TWO NEW PROOFS OF THE FACT THAT TRIANGLE GROUPS ARE DISTINGUISHED BY THEIR FINITE QUOTIENTS

In a 2016 paper by Alan Reid, Martin Bridson and the author, it was shown using the theory of profinite groups that if Γ is a finitely-generated Fuchsian group and Σ is a lattice in a connected Lie group, such that Γ and Σ have exactly the same finite quotients, then Γ is isomorphic to Σ. As a consequence, two triangle groups ∆(r, s, t) and ∆(u, v, w) have the same finite quotients if and only if (u, v, w) is a permutation of (r, s, t). A direct proof of this property of triangle groups was given in the final section of that paper, with the purpose of exhibiting explicit finite quotients that can distinguish one triangle group from another. Unfortunately, part of the latter direct proof was flawed. In this paper two new direct proofs are given, one being a corrected version using the same approach as before (involving direct products of small quotients), and the other being a shorter one that uses the same preliminary observations as in the earlier version but then takes a different direction (involving further use of the ‘Macbeath trick’).


Introduction
For any positive integers r, s and t, the ordinary (r, s, t) triangle group is the abstract group ∆(r, s, t) with presentation x, y, z | x r = y s = z t = xyz = 1 .
In a recent paper [2], it was shown that if Γ is a finitely-generated Fuchsian group and Σ is a lattice in a connected Lie group, such that Γ and Σ have exactly the same finite quotients, then Γ is isomorphic to Σ. As a consequence, two triangle groups ∆(r, s, t) and ∆(u, v, w) have the same finite quotients if and only if (u, v, w) is a permutation of (r, s, t). The main theorems in the first seven sections of [2], including these, were proved using the theory of profinite groups, without reference to explicit finite quotients. A second proof of the fact concerning triangle groups was given in the final section of [2], with the purpose of exhibiting explicit finite quotients that can distinguish one triangle group from another.
Unfortunately, this second proof (given as the proof of Theorem 8.1 in [2]) was flawed. In particular, Lemma 8.11 is incorrect, even when restricted to triples (r, s, t) and (u, v, w) that survive Lemmas 8.3, 8.4, 8.5 and 8.7, Proposition 8. 6 and Corollaries 8.8 and 8.10, as it fails for the triples (15, 105, 126) and (21, 30, 315) with (q 2 , q 2 ) = (7, 45), as well as for similar triples. Also there were gaps in the argument used in the proof of Theorem 8.1. The situation is somewhat more complicated than was indicated. In particular, the use of direct products of two quotients each isomorphic to PSL(2, p) for some prime p did not cover all possibilities remaining after simpler methods were applied. Please note, however, that the main findings of [2] are unaffected by these flaws, as Section 8 of [2] is independent of the earlier sections.
Here we must sincerely thank Frankie Chan, who pointed out the problems with Lemma 8.11, and provided an effective version of Theorem 8.1 of [2] in his recent PhD thesis [3], while developing an algorithm for distinguishing finite quotients between cocompact triangle groups and lattices of constant-curvature symmetric 2-spaces, with specific attention paid to the case where these lattices are Fuchsian groups.
In this paper we give two new direct proofs of Theorem 8.1 of [2], without requiring Lemma 8.11. The first one is rather intricate, showing how the approach involving direct products of quotients isomorphic to PSL(2, p) for some prime p can be undertaken correctly and extended in some exceptional cases (by using quotients that are direct products of the form PSL(2, p 1 ) × PSL(2, p 2 ) × A where A is cyclic, in those cases).
The second proof is quite a lot shorter, using certain smooth quotients of triangle groups ∆(k, l, m), where k, l and m divide either r, s and t, or u, v and w (in some order). This is an approach we considered earlier and were able to use successfully to deal with the vast majority of triple pairs, but completed only with the help of a key observation made by Frankie Chan in [3] for the remaining pairs, and again we owe a debt of gratitude to him for that.
Here we note that both proofs show that any two non-isomorphic hyperbolic triangle groups ∆(r, s, t) and ∆(u, v, w) can be distinguished by finite quotients that are abelian, dihedral, isomorphic to PSL(2, p) for some prime p, or a direct product PSL(2, p 1 ) × PSL(2, p 2 ) × A where p 1 and p 2 are distinct primes and A is cyclic (or trivial), or an extension of a homocyclic abelian group by one of the preceding groups.
Before giving the two new proofs, we repeat and extend some of the background to this topic, in order to make the paper self-contained.

Further Background
Each triangle group ∆(r, s, t) is called spherical, Euclidean or hyperbolic according to whether the quantity 1/r + 1/s + 1/t is greater than, equal to or less than 1, respectively. Note that ∆(r, s, t) is isomorphic to ∆(u, v, w) if and only if the triple (u, v, w) is a permutation of the triple (r, s, t).
The latter categorisation makes the spherical and Euclidean triangle groups easy to distinguish from others by their finite quotients, and so we will focus our attention on the hyperbolic ones, namely those with 1/r + 1/s + 1/t < 1.
