ is finite. By a small modification, we obtain the first negative examples $(G,H)$ to the Neumanns’ Problem with $H$ infinite.

Corollary 1.2**.**

There are negative examples $(G,H)$ to the Neumanns’ Problem, where $H$ is infinite. For instance, $G=\operatorname{PSL}(2,\mathbb{Z})$ and $H=\operatorname{PSL}(2,5)\times G/G^{\prime\prime}$ yield such an example: $H$ is a homomorphic image of $G$, but $H$ admits generating pairs that do not lift to generating pairs for $G$ along any homomorphism.

Similar to Dunwoody, we focus in this paper exclusively on $2$-generated groups and on lifting generating pairs. It remains a challenge to produce negative examples $(G,H)$ to the Neumanns’ Problem for $n\geq 3$, where $G$ and $H$ are $n$-generated and certain generating $n$-tuples do not lift back. It is conceivable that such examples can be found as a byproduct of a systematic study of the Neumanns’ Problem for pairs $(G,H)$, where $G$ is a finite free product of cyclic groups. Theorem 1.1(iv) can be seen as a first minor step toward such a systematic study.

Finally, we think it would be interesting to investigate to what degree the extra condition $q\equiv_{4}3$ in Theorem 1.1(iii) is needed. Our proof relies on this assumption and breaks down, for instance, for $G=C_{2}\ast C_{m}$ and $H=\operatorname{PSL}(2,q)$ with $(m,q)\in\{(7,13),(19,37),(20,41),(21,41)\}$: one can check by direct computer calculation that, in these cases, the trace invariants associated to relevant generating pairs $(\overline{A},\overline{B})$ assume all possible values, apart from $2$. This stands in contrast to the situations that we deal with.

Acknowledgements. The initial ideas for our paper stem from an unpublished manuscript of the second author, which contained a self-contained number-theoretic proof that, for every prime $p\geq 5$, there exist generating pairs of $H=\operatorname{PSL}(2,p)$ which do not lift to $G=\operatorname{PSL}(2,\mathbb{Z})\cong C_{2}\ast C_{3}$. The third author thanks Sandro Mattarei and Alex Bartel for helpful discussions. We acknowledge the referee’s feedback which led to improvements of the presentation.

2. Preliminary set-up

Let $G$ be a $2$-generated group. There is a natural semi-regular action of $\operatorname{Aut}(G)$ on the set

[TABLE]

of all generating pairs of $G$ under which $\alpha\in\operatorname{Aut}(G)$ sends $(g_{1},g_{2})\in\Gamma_{G}$ to $(g_{1}^{\,\alpha},g_{2}^{\,\alpha})$.

We are more interested in the action of the automorphism group $\mathcal{A}=\operatorname{Aut}(F)$ of a free group $F=\langle x,y\rangle$ of rank two on $\Gamma_{G}$ defined as follows. There is a bijection

[TABLE]

where $w(x,y)$ denotes any word in $x,y$. The group $\mathcal{A}$ permutes the elements of $\mathcal{E}$, via $\eta^{\alpha}=\alpha^{-1}\eta$ for $\eta\in\mathcal{E}$ and $\alpha\in\mathcal{A}$, and this action pulls back along $\Phi$ to an action of $\mathcal{A}$ on $\Gamma_{G}$.

It is well-known that $\mathcal{A}$ is generated by the basic Nielsen transformations

[TABLE]

Application of these to an element $(g_{1},g_{2})\in\Gamma_{G}$ via $\Phi$ yields

[TABLE]

We note that, if $\vartheta\colon G\twoheadrightarrow H$ is an epimorphism of groups, then $\vartheta$ induces a map $\vartheta^{*}\colon\Gamma_{G}\to\Gamma_{H},(g_{1},g_{2})\mapsto(g_{1}^{\,\vartheta},g_{2}^{\,\vartheta})$ which commutes with the action of $\mathcal{A}$ on $\Gamma_{G}$ and $\Gamma_{H}$ respectively; i.e., for every $\alpha\in\mathcal{A}$ and all $(g_{1},g_{2})\in\Gamma_{G}$ we have

[TABLE]

In general, neither $\mathcal{A}$ nor the permutation group $P$ induced by $\mathcal{A}$ and $\operatorname{Aut}(G)$ is necessarily going to act transitively on $\Gamma_{G}$.

