Simple groups, generation and probabilistic methods

Transcription

1 Simple groups, generation and probabilistic methods Tim Burness Groups St Andrews University of Birmingham August 6th 2017

2 Overview 1: Introduction: Simple groups and generation 2: Generating subgroups of simple groups 3 3: 2-generation, spread and probabilistic methods 4: The generating graph of a finite group

3 Part 1: Simple groups and generation

4 Let G be a finite group and define d(g) = min{ S : G = S }

5 Let G be a finite group and define d(g) = min{ S : G = S } Note. Subgroups may need many more generators! Example. (Z 2 ) n = (1, 2), (3, 4),..., (2n 1, 2n) < S 2n

6 Let G be a finite group and define d(g) = min{ S : G = S } Note. Subgroups may need many more generators! Example. (Z 2 ) n = (1, 2), (3, 4),..., (2n 1, 2n) < S 2n Example. Let p 3 be a prime and consider G = (Z 2 ) p+1 Z p and H = (Z 2 ) p+1 Then H < G is maximal, d(g) = 2 and d(h) = p + 1 = [G : H] + 1.

7 Let G be a finite group and define d(g) = min{ S : G = S } Note. Subgroups may need many more generators! Example. (Z 2 ) n = (1, 2), (3, 4),..., (2n 1, 2n) < S 2n Example. Let p 3 be a prime and consider G = (Z 2 ) p+1 Z p and H = (Z 2 ) p+1 Then H < G is maximal, d(g) = 2 and d(h) = p + 1 = [G : H] + 1. Lemma. If H G then d(h) [G : H] (d(g) 1) + 1.

8 Theorem (Steinberg, 1962). Finite simple groups are 2-generated. Example. A n = (1, 2, 3), (k, k + 1,..., n) with k = 1 or 2. Example. If q 4 then where F q = α. SL 2 (q) = ( α 0 0 α 1 ) ( 1 1, 1 0 )

9 Theorem (Steinberg, 1962). Finite simple groups are 2-generated. Example. A n = (1, 2, 3), (k, k + 1,..., n) with k = 1 or 2. Example. If q 4 then where F q = α. SL 2 (q) = ( α 0 0 α 1 ) ( 1 1, 1 0 ) G is almost simple if T G Aut(T ) for some non-abelian simple T. Theorem (Dalla Volta & Lucchini, 1995). Every almost simple group is 3-generated.

10 Let P k (G) be the probability that k random elements generate G.

11 Let P k (G) be the probability that k random elements generate G. Theorem. For a non-abelian simple group G: (a) P 2 (G) 53/90 [Menezes, Quick & Roney-Dougal, 2013]

12 Let P k (G) be the probability that k random elements generate G. Theorem. For a non-abelian simple group G: (a) P 2 (G) 53/90 [Menezes, Quick & Roney-Dougal, 2013] (b) P 2 (G) 1 as G [Liebeck & Shalev, 1995]

13 Let P k (G) be the probability that k random elements generate G. Theorem. For a non-abelian simple group G: (a) P 2 (G) 53/90 [Menezes, Quick & Roney-Dougal, 2013] (b) P 2 (G) 1 as G [Liebeck & Shalev, 1995] Main idea. Let M be the set of maximal subgroups of G and suppose x, y G are randomly chosen. If G x, y then x, y H for some H M, and the probability of this event is H 2 / G 2 = [G : H] 2.

14 Let P k (G) be the probability that k random elements generate G. Theorem. For a non-abelian simple group G: (a) P 2 (G) 53/90 [Menezes, Quick & Roney-Dougal, 2013] (b) P 2 (G) 1 as G [Liebeck & Shalev, 1995] Main idea. Let M be the set of maximal subgroups of G and suppose x, y G are randomly chosen. If G x, y then x, y H for some H M, and the probability of this event is H 2 / G 2 = [G : H] 2. Therefore, 1 P 2 (G) H M [G : H] 2 =: Q(G) Now study M and show that Q(G) 0 as G.

15 Part 2: Generating subgroups of simple groups

16 Maximal subgroups Question. Is there a constant c such that d(h) c for every maximal subgroup H of a finite simple group?