As in [2], we define a finite group G to be (k, l, m)-generated if G can be generated by elements a, b and c of (precise) orders k, l and m such that abc = 1. In this case we say that G is a smooth quotient of the triangle group ∆(k, l, m), noting that the corresponding epimorphism from ∆(k, l, m) to G preserves the orders of the 829 canonical generators of ∆(k, l, m), and has torsion-free kernel. For any hyperbolic triple (k, l, m), the set of (k, l, m)-generated groups is non-empty because of the residual finiteness of the hyperbolic triangle group ∆(k, l, m), but in most cases ∆(k, l, m) can also have 'non-smooth' quotients, in which the orders of the canonical generators of ∆(k, l, m) are not preserved.
We will make use of the following important but relatively straightforward observation, the proof of which we leave as an exercise for the reader: Lemma 2.1. For any integers k, l and m, all greater than 1, let ∆ be the triangle group ∆(k, l, m). Then the abelianisation ∆/∆ = ∆/[∆, ∆] of ∆ is isomorphic to the direct product C e × C d , where e = lcm(gcd(k, l), gcd(k, m), gcd(l, m)) and d = gcd(k, l, m), and also de = klm lcm(k, l, m). Moreover, ∆/[∆, ∆] is (k , l , m )generated, where k = gcd(k, lm), l = gcd(l, km) and m = gcd(m, kl).
We also make use of the fact that if the finite group G is (k, l, m)-generated, then G is a group of conformal automorphisms of a compact Riemann surface S of genus g, where 2 − 2g = |G| 1 k + 1 l + 1 m − 1 as a consequence of the Riemann-Hurwitz formula. The kernel K of the corresponding smooth homomorphism from ∆(k, l, m) onto G is the fundamental group of S, and is itself a Fuchsian group, with signature (2g; −). In particular, if g ≥ 1 then K is generated by 2g elements a 1 , b 1 , . . . , a g , b g subject to a single defining relation [a 1 , b 1 ] . . . [a g , b g ] = 1. Now for any positive integer n, the subgroup K K (n) generated by the derived subgroup K and the nth powers of all elements of K is characteristic in K and hence normal in ∆(k, l, m), and the quotient ∆(k, l, m)/K K (n) is then isomorphic to an extension by G of an abelian subgroup K/K K (n) of rank 2g and exponent n (and order n 2g ). This observation is often known as 'the Macbeath trick'.
Hence for any (k, l, m)-generated group G, we can construct an infinite family of smooth quotients of ∆(k, l, m), to help distinguish ∆(k, l, m) from other triangle groups.
In what follows, we will also require some information about the groups PSL(2, p), for p prime. When p is odd, the orders of the elements of PSL(2, p) are precisely the divisors of p, p−1 2 and p+1 2 (see [7,Chapter 3.6] for example). The integers p, p−1 and p+1 2 are pairwise coprime, so the order of any non-trivial element of PSL(2, p) divides exactly one of them.
Next, define the L 2 -set of a triple (k, l, m) to be the (unique) set of pairwise coprime positive integers whose least common multiple is the same as that of {k, l, m} and which has the property that each of k, l and m divides exactly one member of that set. For example, if k, l and m are themselves pairwise coprime, then the L 2set of the triple (k, l, m) is just {k, l, m}, while if gcd(k, lm) = 1 but gcd(l, m) > 1 then its L 2 -set is {k, lcm(l, m))}, and if gcd(k, l) > 1 and gcd(l, m) > 1 then its L 2set is {lcm(k, l, m))}. Note that every maximal prime-power divisor of lcm(k, l, m) divides exactly one member of the L 2 -set.
Accordingly, we have the following corollary of Macbeath's theorem (Theorem 2.2): (a) If (k, l, m) is a hyperbolic triple and p is an odd prime, then the group PSL(2, p) is (k, l, m)-generated if and only if every member of the L 2 -set of the triple (k, l, m) is equal to p or a divisor of p±1 2 . (b) Let (k, l, m) be any triple of integers greater than 1. Then for every integer q > 3 that does not divide any of the members of the L 2 -set of (k, l, m), there exists a smooth finite quotient G of the (k, l, m) triangle group such that G has no element of order q. Similarly, for every integer q > 1 that is coprime to 6 and to every member of the L 2 -set of (k, l, m), there exists a smooth finite quotient G of the (k, l, m) triangle group such that G has no non-trivial element of order dividing q. Moreover, when the triple (k, l, m) is hyperbolic, in both cases G can be taken as PSL(2, p) for some prime p > 5.
(c) If two triples of integers greater than 1 have the same least common multiple but different L 2 -sets, then the corresponding triangle groups have different sets of smooth quotients.
Proof. Part (a) is an immediate consequence of Theorem 2.2.
Finally, for part (c), the L 2 -set of one of the two triples must contain an integer q > 3 that does not divide any of the members of the L 2 -set of the other triple, and then the assertion follows from part (b).