Example 2.1**.**

For instance, if $G=\operatorname{PSL}(2,5)\cong\operatorname{Alt}(5)$ is the alternating group of degree $5$, then $\Gamma_{G}$ consists of $2280$ elements. They fall into nineteen $\operatorname{Aut}(G)$-orbits, each of length $120$, and into two $P$-orbits of length $1080=9\cdot 120$ and $1200=10\cdot 120$ respectively. For details see [5] and [14, §10]. By means of a direct calculation one can see that in the same example $\Gamma_{G}$ splits into three $\mathcal{A}$-orbits of size $600=5\cdot 120$, $600=5\cdot 120$, and $1080=9\cdot 120$, respectively.

But under special circumstances $\mathcal{A}$ acts transitively on $\Gamma_{G}$, e.g., if $G$ is abelian or if $G$ is free. We require the following not so obvious fact.

Fact 2.2**.**

If $G\cong C_{m}\ast C_{n}$ for $m,n\in\mathbb{N}$, then $\Gamma_{G}$ consists of a single $\mathcal{A}$-orbit.

This is a consequence of the Grushko–Neumann Theorem about free groups, which can be proved, for instance, by cancellation arguments due to R. C. Lyndon; see [16, Section 6.3] and [1, Appendix]. From Fact 2.2 and (2.2) we obtain

Corollary 2.3**.**

Let $G=C_{m}\ast C_{n}$ for $m,n\in\mathbb{N}$. Suppose that $H$ is a homomorphic image of $G$, and let $(h_{1},h_{2})\in\Gamma_{H}$. Then there exists $\vartheta\colon G\to H$ and $(g_{1},g_{2})\in\Gamma_{G}$ with $(g_{1}^{\,\vartheta},g_{2}^{\,\vartheta})=(h_{1},h_{2})$ if and only if the $\mathcal{A}$-orbit of $(h_{1},h_{2})$ in $\Gamma_{H}$ contains an element $(\tilde{h}_{1},\tilde{h}_{2})$ such that ${\tilde{h}_{1}}^{\,m}=\tilde{h}_{2}^{\,n}=1$.

As above, let $H$ be a homomorphic image of $G$. We say that a subset $\Omega\subseteq\Gamma_{H}$ is $(m,n)$-free if there exists no $(h_{1},h_{2})\in\Omega$ such that $h_{1}^{\,m}=h_{2}^{\,n}=1$. Furthermore, we refer to $(h_{1},h_{2})\in\Gamma_{H}$ as an $(m,n)$-generating pair if $h_{1}^{\,m}=h_{2}^{\,n}=1$, i.e., if the orders of $h_{1}$ and $h_{2}$ divide $m$ and $n$, respectively.

We make use of a fruitful observation by D. G. Higman; see [13]. Let $\Omega\subseteq\Gamma_{H}$ be an $\mathcal{A}$-orbit, and let

[TABLE]

Since the identities

[TABLE]

hold in any group, the induced Nielsen transformations on $\Gamma_{G}$ show that for any choice of $(h_{1},h_{2})\in\Omega$, we have

[TABLE]

For instance, this shows that the order of $[h_{1},h_{2}]$ is constant for $(h_{1},h_{2})\in\Omega$.

We take interest in the special situation, where $H=\operatorname{PSL}(2,q)$ is a projective special linear group, for a prime power $q$. Every element $h\in H$ has matrix representatives $R(h),-R(h)\in\operatorname{SL}(2,q)$. Let $\operatorname{tr}\colon\operatorname{SL}(2,q)\to\mathbb{F}_{q}$ denote the usual trace map. The group commutator yields a canonical map from $H\times H=\operatorname{PSL}(2,q)\times\operatorname{PSL}(2,q)$ to $\operatorname{SL}(2,p^{k})$: for $(h_{1},h_{2})\in H\times H$, the element

[TABLE]

is independent of the particular choice of representatives. Combining this map with the usual trace map, we obtain the trace invariant

[TABLE]

which is constant on $\mathcal{A}$-orbits $\Omega\subseteq\Gamma_{H}$.

As mentioned in the introduction, the (maximal) subgroups of finite projective special linear groups are fully understood; therefore it is possible to locate generating pairs satisfying additional properties. Using this approach, D. McCullough and M. Wanderley [11] showed for instance that, for $q\geq 13$, every non-trivial element of $\operatorname{PSL}(2,q)$ is a commutator of a generating pair. For convenience we state the following consequence of their work.