17 Maximal subgroups Question. Is there a constant c such that d(h) c for every maximal subgroup H of a finite simple group? Theorem (B, Liebeck & Shalev, 2013). (a) Every maximal subgroup of a simple group is 4-generated. (b) Given ɛ > 0, there exists k = k(ɛ) such that P k (H) > 1 ɛ for any maximal subgroup H of any simple group. In (a), infinitely many examples need 4 generators. The example H = S n 2 < A n shows that (b) is best possible.

18 Maximal subgroups Question. Is there a constant c such that d(h) c for every maximal subgroup H of a finite simple group? Theorem (B, Liebeck & Shalev, 2013). (a) Every maximal subgroup of a simple group is 4-generated. (b) Given ɛ > 0, there exists k = k(ɛ) such that P k (H) > 1 ɛ for any maximal subgroup H of any simple group. In (a), infinitely many examples need 4 generators. The example H = S n 2 < A n shows that (b) is best possible. We get d(h) 6 for any maximal subgroup of an almost simple group. Question. Is the optimal bound d(h) 4?

19 Main ingredients The maximal subgroups H of a given simple group are not known in general, but there are powerful reduction theorems: Alternating groups: O Nan-Scott theorem Classical groups: Aschbacher s theorem Exceptional groups: Results of Liebeck, Seitz and others In particular, either H is known, or H is almost simple

20 Main ingredients The maximal subgroups H of a given simple group are not known in general, but there are powerful reduction theorems: Alternating groups: O Nan-Scott theorem Classical groups: Aschbacher s theorem Exceptional groups: Results of Liebeck, Seitz and others In particular, either H is known, or H is almost simple For H almost simple, d(h) 3 by Dalla Volta & Lucchini. For sporadic groups, we consult the ATLAS. For random generation (part (b)), we apply a deep theorem of Jaikin-Zapirain & Pyber (2011).

21 Example Let H be a maximal subgroup of S n or A n. Lemma. We have d(s k S n k ) = d(agl m (p)) = d(s k S t ) = 2. In particular, d(h) 3 if H is non-diagonal.

22 Example Let H be a maximal subgroup of S n or A n. Lemma. We have d(s k S n k ) = d(agl m (p)) = d(s k S t ) = 2. In particular, d(h) 3 if H is non-diagonal. Suppose H = T k.(out(t ) S k ) is diagonal (k 2, T simple). Then d(h) = max{2, d(out(t ) S k )} by a theorem of Lucchini & Menegazzo (1997), and we get d(h) 4.

23 Example Let H be a maximal subgroup of S n or A n. Lemma. We have d(s k S n k ) = d(agl m (p)) = d(s k S t ) = 2. In particular, d(h) 3 if H is non-diagonal. Suppose H = T k.(out(t ) S k ) is diagonal (k 2, T simple). Then d(h) = max{2, d(out(t ) S k )} by a theorem of Lucchini & Menegazzo (1997), and we get d(h) 4. Example. Suppose T = PΩ + 12 (p2 ), p 3 and k = 2.

24 Example Let H be a maximal subgroup of S n or A n. Lemma. We have d(s k S n k ) = d(agl m (p)) = d(s k S t ) = 2. In particular, d(h) 3 if H is non-diagonal. Suppose H = T k.(out(t ) S k ) is diagonal (k 2, T simple). Then d(h) = max{2, d(out(t ) S k )} by a theorem of Lucchini & Menegazzo (1997), and we get d(h) 4. Example. Suppose T = PΩ + 12 (p2 ), p 3 and k = 2. Then H = T 2.(Out(T ) S 2 ) < A n is maximal (with n = T ) and d(h) = max{2, d(out(t ) S 2 )} = d(d 8 Z 2 Z 2 ) = 4.

25 Application: Primitive groups Let G Sym(Ω) be a finite primitive permutation group with point stabiliser H, so d(g) 1 d(h) [G : H] (d(g) 1) + 1 Question. Is there a constant c such that d(h) d(g) + c?