Main Theorem and Preliminary Steps for its New Proofs
We can now begin to prove the following, in which we use the notation (r, s, t) (u, v, w) to mean that (r, s, t) is a permutation of (u, v, w).
As a first step, we may suppose that (r, s, t) and (u, v, w) are hyperbolic triples. Next, we present a key observation that generalises part of Lemma 8.4 of [2], followed by combination of the remainder of Lemmas 8.3 to 8.5 and 8.7 of [2] and some further helpful properties.
Proof. This follows easily from the assumption that Γ and Σ have exactly the same finite quotients.
Proof. Parts (a) to (c) follow immediately from Lemma 2.1. For part (d), we may suppose without loss of generality that 1 and let G be any (r, s, t)-generated finite quotient of Γ. Then G is also a quotient of Σ. So now let u , v and w be divisors of u, v and w (respectively) chosen such that G is (u , v , w )-generated, and 1 u + 1 v + 1 w is as small as possible subject to those conditions. Then in particular, Next, for any n coprime to |G|, the largest quotient of Γ that is an extension of an abelian group of exponent n by G has order n 2g |G|, where 2−2g = |G|( 1 r + 1 s + 1 t −1), by comments made in the previous section. On the other hand, the largest quotient of Σ that is an extension of an abelian group of exponent n by G is a smooth quotient of the (u , v , w ) triangle group and so has order n 2g |G|, where 2−2g = |G|( 1 u + 1 v + 1 w − 1). Since Γ and Σ have the same quotients, we find that g = g, The final observation in the previous paragraph now gives us 1 The latter also implies that (u , v , w ) = (u, v, w), and hence that G is (u, v, w)generated. The converse holds by the same argument, with the roles of (r, s, t) and (u, v, w) reversed, and this proves (e).
For part (f), let p be any prime divisor of rst, and let p α , p β and p γ be the largest powers of p dividing r, s and t, ordered in such a way that α ≤ β ≤ γ. Then p α must be the largest power of p dividing gcd(r, s, t), while p β is the largest power of p dividing lcm(gcd(r, s), gcd(r, t), gcd(s, t)), and p γ is the largest power of p dividing lcm(r, s, t). Also p α+β+γ is the largest power of p dividing rst, and so p α+β is the largest power of p dividing rst lcm(r,s,t) . Furthermore, either γ = β, or p γ is the largest power of p dividing the denominator of 1 when the latter is expressed in reduced form. (To verify the last claim, note that rs + rt + st is divisible by p α+β but not p α+β+1 when α ≤ β < γ.) Hence the largest powers of p dividing r,s and t are determined by the quantities gcd(r, s, t), rst lcm(r,s,t) and 1 r + 1 s + 1 t . By parts (a), (b) and (d), these three quantities are the same for the triple (u, v, w), and hence the largest powers of p dividing u, v and w are equal to those for r, s and t, in some order. As this holds for all p, the stated identities follow easily.
Part (g) follows from expanding and simplifying v 2 (s−u)(w −s)−s 2 (v −r)(t−v) using the identities rst = uvw and rs + rt + st = uv + uw + vw from part (f), and noting that those two identities are invariant under all permutations of (r, s, t) and/or (u, v, w).
To prove part (h), let us suppose that r < u, so that r = min{r, s, t, u, v, w}. Then since (r, s, t) and (u, v, w) have no common entry, we have only the following five possibilities, with others ruled out by the identity rst = uvw: The first, second, fourth and fifth of these possibilities can be eliminated, however, The analogous argument works also when u < r. Finally, part (i) follows immediately from part (c) of Corollary 2.3.
At this stage we have enough to prove what happens in certain special cases, some of which were covered by Lemma 8.6 and Corollaries 8.8 and 8.10 of [2]. Proposition 3.4. If u, v and w have divisors u , v and w , all greater than 1, such that at least one of r, s and t is coprime to each of 6, u , v and w , then there exists a finite group that is a quotient of ∆(u, v, w) but not a quotient of ∆(r, s, t). In particular, this holds when one of r, s and t is a power of some prime p > 3 such that none of u, v and w is a power of p. Also the analogous statements hold when (r, s, t) and (u, v, w) are interchanged.
Proof. Suppose k ∈ {r, s, t} is coprime to 6, u , v and w . Then by part (b) of Corollary 2.3, there exists a smooth finite quotient G of the triangle group ∆(u , v , w ) such that G has no non-trivial element of order dividing k, and so G is a quotient of ∆(u, v, w) but cannot be a quotient of ∆(r, s, t). The rest follows easily.
Proposition 3.5. Theorem 3.1 holds in each of the following cases (or its equivalent form when r, s and t are permuted, or (r, s, t) and (u, v, w) are interchanged ): (a) (r, s, t) is one of the exceptional triples (2, 5, 5), (3,4,4), (3,3,5), (3,5,5), (5,5,5); (b) (r, s, t) and (u, v, w) have a common entry ; (c) two or more of (r, s, t) are even; (d) one of r, s, t is coprime to each of the other two ; (e) one of r, s, t is a power of 2 ; (f) one or more of r, s, t, u, v and w is equal to gcd(r, s, t).