Theorem 2.4** ([11, Thm. 2.1]).**

Let $H=\operatorname{PSL}(2,q)$, where $q$ is a prime power. The set $\mathcal{T}(H)=\{\tau(h_{1},h_{2})\mid(h_{1},h_{2})\in\Gamma_{H}\}\subseteq\mathbb{F}_{q}$ of trace invariants of generating pairs for $\operatorname{PSL}(2,q)$ satisfies

[TABLE]

Next, we record a basic and well-known fact.

Lemma 2.5**.**

Let $A\in\operatorname{SL}(2,q)\smallsetminus\{I,-I\}$, where $q=p^{k}$ is a power of a prime $p\geq 3$. Then

1. (a)

$\operatorname{tr}(A)=2$* if and only if $A$ has order $p$, and* 2. (b)

$\operatorname{tr}(A)=-2$* if and only if $A$ has order $2p$.*

Proposition 2.6**.**

Let $q$ be a prime power such that $q\equiv_{4}3$. Suppose that $\operatorname{SL}(2,q)=\langle A,B\rangle$, where $A^{4}=I$. Then $\operatorname{tr}([A,B])\neq-2$.

Proof.

For a contradiction, assume that $\operatorname{tr}([A,B])=-2$. Conjugating $A$ and $B$ by a suitable element of $\operatorname{GL}(2,q)$, we may suppose that

[TABLE]

Since $\operatorname{PSL}(2,q)=\langle\overline{A},\overline{B}\rangle$ is not abelian, we deduce that $x\neq 0$. Since $A$ has order $4$ in $\operatorname{SL}(2,q)$, it follows that $\operatorname{tr}(A)=0$, hence we may write

[TABLE]

Since

[TABLE]

also has trace [math], we deduce that $c=0$. From $\det A=1$ we conclude that $a^{2}=-1$, in contradiction to $q\equiv_{4}3$. ∎

Using Theorem 2.4, we deduce from Proposition 2.6 the following key corollary.

Corollary 2.7**.**

Let $q$ be a prime power such that $q\equiv_{4}3$, but $q\neq 3$, and let $m\in\mathbb{N}$. Suppose that $H=\operatorname{PSL}(2,q)$ is $(2,m)$-generated. Then $G=C_{2}\ast C_{m}$ and $H$ yield a negative example $(G,H)$ to the Neumanns’ Problem.

Proof.

By Theorem 2.4, we find a generating pair $(h_{1},h_{2})\in\Gamma_{H}$, where $h_{1}=\overline{A}$ and $h_{2}=\overline{B}$ for $A,B\in\operatorname{SL}(2,q)$, such that $\operatorname{tr}([A,B])=\tau(h_{1},h_{2})=-2$. By Corollary 2.3, it suffices to show that $(h_{1},h_{2})$ represents an $\mathcal{A}$-orbit that is $(2,m)$-free.

Suppose that $(h_{1}^{\prime},h_{2}^{\prime})=(\overline{A^{\prime}},\overline{B^{\prime}})$, for $A^{\prime},B^{\prime}\in\operatorname{SL}(2,q)$, is a $(2,m)$-generating pair for $H$. Then $(A^{\prime},B^{\prime})$ is a generating pair for $\operatorname{SL}(2,q)$ and $A^{\prime 4}=I$. By Proposition 2.6, we have $\operatorname{tr}([A^{\prime},B^{\prime}])\neq-2$. Hence $\tau(h_{1}^{\prime},h_{2}^{\prime})\neq\tau(h_{1},h_{2})$, and $(h_{1}^{\prime},h_{2}^{\prime})$ cannot lie in the $\mathcal{A}$-orbit of $(h_{1},h_{2})$. ∎

3. Proof of the main theorem and corollary

We continue to use the notation introduced in Section 2.

Proof of Theorem 1.1(iii).

This follows directly from Corollary 2.7. ∎

Our proof of Theorem 1.1(i) uses group-theoretic results, due to G. A. Miller.

Proof of Theorem 1.1(i).

Consider the group $H=\operatorname{PSL}(2,q)$, where $q\geq 4$, but $q\neq 9$. By Corollary 2.3, it suffices to show that there is an $\mathcal{A}$-orbit in $\Gamma_{H}$ that is $(2,3)$-free.

First suppose that $p\neq 2$ and $q\not\in\{5,7\}$. Using Theorem 2.4, we find a generating pair $(h_{1},h_{2})\in\Gamma_{H}$ such that $\tau(h_{1},h_{2})=0$. This implies that $[h_{1},h_{2}]\neq 1$ has order $2$. A classical result of G. A. Miller [12, §1] implies that any group generated by an element of order two and an element of order three whose commutator has order two is isomorphic to either $\operatorname{Alt}(4)$ or $\operatorname{Alt}(4)\times C_{2}$. Thus the $\mathcal{A}$-orbit of $(h_{1},h_{2})$ is $(2,3)$-free.