26 Application: Primitive groups Let G Sym(Ω) be a finite primitive permutation group with point stabiliser H, so d(g) 1 d(h) [G : H] (d(g) 1) + 1 Question. Is there a constant c such that d(h) d(g) + c? Theorem. d(h) d(g) + 4

27 Application: Primitive groups Let G Sym(Ω) be a finite primitive permutation group with point stabiliser H, so d(g) 1 d(h) [G : H] (d(g) 1) + 1 Question. Is there a constant c such that d(h) d(g) + c? Theorem. d(h) d(g) + 4 Example. If G has a regular normal subgroup N then H = G/N and thus d(h) = d(g/n) d(g). Example. If G is almost simple then d(h) 6 d(g) + 4. Question. Is d(h) d(g) + 2 the optimal bound?

28 Deeper subgroups The depth of a subgroup H G is the maximal length of a chain of subgroups from H to G. e.g. H is maximal if and only if it has depth 1. We say H is second maximal if it has depth 2, and so on.

29 Deeper subgroups The depth of a subgroup H G is the maximal length of a chain of subgroups from H to G. e.g. H is maximal if and only if it has depth 1. We say H is second maximal if it has depth 2, and so on. Question. Is there a constant c such that d(h) c for all second maximal subgroups H of simple groups?

30 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G

31 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G with soc(g) {L 2 (q), 2 B 2 (q), 2 G 2 (q)}.

32 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G with soc(g) {L 2 (q), 2 B 2 (q), 2 G 2 (q)}. We can take c = 12, unless G is exceptional and H is maximal in a parabolic subgroup of G (here we take c = 70).

33 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G with soc(g) {L 2 (q), 2 B 2 (q), 2 G 2 (q)}. We can take c = 12, unless G is exceptional and H is maximal in a parabolic subgroup of G (here we take c = 70). There is a second maximal subgroup H of a simple group G with d(h) = :

34 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G with soc(g) {L 2 (q), 2 B 2 (q), 2 G 2 (q)}. We can take c = 12, unless G is exceptional and H is maximal in a parabolic subgroup of G (here we take c = 70). There is a second maximal subgroup H of a simple group G with d(h) = : Take q = and

35 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G with soc(g) {L 2 (q), 2 B 2 (q), 2 G 2 (q)}. We can take c = 12, unless G is exceptional and H is maximal in a parabolic subgroup of G (here we take c = 70). There is a second maximal subgroup H of a simple group G with d(h) = : Take q = and G = PSL 2 (q)

36 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G with soc(g) {L 2 (q), 2 B 2 (q), 2 G 2 (q)}. We can take c = 12, unless G is exceptional and H is maximal in a parabolic subgroup of G (here we take c = 70). There is a second maximal subgroup H of a simple group G with d(h) = : Take q = and G = PSL 2 (q) > (Z 2 ) Z q 1

37 Theorem (B, Liebeck & Shalev, 2016). There is a constant c s.t. d(h) c for all second maximal subgroups H of almost simple groups G with soc(g) {L 2 (q), 2 B 2 (q), 2 G 2 (q)}. We can take c = 12, unless G is exceptional and H is maximal in a parabolic subgroup of G (here we take c = 70). There is a second maximal subgroup H of a simple group G with d(h) = : Take q = and G = PSL 2 (q) > (Z 2 ) Z q 1 > (Z 2 ) = H

38 Main ingredients Let H < M < G be second maximal with G almost simple. If M is almost simple then d(h) 6 by [BLS, 2013]

39 Main ingredients Let H < M < G be second maximal with G almost simple. If M is almost simple then d(h) 6 by [BLS, 2013] If core M (H) = m M Hm = 1, then M acts faithfully and primitively on the cosets M/H, so by [BLS, 2013] d(h) d(m) Remaining cases: Study the possibilities for H using the reduction theorems of O Nan-Scott, Aschbacher, Liebeck-Seitz and others.

40 Special primes Question. Is there a constant c such that d(h) c for all second maximal subgroups H of almost simple groups? Theorem. This is equivalent to the following open problem: Are there only finitely many primes k for which there is a prime power q such that (q k 1)/(q 1) is prime?

41 Special primes Question. Is there a constant c such that d(h) c for all second maximal subgroups H of almost simple groups? Theorem. This is equivalent to the following open problem: Are there only finitely many primes k for which there is a prime power q such that (q k 1)/(q 1) is prime? The answer is believed to be no, but existing methods in Number Theory are very far from a proof. Note. The answer is no if there are infinitely many Mersenne primes.