Proof. First, case (a) follows from parts (a)  For case (b), suppose for example that t = w. Then rs = rst t = uvw w = uv, and then since rs Hence r and s are the zeroes of the same quadratic x 2 − bx + c as u and v (namely with b = u + v and c = uv), so {r, s} = {u, v}, and then (r, s, t) (u, v, w). The same argument works for all other coincidences between entries of (r, s, t) and (u, v, w).
Case (c) follows from the fact that for every integer m ≥ 2, the only triangle group having the dihedral group D m of order 2m as a smooth quotient is ∆(2, 2, m). For suppose that two or more of r, s and t are even. Then lcm(gcd(r, s), gcd(r, t), gcd(s, t)) is even, and therefore so is lcm(gcd(u, v), gcd(u, w), gcd(v, w)), and hence two or more of u, v and w are even. Also if all three of r, s and t are even, then gcd(r, s, t) is even, and then so is gcd(u, v, w), and hence all three of u, v and w are even. Now let m = max(r, s, t, u, v, w) if all three of r, s and t are even, or otherwise let m be the largest odd integer among r, s, t, u, v and w. Then the dihedral group D m is a (2, 2, m)-generated quotient of Γ or Σ, and hence must also be a (2, 2, m)-generated quotient of the other. By definition of m, and the fact mentioned at the start of this paragraph, it follows that m appears in both triples (r, s, t) and (u, v, w), and hence by case (b) we know that (r, s, t) (u, v, w).
In case (d), suppose first that gcd(r, st) = 1. Then the L 2 -set of (r, s, t) is either {r, s, t} or {r, st}, and is the same as the L 2 -set of (u, v, w), by part (i) of Lemma 3.3. It follows that each of u, v and w divides r or st. If one of them divides r and the other two divide st, then since gcd(r, st) = 1, one of them is equal to r and then (r, s, t) (u, v, w) by case (b). On the other hand, suppose two of them divide r, while the other one divides st and hence is coprime to the other two. Then A = lcm(gcd(u, v), gcd(u, w), gcd(v, w)) divides r, while B = lcm(gcd(r, s), gcd(r, t), gcd(s, t)) = gcd(s, t) which divides st, and then because gcd(r, st) = 1, and A = B (by part (c) of Lemma 3.3), we find A = B = 1, and so gcd(s, t) = 1. Thus r, s and t are pairwise coprime, and it follows that the L 2 -sets of (r, s, t) and (u, v, w) are both equal to {r, s, t}, and therefore (r, s, t) (u, v, w). The other two possibilities gcd(s, rt) = 1 and gcd(t, sr) = 1 can be handled in the same way.
Case (e) now follows immediately from cases (c) and (d). Finally, for case (f), suppose that d = gcd(r, s, t) = r, say, with r ≤ s ≤ t and u ≤ v ≤ w. If (r, s, t) and (u, v, w) have no common entry, then r < u ≤ v < s ≤ t < w by part (h) of Lemma 3.3, and then uvw = rst = dst, so st = uvw d , and then because t ≥ s ≥ v + 1 it follows that rs while on the other hand, if u < 3d then u = 2d, and therefore 2d(v + 1) = u(v + 1) > uv and uvw d = 2vw ≥ uw + vw, which together imply that rs + rt + st ≥ 2d(v + 1) + uvw d > uv + uw + vw, another contradiction. Thus (r, s, t) and (u, v, w) have a common entry, and case (b) applies.

The First New Proof, using Direct Products
For our first new proof of Theorem 3.1, we may suppose that (r, s, t) and (u, v, w) are non-exceptional hyperbolic triples for which Γ = ∆(r, s, t) and Σ = ∆(u, v, w) have exactly the same finite quotients, and hence satisfy the conclusions of Lemma 3.3 but not the principal hypothesis of Proposition 3.4, and do not satisfy any of the sufficient conditions (b) to (f) given in Proposition 3.5.
In particular, we suppose that (r, s, t) and (u, v, w) have no entry in common, no entry that is a power of 2, no entry equal to gcd(r, s, t), and no more than one even entry each, and also that no element of one of the triples is coprime to each of the other two, and no element of one of the triples is coprime to 6 and to proper divisors of the elements of the other triple. Furthermore, we let M = lcm(r, s, t) = lcm(u, v, w), the maximal prime-power divisors of which will be a key to what follows.