Now suppose that $p=2$ or $q\in\{5,7\}$. Using Theorem 2.4, we find a generating pair $(h_{1},h_{2})\in\Gamma_{H}$ such that $\tau(h_{1},h_{2})\in\{1,-1\}$. This implies that $[h_{1},h_{2}]\neq 1$ has order $3$. By [12, §3], any group generated by an element of order two and an element of order three whose commutator has order three is solvable. (Indeed, the finitely presented group $K=\langle x,y\mid x^{2}=y^{3}=[x,y]^{3}=1\rangle$ has derived length $3$; one has $\lvert K:K^{\prime\prime}\rvert=54$ and $K^{\prime\prime}\cong C_{\infty}\times C_{\infty}$.) Thus the $\mathcal{A}$-orbit of $(h_{1},h_{2})$ is $(2,3)$-free. ∎

For completeness we remark that it is easy to check that every generating pair of $\operatorname{PSL}(2,2)\cong\operatorname{Sym}(3)$ and $\operatorname{PSL}(2,3)\cong\operatorname{Alt}(4)$ respectively lifts to a generating pair of $C_{2}\ast C_{3}\cong\operatorname{PSL}(2,\mathbb{Z})$ along a suitable epimorphism.

Proof of Theorem 1.1(ii).

Consider the group $H=\operatorname{PSL}(2,q)$, where $q=p^{k}$ is a power of a prime $p\geq 3$ such that $q\geq 7$, but $q\neq 9$. By Corollary 2.3, it suffices to show that there is an $\mathcal{A}$-orbit of $\Gamma_{H}$ that is $(2,p)$-free. If $q\equiv_{4}3$, we apply Corollary 2.7, as in the proof of part (iii).

Now suppose that $q\equiv_{4}1$. The argument we provide actually works for all prime powers $q\geq 11$. Suppose that $\operatorname{PSL}(2,q)=\langle h_{1},h_{2}\rangle$ with $h_{1}^{\,2}=h_{2}^{\,p}=1$, where $h_{1}=\overline{A}$ and $h_{2}=\overline{B}$ for $A,B\in\operatorname{SL}(2,q)$. Then $\langle A,B\rangle=\operatorname{SL}(2,q)$, and since there is only one conjugacy class of elements of order $4$ in $\operatorname{SL}(2,q)$, we may suppose that

[TABLE]

We write

[TABLE]

Since $B$ has order $p$ or $2p$, Lemma 2.5 shows that the trace of $\operatorname{tr}(B)\in\{2,-2\}$. Put $s=b-c\in\mathbb{F}_{q}$. We have

[TABLE]

Thus $(h_{1},h_{2})\in\Gamma_{H}$ has trace invariant

[TABLE]

By Theorem 2.4, there is an $\mathcal{A}$-orbit in $\Gamma_{H}$ such that the associated trace invariant is not of the form $s^{2}+2$ for $s\in\mathbb{F}_{q}$. Such an $\mathcal{A}$-orbit is $(2,p)$-free. ∎

In preparation of the proof of Theorem 1.1(iv), we establish two lemmata that could also be checked directly, using a computer.

Lemma 3.1**.**

If $\operatorname{SL}(2,5)=\langle A,B\rangle$, where $A^{3},B^{3}\in\{I,-I\}$, then $\operatorname{tr}([A,B])\neq-2$.

Proof.

By Lemma 2.5, it suffices to show that $[A,B]$ does not have order $10$. For this it suffices to show that $[\overline{A},\overline{B}]\in\operatorname{PSL}(2,5)$ does not have order $5$.

Recall that $\operatorname{PSL}(2,5)\cong\operatorname{Alt}(5)$. Hence it is enough to prove that in $\operatorname{Alt}(5)$ no commutator of two $3$-cycles yields a $5$-cycle. Using conjugation in $\operatorname{Sym}(5)$, it suffices to consider a single commutator:

[TABLE]

Lemma 3.2**.**

If $\operatorname{SL}(2,7)=\langle A,B\rangle$, where $A^{3},B^{3}\in\{I,-I\}$, then $\operatorname{tr}([A,B])\not\in\{3,-3\}$.

Proof.