42 Third maximals Lemma. For each c N, there exists a third maximal subgroup H of an almost simple group such that d(h) > c.

43 Third maximals Lemma. For each c N, there exists a third maximal subgroup H of an almost simple group such that d(h) > c. Proof. Let p 5 be a prime such that p ±3 (mod 8).

44 Third maximals Lemma. For each c N, there exists a third maximal subgroup H of an almost simple group such that d(h) > c. Proof. Let p 5 be a prime such that p ±3 (mod 8). Then G = S 2(p+1) > S 2 S p+1

45 Third maximals Lemma. For each c N, there exists a third maximal subgroup H of an almost simple group such that d(h) > c. Proof. Let p 5 be a prime such that p ±3 (mod 8). Then G = S 2(p+1) > S 2 S p+1 > (S 2 ) p+1.pgl 2 (p)

46 Third maximals Lemma. For each c N, there exists a third maximal subgroup H of an almost simple group such that d(h) > c. Proof. Let p 5 be a prime such that p ±3 (mod 8). Then G = S 2(p+1) > S 2 S p+1 > (S 2 ) p+1.pgl 2 (p) > (S 2 ) p+1.s 4 = H

47 Third maximals Lemma. For each c N, there exists a third maximal subgroup H of an almost simple group such that d(h) > c. Proof. Let p 5 be a prime such that p ±3 (mod 8). Then G = S 2(p+1) > S 2 S p+1 > (S 2 ) p+1.pgl 2 (p) > (S 2 ) p+1.s 4 = H is a third maximal subgroup and p + 1 = d((s 2 ) p+1 ) [H : (S 2 ) p+1 ] (d(h) 1) + 1, so d(h) p

48 Part 3: 3 2-generation, spread and probabilistic methods

49 G is 3 2-generated if for any x G \ {1} there exists y G s.t. G = x, y.

50 G is 3 2-generated if for any x G \ {1} there exists y G s.t. G = x, y. Theorem (Guralnick & Kantor, 2000). Every finite simple group is 3 2 -generated.

51 G is 3 2-generated if for any x G \ {1} there exists y G s.t. G = x, y. Theorem (Guralnick & Kantor, 2000). Every finite simple group is 3 2 -generated. G has spread k if for any x 1,..., x k G \ {1} there exists y G such that G = x i, y for all i. Let s(g) 0 be the exact spread of G.

52 G is 3 2-generated if for any x G \ {1} there exists y G s.t. G = x, y. Theorem (Guralnick & Kantor, 2000). Every finite simple group is 3 2 -generated. G has spread k if for any x 1,..., x k G \ {1} there exists y G such that G = x i, y for all i. Let s(g) 0 be the exact spread of G. Theorem (Brenner & Wiegold, 1975). For n 4, s(a 2n ) = 4

53 G is 3 2-generated if for any x G \ {1} there exists y G s.t. G = x, y. Theorem (Guralnick & Kantor, 2000). Every finite simple group is 3 2 -generated. G has spread k if for any x 1,..., x k G \ {1} there exists y G such that G = x i, y for all i. Let s(g) 0 be the exact spread of G. Theorem (Brenner & Wiegold, 1975). For n 4, s(a 2n ) = s(a 19 ) For q 4 even, s(psl 2 (q)) = q 2

54 Spread for simple groups Theorem (Guralnick & Shalev, 2003). Let (G n ) be a sequence of simple groups with G n. Then either s(g n ) ; or (G n ) has a subsequence of alternating groups of degree divisible by a fixed prime, or odd-dimensional orthogonal groups over a field of fixed size.

55 Spread for simple groups Theorem (Guralnick & Shalev, 2003). Let (G n ) be a sequence of simple groups with G n. Then either s(g n ) ; or (G n ) has a subsequence of alternating groups of degree divisible by a fixed prime, or odd-dimensional orthogonal groups over a field of fixed size. Theorem (Breuer, Guralnick & Kantor, 2008). s(g) 2 for every simple group G, with equality iff G = A 5, A 6, Ω + 8 (2) or Ω 2m+1(2) with m 3

56 For x, y G, let Q(x, y) = {z y G : G x, z } y G be the probability that x and a random conjugate of y do not generate G.