For interested readers, we also give an indication of the relative numbers of triple-pairs in each situation considered, among the set T of 542970 distinct triplepairs {(r, s, t), (u, v, w)} satisfying the hypotheses two paragraphs above, with 2 ≤ r < u ≤ v < s ≤ t < w and rst = uvw ≤ 12000000d 3 , where d = gcd(r, s, t) = gcd(u, v, w). (Here we note that dropping the condition given in Proposition 3.4 would add another 830030 triple-pairs, which gives an indication of the importance of that condition.) We consider four separate cases, which depend on the distribution of the maximal prime-power divisors of M among the divisors of the integers r, s, t, u, v and w, and we derive contradictions in all four cases. To do this, we consider a wider class of finite quotients of Γ and Σ, namely direct products of the form G = Q 1 × Q 2 or Q 1 × Q 2 × A, where each Q i is PSL(2, p i ) for some prime p i , and A is abelian (indeed cyclic of order d = gcd(r, s, t)).
In the first three of our four cases, Q 1 and Q 2 will be determined by a 'factorisation' of one of the triples, say (r, s, t), as a kind of product of two triples (r 1 , s 1 , t 1 ) and (r 2 , s 2 , t 2 ) of integers greater than 1, with the following properties: (a) lcm(r 1 , r 2 ) = r and lcm(s 1 , s 2 ) = s and lcm(t 1 , t 2 ) = t, (b) Q 1 and Q 2 are smooth quotients of the triangle groups ∆(r 1 , s 1 , t 1 ) and ∆(r 2 , s 2 , t 2 ), respectively, so that Q 1 × Q 2 is a smooth quotient of Γ = ∆(r, s, t), but (c) at least one of u, v and w is not the order of some element of Q 1 × Q 2 , and therefore (d) Q 1 × Q 2 is not a smooth quotient of Σ = ∆(u, v, w).
In the fourth case, we do the same but using three triples (r 1 , s 1 , t 1 ), (r 2 , s 2 , t 2 ) and (r 3 , s 3 , t 3 ), and a smooth abelian quotient A of ∆(r 3 , s 3 , t 3 ), such that lcm(r 1 , r 2 , r 3 ) = r and lcm(s 1 , s 2 , s 3 ) = s and lcm(t 1 , t 2 , t 3 ) = t, but at least one of u, v and w is not the order of some element of Q 1 × Q 2 × A. In this case Q 1 × Q 2 × A is a smooth quotient of Γ but not one of Σ. (In fact we take A = C d and (r 3 , s 3 , s, t).) The first two cases cover the vast majority of the triple-pairs in our set T of 'small' triples, but each of them has a special sub-case which despite involving no triple-pairs from T , appeared to need special treatment. (It may be possible to simplify the proof for those two cases by a more thorough application of the condition rs + rt + st = uv + uw + vw.) Case (1): Suppose that some maximal prime-power divisor of M = lcm(r, s, t) greater than 3 divides just one of r, s and t, and hence also divides just one of u, v and w.
In this case, let q be the largest such maximal prime-power divisor of M . By swapping the triples (r, s, t) and (u, v, w) and/or re-ordering each one if necessary, we may suppose that q divides both t and w but divides none of r, s, u and v, and that t q < w q , noting that t = w because (r, s, t) and (u, v, w) have no entry in common.
When this happens, let m = q, let p be the prime divisor of m, so that m = p γ , say, and let m = M q . Then m m = M and gcd(m, m ) = 1, with m = q > 1 and also m ≥ w q > 1. Furthermore, t = p, for otherwise t = m = q, and so t is coprime to each of r and s.
Next, let (r 1 , s 1 , t 1 ) and (r 2 , s 2 , t 2 ) be the triples defined for each x ∈ {r, s, t} as follows: Then clearly x 1 > 1 and x 2 > 1 and lcm(x 1 , x 2 ) = x for all x ∈ {r, s, t}. Moreover, each x i is a divisor of either m or m , and in particular, t 1 = m and t 2 = t m or p, but on the other hand, none of r 2 , s 2 and t 2 is divisible by m. (To see this, note that if x divides m, and p = x 2 = m, then x = m = p and so t = x = p, but we showed above that t = p.) Hence the L 2 -set of (r 1 , s 1 , t 1 ) is {m, b} or {m, b, c} where b and c are divisors of m , while the prime-power m = p γ divides no member of the L 2 -set of (r 2 , s 2 , t 2 ).
In fact, the L 2 -set L of (r 2 , s 2 , t 2 ) has one of the following forms, where α satisfies 1 ≤ α < γ, and k and l are prime divisors of m : Now if w m divides no element of the L 2 -set of (r 2 , s 2 , t 2 ), then primes p 1 and p 2 can be found such that Q 1 = PSL(2, p 1 ) and Q 2 = PSL(2, p 2 ) are (r 1 , s 1 , t 1 )-and (r 2 , s 2 , t 2 )-generated, but Q 2 contains no element of order divisible by m or w m . It then follows that Q 1 × Q 2 is (r, s, t)-generated, but Q 1 × Q 2 contains no element of order divisible by m w m = w, and hence Q 1 × Q 2 cannot be (u, v, w)-generated, a contradiction.
