aa r X i v : . [ m a t h . G R ] J u l THE SYM(3) CONJECTURE AND ALT(8)
CECIL ANDREW ELLARD
Abstract.
We give an alternate computer-free proof of a result of Z. Arad,M. Muzychuk, and A. Oliver: if G is a minimal counterexample to the Sym (3)conjecture, then
Soc ( G ) ′ cannot be isomorphic to Alt (8). The
Sym (3)
Conjecture
The
Sym (3) conjecture states that if G is a non-trivial finite group whose con-jugacy classes have distinct cardinalities, then G is isomorphic to Sym (3), thesymmetric group of order 6. With additional assumptions on G , the conclusion isknown to be true: for example when G is supersolvable [Markel, 1973] or when G is solvable [Zhang, 1994]. Let H be the hypothesis of the Sym (3) conjecture:( H ) The group G is a finite, non-trivial group whose conjugacy classes have distinctcardinalities.The authors Z. Arad, M. Muzychuk, and A. Oliver [2004] proved that if G sat-isfies H , then either Soc ( G ) ′ = 1 or Soc ( G ) ′ is isomorphic to one of the following:1. Alt (5) a , for 1 ≤ a ≤ , a = 22. Alt (8)3.
P SL (3 , e , for 1 ≤ e ≤ Alt (5) × P SL (3 , e , for 1 ≤ e ≤ G is a minimal counterexample to the Sym (3) con-jecture, then the list of possibilities for
Soc ( G ) ′ can be shortened; either Soc ( G ) ′ = 1or Soc ( G ) ′ is isomorphic to one of the following:1. Alt (5) a , for 3 ≤ a ≤ P SL (3 , e , for 1 ≤ e ≤ Sym (3) conjecture,
Soc ( G ) ′ cannot be isomorphic to Alt (8). Their proof makes use of computer programs in thealgebra system GAP. The purpose of this note is to give an alternate computer-freeproof of their result for
Alt (8):
Theorem:
Let G be a minimal counterexample to the Sym (3) conjecture. Then
Soc ( G ) ′ is not isomorphic to Alt (8).In what follows, we will let
Sym ( n ) be the symmetric group on n letters, and Alt ( n ) the alternating group on n letters. For any group G , we will let Soc ( G ) be the socle of G (the subgroup generated by the minimal normal subgroups of G ), Z ( G ) the center of G , G ′ the derived subgroup of G , Aut ( G ) the automorphismgroup of G , Inn ( G ) inner automorphism group of G , and Out ( G ) the outer auto-morphism group of G . If X is a subset of G , then C G ( X ) will be the centralizer of X in G , and N G ( X ) will be the normalizer of X in G . If g ∈ G , we will let g G de-note the set of all elements of G that are conjugate in G to g . In other words, g G isthe conjugacy class in G containing g . Euler’s totient function will be denoted by φ .A finite group is said to be a rational group if every ordinary (i.e. complex) characteris rational-valued. That is, for every ordinary character χ of G and for every g ∈ G ,we have χ ( g ) ∈ Q . This is equivalent to only requiring that the irreducible ordinarycharacters of G are rational-valued. We have the following characterization ofrational groups (see [Isaacs], [Serre], or [Ellard, 2013]): Lemma 1.1.
Let G be a finite group. Then the following are equivalent:(i) G is a rational group(ii) For all g ∈ G , all generators of h g i are conjugate in G .(iii) For all g ∈ G , [ N G ( h g i ): C G ( h g i )] = φ ( ord ( g )) . The finite symmetric groups
Sym ( n ) for n ≥ H is a rational group: Lemma 1.2.