57 For x, y G, let Q(x, y) = {z y G : G x, z } y G be the probability that x and a random conjugate of y do not generate G. Lemma. Suppose there exists y G and k N s.t. Q(x, y) < 1/k for all non-trivial x G. Then s(g) k.

58 For x, y G, let Q(x, y) = {z y G : G x, z } y G be the probability that x and a random conjugate of y do not generate G. Lemma. Suppose there exists y G and k N s.t. Q(x, y) < 1/k for all non-trivial x G. Then s(g) k. Proof. For x 1,..., x k G \ {1}, let E i be the event that G = x i, z for a random conjugate z y G. Set E = E 1 E k.

59 For x, y G, let Q(x, y) = {z y G : G x, z } y G be the probability that x and a random conjugate of y do not generate G. Lemma. Suppose there exists y G and k N s.t. Q(x, y) < 1/k for all non-trivial x G. Then s(g) k. Proof. For x 1,..., x k G \ {1}, let E i be the event that G = x i, z for a random conjugate z y G. Set E = E 1 E k. Then P(E) = 1 P(Ē 1 Ē k ) 1 and thus P(E) > 1 k 1 k = 0. k P(Ē i ) = 1 i=1 k Q(x i, y) i=1

60 For H < G and x G, let fpr(x, G/H) = x G H x G be the fixed point ratio of x w.r.t the action of G on G/H. For y G, let M(y) be the set of maximal subgroups of G containing y.

61 For H < G and x G, let fpr(x, G/H) = x G H x G be the fixed point ratio of x w.r.t the action of G on G/H. For y G, let M(y) be the set of maximal subgroups of G containing y. Key Lemma. Suppose there exists y G and k N such that fpr(x, G/H) < 1 k H M(y) for all x G of prime order. Then s(g) k.

62 For H < G and x G, let fpr(x, G/H) = x G H x G be the fixed point ratio of x w.r.t the action of G on G/H. For y G, let M(y) be the set of maximal subgroups of G containing y. Key Lemma. Suppose there exists y G and k N such that fpr(x, G/H) < 1 k H M(y) for all x G of prime order. Then s(g) k. Example. G = A 19, y = 19 = M(y) = {H}, H = AGL 1 (19) G. Then max 1 x G fpr(x, G/H) = = s(g)

63 Example Claim. If G = A n and n 8 is even, then s(g) 3. Set t = n/2 gcd(2, n/2 1) and y = (1,..., t)(t + 1,..., n) G.

64 Example Claim. If G = A n and n 8 is even, then s(g) 3. Set t = n/2 gcd(2, n/2 1) and y = (1,..., t)(t + 1,..., n) G. M(y) = {L} with L = (S t S n t ) G: Let H M(y). Then H intransitive = H = L H imprimitive ruled out by cycle-shape of y H primitive ruled out by Marggraff s Theorem (1889) since y contains a t-cycle and t < n/2

65 Example Claim. If G = A n and n 8 is even, then s(g) 3. Set t = n/2 gcd(2, n/2 1) and y = (1,..., t)(t + 1,..., n) G. M(y) = {L} with L = (S t S n t ) G: Let H M(y). Then H intransitive = H = L H imprimitive ruled out by cycle-shape of y H primitive ruled out by Marggraff s Theorem (1889) since y contains a t-cycle and t < n/2 Easy combinatorial argument with t-sets = fpr(x, G/L) < 1/3 for all x G of prime order Now apply the Key Lemma

66 Conjecture (Breuer, Guralnick & Kantor, 2008). A finite group G is 3 2-generated iff G/N is cyclic for every non-trivial normal subgroup N of G. Note. G/N non-cyclic = no element in N belongs to a generating pair

67 Conjecture (Breuer, Guralnick & Kantor, 2008). A finite group G is 3 2-generated iff G/N is cyclic for every non-trivial normal subgroup N of G. Note. G/N non-cyclic = no element in N belongs to a generating pair G soluble: Brenner & Wiegold, 1975 G simple: Guralnick & Kantor, 2000