Assume the contrary. Then since w m divides neither p α nor t m , it must divide k or l, and so we may suppose that w = mk, where k is a prime divisor of m but not one of t m and hence not one of t. Also because k ∈ L we may suppose from the definition of (r 2 , s 2 , t 2 ) that at least one of r and s divides m and is divisible by k (but not by p). Moreover, if every x ∈ {r, s} with this property were divisible by another prime divisor k of m , then we could re-define x 2 as k , and thereby alter L so it contains no element divisible by k = w m . Hence we may suppose that one of r and s, say r, is a power of k.
Next, because k is coprime to t, but r is not coprime to both s and t, we find that s is divisible by k as well. Also gcd(r, s) is divisible by k, so k is odd (because r and s cannot both be even), and gcd(u, v, w) = gcd(r, s, t) = 1 because r is coprime to t, and then since gcd(r, s) = k we find by part (c) of Lemma 3.3 that exactly two of u, v and w are divisible by k. Hence we may suppose without loss of generality that v is divisible by k, but u is not. In particular, if we let k λ be the largest power of k dividing M = lcm(r, s, t), then the k-part of v is k λ , and the k-parts of r and s are k and k λ in some order. Now assume for the moment that s is coprime to p (and hence to m), and let f = gcd(s, t). Then f > 1, because t is coprime to k and hence to r, and so cannot be coprime to s. But then f is odd (as it divides both s and t), so f ≥ 3, and f is coprime to p (as f divides s), and so km = w > t ≥ mf ≥ 3m, which implies that k ≥ 5. Moreover, f must divide two of u, v and w, but is coprime to w, so f divides v, which then cannot be a power of k. Hence Proposition 3.4 applies to r, because r (= k or k λ ) is coprime to 6 while none of u, v and w is a power of k. This contradiction shows that gcd(p, s) > 1, and puts us in case (b) of the possibilities for the set L.
In particular, s must be divisible by p α , and so p is odd (because t is divisible by p as well), and also p α divides exactly one of u and v (because it divides s and t and w but not r).
Thus (r, s, t) = (k, k λ p α s , mt ) or (k λ , kp α s , mt ) for some positive integers s and t , both coprime to k and m, and (u, v, w) = (p α u , k λ v , km) or (u , p α k λ v , km) for some positive integers u and v , both coprime to k and m. Also k 1+λ p α s mt = rst = uvw = p α k 1+λ u v m and therefore s t = u v .
Case (2): Suppose that some maximal prime-power divisor q of M = lcm(r, s, t) divides just one of r, s and t, and just one of u, v and w, but the largest such q is 2 or 3.
Here we may suppose that q divides t and w, but divides none of r, s, u and v, while every maximal prime-power divisor of M greater than 3 divides at least two of r, s and t, and least two of u, v and w. Also neither t nor w can be equal to q, for otherwise case (d) of Proposition 3.5 would hold, and so each of them is a proper multiple of q.
We claim that t and w are not divisible by the same maximal prime-power divisors of M. Assume the contrary, and that (say) t < w. Then w has a non-trivial prime-power divisor k = q such that k does not divide t. If k > q then k cannot be a maximal prime-power divisor of M (for otherwise by assumption k would divide t), and so k strictly divides some maximal prime-power divisor of M , which then divides both u and v. But this implies that k divides all three of u, v and w, and so divides gcd(u, v, w) = gcd(r, s, t), a contradiction (again since k does not divide t). Hence the only possibility is k = 2, giving q = 3, and then (t, w) = (3c, 6c} for some odd c. Also one of r and s must be even, say s, so s = 2b for some b coprime to 6. Now 6rbc = rst = uvw = 6uvc, and so rb = uv, which is coprime to 6, and yet 2rb ≡ rs ≡ rs + rt + st ≡ uv + uw + vw ≡ uv ≡ rb mod 3, from which it follows that 3 divides 2rb − rb = rb, a contradiction. Hence we may suppose that there exists a maximal prime-power divisor m of M that divides w but not t. Then since m does not divide t, it divides r and s, so cannot be even, and as it does not divide gcd(r, s, t) = gcd(u, v, w), it divides just one of u and v, say v. It follows that m = 3, for otherwise q = 2 and then u is coprime to 6 while r, s and t are divisible by 3, 3 and 2, so Proposition 3.4 applies to u. Thus m is odd and m > 3.
With this choice of m, again let m = M m , so m m = M and gcd(m, m ) = 1. This time m is divisible by w m and hence by q, and so gcd(t, m ) is divisible by q, giving gcd(t, m ) > 1. Now let p be the prime divisor of m, and let (r 1 , s 1 , t 1 ) = (m, m, gcd(t, m )), and let r 2 = p or r m , depending on whether or not r = m, and let s 2 = p or s m , depending on whether or not s = m, and let t 2 be gcd(t, m) or the smallest prime divisor of t q , depending on whether or not gcd(t, m) > 1. Then x 1 > 1 and x 2 > 1 and lcm(x 1 , x 2 ) = x for all x ∈ {r, s, t}. Note also that r 2 = p and s 2 = p when m = p, because otherwise r or s is equal to p and then Proposition 3.4 applies to r or s.
