REVIEW 3 minor 16 references
Generating pairs of projective special linear groups that fail to lift
T0 review · 0 major / 3 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read PSL(2,q) has generating pairs that cannot be lifted.
desk verdict A solid construction paper: infinite families of free-product/PSL(2,q) negative examples for the Neumanns' problem, with a correct proof and only minor presentation gaps. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing object is the trace invariant. For a generating pair $(h_1,h_2)$ of $H=\mathrm{PSL}(2,q)$, choose matrix representatives in $\mathrm{SL}(2,q)$, form their commutator, and take its trace; the result $\tau(h_1,h_2)\in\mathbb{F}_q$ is independent of the choice of representatives and is constant on the orbit of the pair under the automorphism group of the free group on two generators. Because generating pairs of a free product $C_m*C_n$ form a single such orbit, a pair of $H$ lifts to $G$ exactly when its orbit contains a pair $(h_1,h_2)$ with $h_1^m=h_2^n=1$. The proofs use a classification of which trace values occur for generating pairs of $\mathrm{PSL}(2,q)$ together with classical facts about groups generated by two elements of prescribed orders whose commutator has order 2 or 3. The trace invariant is the bridge that turns a question about all homomorphisms into a finite check about one number.
What would settle it
Choose $q=11$ and exhaustively list all generating pairs $(h_1,h_2)$ of $\mathrm{PSL}(2,11)$ with $h_1^2=h_2^3=1$, computing $\tau(h_1,h_2)=\operatorname{tr}([h_1,h_2])$. The theorem predicts that no such pair has trace invariant $0$ and that none has commutator of order 2 or 3, since that would force the group they generate to be solvable; finding one would refute Theorem 1.1(i).
Extended reading notes
Core claim
The central claim, stated as Theorem 1.1, is that the following pairs $(G,H)$ are negative examples to the lifting problem: (i) $G=C_2*C_3$, $H=\mathrm{PSL}(2,q)$ for $q\ge 4$, $q\ne 9$; (ii) $G=C_2*C_p$, $H=\mathrm{PSL}(2,q)$ with $q=p^k$, $p\ge 3$, $q\ge 7$, $q\ne 9$; (iii) $G=C_2*C_m$, $H=\mathrm{PSL}(2,q)$ with $q\equiv 3 \pmod 4$, $q\ne 3$, and an additional divisibility condition linking $m$ to $p$, $(q+1)/2$, or $(q-1)/2$; and (iv) $G=C_3*C_3$, $H=\mathrm{PSL}(2,q)$ for $q\ge 5$. In each case $H$ is a homomorphic image of $G$, but there exist generating pairs $(h_1,h_2)$ of $H$ such that for every homomorphism $\vartheta\colon G\to H$ and every generating pair $(g_1,g_2)$ of $G$, the images $(g_1^\vartheta,g_2^\vartheta)$ are not equal to $(h_1,h_2)$. Corollary 1.2 transfers the phenomenon to infinite $H$ by taking a direct product with the infinite group $G/G''$; the projection back to the finite factor would turn any lift of the product pair into a lift of the original non-liftable pair.
Load-bearing premise
The argument rests on the fact that all generating pairs of a free product $C_m*C_n$ form a single orbit under the automorphism group of the free group on two generators; if that transitivity failed, a pair with a forbidden trace invariant could still lift through a pair from a different orbit.
Editorial extensions
If this is right
- For every prime $p\ge 5$, the quotient $\mathrm{PSL}(2,p)$ of the modular group $C_2*C_3$ has generating pairs that no homomorphism from the modular group can lift; the finite-quotient picture of the modular group is therefore not tuple-surjective.
- Non-lifting is not a pathology of one crafted pair of solvable groups: it occurs for infinitely many pairs in which $G$ is a free product of cyclic groups and $H$ is one of the most studied finite simple groups.
- The first negative examples with infinite $H$ follow formally from the finite ones: $H=\mathrm{PSL}(2,5)\times G/G''$ is a quotient of $G=C_2*C_3$ but has non-liftable generating pairs.
- The obstruction is visible in a single scalar invariant, so the same orbit-and-trace test can be run for any finite quotient for which the trace values on generating pairs are known.
Reading between the lines
- One could extend the same orbit-and-trace test to other non-solvable quotients of free products of cyclic groups: whenever the set of trace invariants is known, the existence of non-liftable pairs reduces to a finite computation, potentially producing negative examples outside the $\mathrm{PSL}(2,q)$ family.
- The residue condition $q\equiv 3 \pmod 4$ in part (iii) is likely an artifact of the proof; the authors note small cases where the invariant takes all but one value, so a refined invariant or a different choice of orbit might remove the condition.
- The general recipe behind the proofs suggests that any orbit whose commutator has an order forcing solvability of two-generator quotients with the prescribed element orders will give a non-lifting example in a non-solvable target, a criterion that could be tested on other finite non-solvable groups.
- For $n\ge 3$, no negative examples are known; since generating $n$-tuples of a free product do not form a single orbit under $\mathrm{Aut}(F_n)$, new ideas are needed, but partial results might come from studying orbits of $n$-tuples with fixed trace-like invariants.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper addresses the Neumann problem on lifting generating tuples along homomorphisms. For n=2, the authors exhibit infinitely many pairs (G,H), where G is a free product of finite cyclic groups and H=PSL(2,q), such that H is a homomorphic image of G but some generating pair of H is not the image of any generating pair of G under any homomorphism. Theorem 1.1 splits into four cases: G=C2*C3 with H=PSL(2,q) for q≥4 and q≠9; G=C2*C_p with H=PSL(2,p^k); G=C2*C_m with q≡3 mod4 under explicit arithmetic conditions on m; and G=C3*C3 with H=PSL(2,q) for q≥5. Corollary 1.2 produces the first negative examples in which H is infinite, for instance H=PSL(2,5)×G/G'' with G=PSL(2,Z). The method reduces the lifting question, via the Grushko–Neumann theorem, to whether a given Aut(F_2)-orbit of generating pairs of H contains an (m,n)-generating pair; a trace invariant of the commutator is then used to separate the desired orbit from all (m,n)-generating pairs.