68 Conjecture (Breuer, Guralnick & Kantor, 2008). A finite group G is 3 2-generated iff G/N is cyclic for every non-trivial normal subgroup N of G. Note. G/N non-cyclic = no element in N belongs to a generating pair G soluble: Brenner & Wiegold, 1975 G simple: Guralnick & Kantor, 2000 Guralnick: A reduction to almost simple groups. Let G = T, x be an almost simple group with socle T :

69 Conjecture (Breuer, Guralnick & Kantor, 2008). A finite group G is 3 2-generated iff G/N is cyclic for every non-trivial normal subgroup N of G. Note. G/N non-cyclic = no element in N belongs to a generating pair G soluble: Brenner & Wiegold, 1975 G simple: Guralnick & Kantor, 2000 Guralnick: A reduction to almost simple groups. Let G = T, x be an almost simple group with socle T : T alternating or sporadic: Breuer et al., 2008 T = PSLn (q): B & Guest, 2013 T = PSpn (q) or Ω n (q): Harper, 2017

70 Scott s talk Scott Harper (Bristol) 3 2-generation of finite groups Monday, 14:30 15:00 Poynting Building, SLT

71 Part 4: The generating graph of a finite group

72 Let G be a finite group. The generating graph Γ(G) has vertex set G \ {1} and two vertices x, y are joined by an edge if and only if G = x, y. Example. D 8 = a, b a 4 = b 2 = 1, ab = ba 1 : a 2 a a 3 a 3 b b ab a 2 b

73 Γ(G) encodes some interesting generation properties of G, e.g. d(g) 2 Γ(G) is non-empty s(g) 1 Γ(G) has no isolated vertices s(g) 2 = Γ(G) is connected with diameter at most 2

74 Γ(G) encodes some interesting generation properties of G, e.g. d(g) 2 Γ(G) is non-empty s(g) 1 Γ(G) has no isolated vertices s(g) 2 = Γ(G) is connected with diameter at most 2 What is the (co)-clique number of Γ(G)? What is the chromatic number of Γ(G)? Does Γ(G) contain a Hamiltonian cycle? etc. etc.

75 Γ(G) encodes some interesting generation properties of G, e.g. d(g) 2 Γ(G) is non-empty s(g) 1 Γ(G) has no isolated vertices s(g) 2 = Γ(G) is connected with diameter at most 2 What is the (co)-clique number of Γ(G)? What is the chromatic number of Γ(G)? Does Γ(G) contain a Hamiltonian cycle? etc. etc. Example. If G = A 5 then Γ(G) has 59 vertices and 1140 edges. It has clique number 8, coclique number 15 and chromatic number 9.

76 The generating graph of A5

77 Simple groups Theorem. Let G be a non-abelian finite simple group. Γ(G) has no isolated vertices. [Guralnick & Kantor, 2000] Γ(G) is connected and has diameter 2. [Breuer, Guralnick & Kantor, 2008] Γ(G) contains a Hamiltonian cycle if G is sufficiently large. [Breuer, Guralnick, Lucchini, Maróti & Nagy, 2010]

78 Simple groups Theorem. Let G be a non-abelian finite simple group. Γ(G) has no isolated vertices. [Guralnick & Kantor, 2000] Γ(G) is connected and has diameter 2. [Breuer, Guralnick & Kantor, 2008] Γ(G) contains a Hamiltonian cycle if G is sufficiently large. [Breuer, Guralnick, Lucchini, Maróti & Nagy, 2010] Conjecture (BGLMN, 2010). Let G be a finite group with G 4. Then Γ(G) is Hamiltonian iff G/N is cyclic for every non-trivial normal subgroup N of G. Note. Proved in [BGLMN, 2010] for soluble groups.

79 A generalised conjecture Conjecture. The following are equivalent, for any finite group G with G 4: (a) G has spread 1. (b) G has spread 2. (c) Γ(G) has no isolated vertices. (d) Γ(G) is connected. (e) Γ(G) is connected with diameter at most 2. (f) Γ(G) contains a Hamiltonian cycle. (g) G/N is cyclic for every non-trivial normal subgroup N. This would imply that no finite group has exact spread 1. The conjecture is true for soluble groups.