Next, the L 2 -set of (r 1 , s 1 , t 1 ) is {m, gcd(t, m )}, while the L 2 -set of (r 2 , s 2 , t 2 ) contains no element divisible by m or q, and hence no element of order w m = q w qm , since none of r 2 , s 2 and t 2 is divisible by m or q. It follows that unless w m = 2 or 3, there exist primes p 1 and p 2 such that Q 1 = PSL(2, p 1 ) and Q 2 = PSL(2, p 2 ) are (r 1 , s 1 , t 1 )-and (r 2 , s 2 , t 2 )-generated, respectively, but Q 1 contains no element of order mq, and Q 2 contains no element of order divisible by m or w m . When that happens, Q 1 × Q 2 is (r, s, t)-generated but has no element of order m w m = w, and so Q 1 × Q 2 cannot be (u, v, w)-generated, a contradiction.
(For example, when (r, s, t) = (175, 1225, 1470) and (u, v, w) = (245, 3675, 350), with M = 2 · 3 · 25 · 49, we take q = 2, m = 25 and m = 294, and then (r 1 , s 1 , t 1 ) = (25, 25, 294) and (r 2 , s 2 , t 2 ) = (7,49,5), with L 2 -sets {25, 294} and {5, 49}, and we can choose p 2 so that PSL(2, p 2 ) has elements of order 5 and 49 but no element of order m = 25 or w m = 14.) Case (2) special sub-case: To complete case (2), we suppose that w m = 2 or 3. Then we have w = qm = 2m or 3m, while every maximal prime-power divisor of M not in {2, q, m} divides both u and v. So now let B be the product of those other maximal prime-power divisors of M . Then B is odd and divides both u and v, while q does not divide u or v, and since m divides at least two of u, v and w, we may suppose without loss of generality that v = mB and u = p α B for some α ≥ 0.
It follows that (r, s, t) = (mr , ms , qp α t ) for some divisors r , s and t of B, while (u, v, w) = (p α B, mB, qm). Then from qp α m 2 r s t = rst = uvw = qp α m 2 B 2 we find that r s t = B 2 . Also each of r , s and t must be greater than 1, since for example if r = 1 then s t = B 2 and so s = t = B, but then s = mB = v, which is impossible. Moreover, p α < m, for otherwise t = qp α t ≥ qm = w. Now if α > 0, then we can take (u 1 , v 1 , w 1 ) = (B, B, q) and (u 2 , v 2 , w 2 ) = (p α , m, m), which have L 2 -sets {q, B} and {m} respectively, and then find primes p 1 and p 2 such that Q 1 = PSL(2, p 1 ) and Q 2 = PSL(2, p 2 ) are (u 1 , v 1 , w 1 )-and (u 2 , v 2 , w 2 )-generated, but so that Q 1 has no element of order qt , and Q 2 has no element of order p α t or p α q. It then follows that Q 1 × Q 2 is (u, v, w)-generated, but has no element of order qp α t = t and so cannot be (r, s, t)-generated, a contradiction.
Thus α = 0, which gives u = B and t = qt . In particular, u = B is coprime to qm = w, and so gcd(r, s, t) = gcd(u, v, w) = 1. (It also follows that B is divisible by 3, for otherwise Proposition 3.4 applies to u, with r, s and t being divisible by m, m and q, and hence in particular, q = 2; but we do not need to know these things.) Next, consider the maximal prime-power divisors of M that divide B. Every such divisor k is coprime to at least one of r , s and t , because 1 = gcd(r, s, t) = gcd(r , s , t ), and so must divide exactly two of them. It follows that there exist pairwise coprime odd positive integers f , g and h such that (r , s , t ) = (f g, f h, gh), namely f = gcd(r , s ), g = gcd(r , t ) and h = gcd(s , t ), so that B = f gh. Moreover, g > 1, for otherwise we find that v = mB = mf gh = mf h = ms = s, and similarly h > 1. Also f > 1, for otherwise (r, s, t) = (mg, mh, qgh) and (u, v, w) = (gh, mgh, qm) and then since at least two of m, g and h are coprime to 3 and hence to 6, we find that Proposition 3.4 applies to r or s or u.
But now it follows that we can take (r 1 , s 1 , t 1 ) = (mf, mf, gh) and (r 2 , s 2 , t 2 ) = (g, h, q), which have L 2 -sets {mf, gh} and {q, g, h} respectively, and then find primes p 1 and p 2 such that Q 1 = PSL(2, p 1 ) and Q 2 = PSL(2, p 2 ) are (r 1 , s 1 , t 1 )and (r 2 , s 2 , t 2 )-generated, but Q 1 has no element of order mf g or mf h, and Q 2 has no element of order mf or gh. Then Q 1 × Q 2 is (r, s, t)-generated, but has no element of order mf gh = mB = v and so cannot be (u, v, w)-generated, a final contradiction.