Significance. If the main theorem is correct, this is a substantial advance on the Neumann problem: it replaces Dunwoody's single engineered soluble example with infinite families in which H is a simple projective special linear group, and it gives the first negative examples with H infinite. The method is transparent and reproducible: the key trace computations are explicit, the finite exceptional cases q=5,7 are checked by hand, and the appeal to external results (Grushko–Neumann, Dickson–Macbeath generation theorems, the McCullough–Wanderley classification of trace invariants, and Miller's commutator-order theorems) is clearly labelled. In particular, the load-bearing premise Fact 2.2 is a standard theorem rather than a circular assumption, and the reduction in Corollary 2.3 is sound. I found no load-bearing error in the central derivation.
minor comments (3)
- [§3, proof of Theorem 1.1(ii)] In the displayed matrix for B, the entries are declared to lie in F_p, but B is an element of SL(2,q) with q=p^k; the subsequent line puts s=b-c in F_q, so the entries should be in F_q. This is only a typo and does not affect the computation.
- [§2, Proposition 2.6] The proof begins with 'Since A has order 4 in SL(2,q)', but the hypothesis is only A^4=I, which also allows A=-I of order 2. In the order-2 case the conclusion is trivial because then [A,B]=I and tr([A,B])=2, so the statement is correct; however, the proof as written should either restrict to the order-4 case used in Corollary 2.7 or mention the order-2 case separately.
- [§3, proof of Theorem 1.1(iii)] The proof says only 'This follows directly from Corollary 2.7.' Since Corollary 2.7 assumes that H is (2,m)-generated, the proof should explicitly justify that the hypotheses on m imply this: if p divides m, use an element of order p; otherwise, a common divisor d≥3 of m with (q±1)/2 gives a possible element order d, and the Langer–Rosenberger theorem supplies (2,d)-generation, which implies (2,m)-generation.
Circularity Check
No significant circularity: the obstruction rests on external trace classification and Nielsen-transitivity theorems, not on the paper's own conclusion.
full rationale
The derivation is self-contained rather than circular. The lifting problem is reduced via Corollary 2.3 to the existence of an A-orbit in Γ_H that is (m,n)-free, and this reduction uses Fact 2.2 (transitivity of Aut(F_2) on Γ_{C_m*C_n}), an external theorem due to Grushko–Neumann/Lyndon that does not assume the target result. The trace invariant τ is constant on A-orbits, and Theorem 2.4 is quoted from the independent classification of McCullough–Wanderley giving T(H) for PSL(2,q). The individual cases then show by direct computation or by classical results of Miller that no (m,n)-generating pair can have the selected trace invariant: for (2,p) pairs with q ≡ 1 mod 4, τ has the special form s^2+2 and Theorem 2.4 supplies a generating pair outside that set; for (2,3) pairs, Miller's theorems force the commutator to have order 2 or 3, so the generated group is Alt(4), Alt(4)×C_2, or solvable, never PSL(2,q) in the stated ranges; for (3,3) pairs, Lemmas 3.1 and 3.2 handle q=5,7 and Miller rules out τ=0 for q∉{5,7}. None of these steps fits a parameter, renames the conclusion, or invokes a uniqueness theorem from the present authors. The only self-referential item is the acknowledgement of an unpublished manuscript by the second author, which is historical and not load-bearing; the proof is written out in full. The apparent typo in the proof of Theorem 1.1(ii), where entries of B are written in F_p instead of F_q, is harmless because s is subsequently taken in F_q and all trace computations are valid over F_q. Hence no circularity is identified.
Assumptions & free parameters
assumptions (6)
- standard math Fact 2.2: Aut(F2) acts transitively on the generating pairs of Cm*Cn (Grushko-Neumann).
- standard math Theorem 2.4 (McCullough-Wanderley): exact set T(H) of commutator trace invariants for generating pairs of PSL(2,q).
- standard math Langer-Rosenberger: PSL(2,q) is (m,n)-generated whenever m,n are possible element orders, excluding q in {2,4,5,9}.
- standard math Miller's classification: groups generated by elements of orders 2 and 3, or two order-3 elements, with commutator of order 2 or 3 are finite or solvable of limited shape.
- standard math Kurosh subgroup theorem: for G=Cm*Cn, G'/G'' is a free abelian group of rank (m-1)(n-1), in particular infinite.
- standard math Dickson's classification of subgroups of PSL(2,q), as mediated by Macbeath, underlies the generation statements.
Cite this review
Pith. "Pith review of Generating pairs of projective special linear groups that fail to lift." pith.science (2026). https://pith.science/paper/W2LFUR5B
@misc{pith2026190805456,
author = {Pith},
title = {Pith review of: Generating pairs of projective special linear groups that fail to lift},
year = {2026},
howpublished = {\url{https://pith.science/paper/W2LFUR5B}},
note = {Machine review of arXiv:1908.05456}
}
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.
Reference graph
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Reviewed August 14, 2026 · model on record in the stance chip above.
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