Elements $C\in\operatorname{SL}(2,7)$ with $\operatorname{tr}(C)\in\{3,-3\}$ have order $8$, and the image of such an element in $\operatorname{PSL}(2,7)$ has order $4$. Hence it suffices to show that $[\overline{A},\overline{B}]\in\operatorname{PSL}(2,7)$ does not have order $4$.

Recall that $\operatorname{PSL}(2,7)$ acts $2$-transitively on the projective line $\mathbf{P}^{1}(\mathbb{F}_{7})=\mathbb{F}_{7}\cup\{\infty\}$. The stabiliser of any pair of points is cyclic of order $3$, and the stabiliser of any $2$-element set is dihedral of order $6$. For instance, the stabiliser of $\{0,\infty\}$ is generated by

[TABLE]

Elements of order $4$ act fixed-point-freely on $\mathbf{P}^{1}(\mathbb{F}_{7})$. The elements $\overline{A},\overline{B}$ induce fractional linear transformations $u,v$ on $\mathbf{P}^{1}(\mathbb{F}_{7})$, with $\lvert\mathrm{Fix}(u)\rvert=\lvert\mathrm{Fix}(v)\rvert=2$. Using conjugation and replacing $u$ by $u^{-1}$, if necessary, we may assume that $u=u_{0}$.

For a contradiction, assume that $[u,v]$ has order $4$ so that $\mathrm{Fix}([u,v])=\varnothing$. We conclude that $\mathrm{Fix}(u)=\{0,\infty\}$ and $\mathrm{Fix}(u).v$ are disjoint; likewise $\mathrm{Fix}(v)$ and $\mathrm{Fix}(v).u$ are disjoint. The elements of order $3$ form a single conjugacy class in $\operatorname{PSL}(2,7)$, and we write $v=u^{w}$ for suitable $w\in\operatorname{PSL}(2,7)$. Conjugating by $g\in\langle t,u\rangle$ and replacing $u$ by $u^{-1}$, if necessary, we may assume that $0.w=1$. This leaves three options for $\infty.w$, namely $3$, $6$ and $5$. Accordingly, $v$ is one of the transformations

[TABLE]

or one of their inverses. Replacing $v$ by its inverse, if necessary, we may assume that $v$ is one of $v_{1},v_{2},v_{3}$. Direct computations show that

[TABLE]

Proof of Theorem 1.1(iv).

Consider the group $H=\operatorname{PSL}(2,q)$, where $q$ is a prime power such that $q\geq 5$. By Corollary 2.3, it suffices to show that there is an $\mathcal{A}$-orbit of $\Gamma_{H}$ that is $(3,3)$-free.

If $q=5$, Theorem 2.4 and Lemma 3.1 imply that at least one $\mathcal{A}$-orbit in $\Gamma_{H}$ is $(3,3)$-free. If $q=7$, Theorem 2.4 and Lemma 3.2 show that there are $(3,3)$-free $\mathcal{A}$-orbits in $\Gamma_{H}$. Now suppose that $q\not\in\{5,7\}$. We show below that, if $(h_{1},h_{2})\in\Gamma_{H}$ and $h_{1}^{3}=h_{2}^{3}=1$, then $\tau(h_{1},h_{2})\neq 0$. Thus Theorem 2.4 yields that at least one $\mathcal{A}$-orbit in $\Gamma_{H}$ is $(3,3)$-free.

Let $(h_{1},h_{2})\in\Gamma_{H}$ be such that $h_{1}^{3}=h_{2}^{3}=1$. For a contradiction, assume that $\tau(h_{1},h_{2})=0$. Then $[h_{1},h_{2}]\neq 1$ would have order $2$. But a classical result of G. A. Miller [12, §2] implies that any group generated by two elements of order three whose commutator has order two is soluble, in contradiction to $H=\operatorname{PSL}(2,q)$ being non-soluble. (Indeed, the finitely presented group $K=\langle x,y\mid x^{3}=y^{3}=[x,y]^{2}=1\rangle$ has order $288$ and derived length $3$; one has $K/K^{\prime}\cong C_{3}\times C_{3}$ and $K^{\prime\prime}\cong C_{2}$.) ∎

Proof of Corollary 1.2.

Suppose that $H_{0}=\langle y_{1},y_{2}\rangle\cong\operatorname{PSL}(2,q)$ is a homomorphic image of $G=\langle x_{1},x_{2}\rangle\cong C_{m}\ast C_{n}$ under $x_{1}\mapsto y_{1}$ and $x_{2}\mapsto y_{2}$, where $q$ is a prime power such that $q\geq 5$ and $m,n\in\mathbb{N}$. Suppose further that $(h_{1},h_{2})\in\Gamma_{H_{0}}$ does not lift to a generating pair $(g_{1},g_{2})\in\Gamma_{G}$ under any homomorphism $G\to H_{0}$.