Remarks: Cases (1) and (2) considered above apply to 535695 and 6507 of the 542695 triple-pairs in the set T given earlier, leaving just 768 of those triple-pairs to be covered.
For the remaining possibilities, we may suppose that every maximal prime-power divisor k of M divides two or more of r, s and t, and two or more of u, v and w. In particular, every such k is odd, and therefore M is odd, so each of r, s, t, u, v and w is odd. Now for the moment, assume that gcd(r, s, t) = gcd(u, v, w) = 1. Then each prime divisor k of r must be coprime to one of s and t, say s, and furthermore, if m = k λ is the maximum power of k dividing M , then m must divide r and t but be coprime to s. The analogous argument works for each of s, t, u, v and w, and hence every one of r, s, t, u, v and w is a product of maximal prime-power divisors of M . Next, let x be the smallest one of r, s, t, u, v and w that is not divisible by 3, and suppose without loss of generality that x ∈ {u, v, w}. Then x is coprime to 6, and cannot be a multiple of any of r, s and t, and so each of r, s and t is divisible by some odd prime that does not divide x, but in that case Proposition 3.4 applies to x, a contradiction.
Thus gcd(r, s, t) = gcd(u, v, w) > 1, and in particular, no two of r, s and t are coprime, and the same holds for u, v and w.
Case (3): Suppose that no maximal prime-power divisor q of M = lcm(r, s, t) divides just one of r, s and t (or just one of u, v and w), but that some such q divides exactly two of r, s and t, and is coprime to the third.
In this case, q is coprime to gcd(r, s, t) = gcd(u, v, w), and hence also q divides two of u, v and w and is coprime to the third. By swapping the triples (r, s, t) and (u, v, w) and/or re-ordering each one if necessary, we may suppose that gcd(r, q) = gcd(u, q) = 1, so that s, t, v and w are the elements of {r, s, t, u, v, w} divisible by q. Also we may suppose also that w is the largest of these. Now let m = q and m = M q , and define the triples (r 1 , s 1 , t 1 ) and (r 2 , s 2 , t 2 ) as follows: • s 1 = m and s 2 = s m = gcd(s, m ), • t 1 = t m = gcd(t, m ) and t 2 = m, and • r 1 = gcd(r, t 1 ) and r 2 = gcd(r, s 2 ).
It follows that there exist primes p 1 and p 2 such that Q 1 = PSL(2, p 1 ) and Q 2 = PSL(2, p 2 ) are (r 1 , s 1 , t 1 )-and (r 2 , s 2 , t 2 )-generated, but each of Q 1 and Q 2 has no element of order k such that k divides M = mm and is strictly divisible by m.
In particular, Q 1 × Q 2 is (r, s, t)-generated. But on the other hand, the only orders of elements of Q 1 × Q 2 that are divisors of M and strictly divisible by m are divisors of ms 2 (= s) or t 1 m (= t), or perhaps 3m, and then because w is greater than 3m, s and t, it follows that Q 1 × Q 2 has no element of order w, and so Q 1 × Q 2 is not (u, v, w)-generated, a contradiction.
(For example, when (r, s, t) = (105, 585, 819) and (u, v, w) = (315, 117, 1365), with M = 9 · 5 · 7 · 13, we can take m = 13 and m = 315, and then (r 1 , s 1 , t 1 ) = (21, 13, 63) and (r 2 , s 2 , t 2 ) = (15, 45, 13), with L 2 -sets {13, 63} and {13, 45}, and we can choose p 1 and p 2 so that Q 1 × Q 2 has no element of order w = 13 · 105.) Remarks: This case applies to 766 of the 542695 triple-pairs in our set T , leaving just 768 − 766 = 2 triple-pairs in T that need to be covered by case (4). A computation using Magma [1] shows that in these two cases, there is no direct product of the form PSL(2, p 1 ) × PSL(2, p 2 ) that is a smooth quotient of one of ∆(r, s, t) and ∆(u, v, w) but not the other, and thereby explains why we need to consider direct products of three quotients for some triple-pairs. Case (4): Suppose that every maximal prime-power divisor of M = lcm(r, s, t) divides exactly two of r, s and t but is not coprime to the third (and hence also divides exactly two of u, v and w but is not coprime to the third).
In this case, every prime divisor of M divides all six of r, s, t, u, v and w, and hence divides d = gcd(r, s, t) = gcd(u, v, w) > 1. On the other hand, none of those six is equal to d, because otherwise case (j) of Proposition 3.5 applies. Also by swapping and/or re-ordering the triples (r, s, t) and (u, v, w), we may suppose that w = max({r, s, t, u, v, w}). Now let D be the product of all maximal prime-power divisors of M that divide d, let X be the set of all maximal prime-power divisors of M that do not divide d, and let E be their product. Then D divides d, and M = DE with gcd(D, E) = 1. Also every q ∈ X divides exactly two of r, s and t but does not divide the third, and divides exactly two of u, v and w but does not divide the third. Moreover, no