Let G be a group which satisfies hypothesis H . Then G is a rationalgroup.Proof. Let G be a group which satisfies hypothesis H , and let g ∈ G . By the abovelemma, it suffices to show that all generators of h g i are conjugate in G . Assumethat g and g are elements of G which generate h g i . We wish to show that g and g are conjugate. Note that C G ( h g i ) = C G ( h g i ). Therefore | g G | = [ G : C G ( g )] = [ G : C G ( h g i )] = [ G : C G ( h g i )] = [ G : C G ( g )] = | g G | and therefore the conjugacy classes g G and g G have the same cardinalities. So byhypothesis H , g G and g G are the same conjugacy class, so g and g are conjugatein G . (cid:3) We also will make use of the fact that if H and K are finite groups, then H × K is rational if and only if both H and K are rational. This is true because theirreducible ordinary characters of H × K are precisely the functions of the form( h, k ) χ ( h ) · χ ( k ) where χ and χ are irreducible ordinary characters of H and K , respectively. [Isaacs]The term Frattini’s Argument refers to the theorem stating that if G is a finitegroup and if N is a normal subgroup of G , and if P is a Sylow p -subgroup of N ,then G = N G ( P ) N . This theorem is often generalized as follows: if G is a finitegroup acting transitively on a set Ω, and if N is a normal subgroup of G whichis also transitive on Ω, then for any ω ∈ Ω, G = Stab G ( ω ) N . One can go a littlefurther with the following: HE SYM(3) CONJECTURE AND ALT(8) 3
Lemma 1.3. (Extended Frattini Argument) Let G be a finite group acting (notnecessarily transitively) on a set Ω , let N be a normal subgroup of G and let ω ∈ Ω .Then:(i) | ω G | / | ω N | = [ G : Stab G ( ω ) N ] (ii) | ω G | / | ω N | = 1 iff G = Stab G ( ω ) N (iii) | ω G | / | ω N | = [ G : N ] iff Stab G ( ω ) ≤ N (iv) | ω G | / | ω N | < [ G : N ] iff Stab G ( ω ) is not a subgroup of N Proof. (i) | Stab G ( ω ) N | = | Stab G ( ω ) || N | / | Stab G ( ω ) ∩ N | = | Stab G ( ω ) || N | / | Stab N ( ω ) | So | Stab G ( ω ) N | / | Stab G ( ω ) | = | N | / | Stab N ( ω ) | Therefore, [
Stab G ( ω ) N : Stab G ( ω )] = [ N : Stab N ( ω )]So | ω G | = [ G : Stab G ( ω )] = [ G : Stab G ( ω ) N ][ Stab G ( ω ) N : Stab G ( ω )] =[ G : Stab G ( ω ) N ][ N : Stab N ( ω )] = [ G : Stab G ( ω ) N ] | ω N | .Therefore, | ω G | / | ω N | = [ G : Stab G ( ω ) N ]. This proves (i).(ii) | ω G | / | ω N | = 1 iff [ G : Stab G ( ω ) N ] = 1 iff G = Stab G ( ω ) N . This proves (ii).(iii) | ω G | / | ω N | = [ G : N ] iff [ G : Stab G ( ω ) N ] = [ G : N ] iff Stab G ( ω ) N = N iff Stab G ( ω ) ≤ N . This proves (iii).(iv) | ω G | / | ω N | < [ G : N ] iff [ G : Stab G ( ω ) N ] < [ G : N ] iff | N | < | Stab G ( ω ) N | iff Stab G ( ω ) is not a subgroup of N . This proves (iv). (cid:3) A particular instance of this Extended Frattini Argument is the following:
Lemma 1.4.
Let G be a finite group, let N be a normal subgroup of G , and let n ∈ N . Then:(i) | n G | / | n N | = [ G : C G ( n ) N ] (ii) | n G | / | n N | = 1 iff G = C G ( n ) N (iii) | n G | / | n N | = [ G : N ] iff C G ( n ) ≤ N (iv) | n G | / | n N | < [ G : N ] iff C G ( n ) is not a subgroup of N Proof.
Let ω = n , let Ω = N , and apply Lemma 1.3 (cid:3) The group
Alt (8) is a simple subgroup of index 2 in
Sym (8), and so we have | Alt (8) | = ·
8! = 20 , Alt (8) has two classes of involutions: the first withcycle structure ( ab )( cd ) and cardinality 210 and the second with cycle structure( ab )( cd )( ef )( gh ) and cardinality 105. Alt (8) also has two classes of elements of order3; cycle structures ( abc ) and ( abc )( def ) with cardinalities 112 and 1 ,
120 respectively.