Observe that $G$ modulo its second derived subgroup $G^{\prime}$, is infinite: indeed, $G/G^{\prime}=\langle\overline{x_{1}},\overline{x_{2}}\rangle\cong C_{m}\times C_{n}$ and the Kuroš Subgroup Theorem implies $G^{\prime}/G^{\prime\prime}\cong C_{\infty}^{\,(m-1)(n-1)}$. The group $H_{0}$ is perfect, hence $H_{0}^{\prime\prime}=H_{0}$. Therefore $H=H_{0}\times G/G^{\prime\prime}$ is a homomorphic image of $G$ under $x_{1}\mapsto y_{1}\overline{x_{1}}$ and $x_{2}\mapsto y_{2}\overline{x_{2}}$. Furthermore $(h_{1}\overline{x_{1}},h_{2}\overline{x_{2}})$ is a generating pair of $H$ that does not lift to a generating pair $(g_{1},g_{2})\in\Gamma_{G}$ under any homomorphism $\vartheta\colon G\to H$, because otherwise $(h_{1},h_{2})$ would lift to $(g_{1},g_{2})$ under the composition $\vartheta\pi_{1}\colon G\to H_{0}$, where $\pi_{1}\colon H\to H_{0}$ is the projection onto the first factor. ∎

4. Complements

In this section we give two complementary results. First we show that Theorem 1.1(ii) does not extend to $q=5$.

Proposition 4.1**.**

Let $G=C_{2}\ast C_{5}$ and $H=\operatorname{PSL}(2,5)$. Then every generating pair of $H$ lifts to a generating pair of $G$ along some epimorphism.

Proof.

In view of Corollary 2.3, it suffices to show that every $\mathcal{A}$-orbit in $\Gamma_{H}$ contains a $(2,5)$-generating pair. The number $\mathcal{A}$-orbits is known to be $3$; compare Example 2.1.

Consider in $\operatorname{SL}(2,5)$ the elements

[TABLE]

It is straightforward to check that $(\overline{A},\overline{C})$, $(\overline{A},\overline{D})$ and $(\overline{B},\overline{D})$ are $(2,5)$-generating pairs for $H$. We claim that they are representatives for the three distinct $\mathcal{A}$-orbits in $\Gamma_{H}$. Indeed,

[TABLE]

so that

[TABLE]

Thus it remains to show that $(\overline{A},\overline{C})$ and $(\overline{B},\overline{D})$ lie in distinct $\mathcal{A}$-orbits. This follows from the fact that $[A,C]$ is not conjugate to any of the matrices $[B,D]$, $[D,B]$, $-[B,D]$, $-[D,B]$. (The latter two can be ruled out by their trace, the former two are ruled out by direct computation.) ∎

Finally, we justify that the infinite dihedral group $G=C_{2}\ast C_{2}$ does not yield negative examples to the Neumanns’ Problem

Proposition 4.2**.**

Let $G=C_{2}\ast C_{2}$ and let $H=\langle h_{1},h_{2}\rangle$ be a homomorphic image of $G$. Then there exists a homomorphism $\vartheta\colon G\to H$ and a generating pair $(g_{1},g_{2})\in\Gamma_{G}$ such that $(g_{1}^{\,\vartheta},g_{2}^{\,\vartheta})=(h_{1},h_{2})$.

Proof.

The infinite dihedral group $G=\langle x,y\mid x^{2}=y^{2}=1\rangle$ is just infinite; its non-trivial proper homomorphic images are exactly the finite dihedral groups $D_{2m}$, for $m\in\mathbb{N}$. Avoiding trivial and easy cases, we may suppose that $H=D_{2m}$ for $m\geq 3$.

The group $D_{2m}$ consists of $m$ shifts and $m$ reflections. At least one of $h_{1}$ and $h_{2}$ is a reflection. Let’s suppose $h_{1}$ is a reflection. If $h_{2}$ is also a reflection, then there is a homomorphism mapping $g_{1}=x$ to $h_{1}$ and $g_{2}=y$ to $h_{2}$. If $h_{2}$ is a shift (necessarily of order $m$), then there is a homomorphism mapping $g_{1}=x$ to $h_{1}$ and $g_{2}=xy$ to $h_{2}$. ∎