Alt (8) has two classes of elements of order 4 with cycle structures ( abcd )( ef gh )and ( abcd )( ef ) and cardinalities 1 ,
260 and 2 ,
520 respectively. And finally,
Alt (8)has two classes of elements of order 6 with cycle structures ( abcdef )( gh ) and( abc )( de )( f g ) and cardinalities 3 ,
360 and 1 ,
680 respectively.
CECIL ANDREW ELLARD
Note that
Alt (8) has only one class of cardinality 1 , a insuch a class would satisfy | C Alt (8) ( a ) | = 12, and therefore the order of a would bea divisor of 12. The order of a cannot be 1, and it cannot be 12 since Alt (8) hasno element of order 12. Therefore the order of a is 2 , , , or 6. But among theseorders, from the previous paragraph, only one has a class of cardinality 1 , abc )( de )( f g ) and order 6. Similarly, Alt (8)has only one class of cardinality 2 , a in such a class wouldsatisfy | C Alt (8) ( a ) | = 8, and therefore the order of a would be a divisor of 8. Theorder of a cannot be 1, and it cannot be 8 since Alt (8) has no element of order8. So the order of a is 2 or 4. But among these orders, only one has a class ofcardinality 2 , abcd )( ef ) and order 4. Lemma 1.5.
Alt (8) is not a rational group.Proof.
Let a = (1234567), an element of order 7 in Alt (8). By the previous char-acterization of rational groups, it suffices to show that[ N Alt (8) ( h a i ): C Alt (8) ( h a i )] = φ ( ord ( a )).We have C Alt (8) ( h a i ) = h a i , and so | C Alt (8) ( h a i ) | = 7. Therefore,[ Alt (8): C Alt (8) ( h a i )] = ( ≡ mod h a i is a Sylow 7-subgroup of Alt (8) and so by a Sylow theorem we have[
Alt (8): N Alt (8) ( h a i )] ≡ mod N Alt (8) ( h a i ): C Alt (8) ( h a i )] ≡ mod N Alt (8) ( h a i ): C Alt (8) ( h a i )] = 6 = φ (7) = φ ( ord ( a )). (cid:3) Note that since [ N Alt (8) ( h a i ): C Alt (8) ( h a i )] is a divisor of φ ( ord ( a )) = 6 and con-gruent to 3 ( mod Alt (8) does not induce an automor-phism of order 2 by conjugation on h a i . Therefore, a is not conjugate to a − in Alt (8).Since
Alt (8) is not a rational group, it follows that
Alt (8) cannot satisfy hypothesis H . Also, Sym (8) does not satisfy hypothesis H ; in fact the permutations (123456)and (123456)(78) have the same centralizer in Sym (8) (and therefore their conju-gacy classes have the same cardinality) but since they have different cycle structures,they are not conjugate in
Sym (8). Finally, we note that
Aut ( Alt (8)) ∼ = Sym (8),and therefore | Out ( Alt (8)) | = 2.We can now prove the main theorem: Theorem 1.6.
Let G be a minimal counterexample to the Sym (3) conjecture. Then
Soc ( G ) ′ is not isomorphic to Alt (8) .Proof.
Let G be a minimal counterexample to the
Sym (3) conjecture; so G satisfiesthe hypothesis H , but G is not isomorphic to Sym (3) and G is a group of least car-dinality satisfying hypothesis H but not isomorphic to Sym (3). We wish to prove
HE SYM(3) CONJECTURE AND ALT(8) 5 that
Soc ( G ) ′ is not isomorphic to Alt (8). Assume to the contrary that
Soc ( G ) ′ isisomorphic to Alt (8). We wish to derive a contradiction.Let A = Soc ( G ) ′ and let B = C G ( A ). A = Soc ( G ) ′ char Soc ( G ) E G , and so A E G . Also, B = C G ( A ) E N G ( A ) = G , and so B E G . Therefore A ∩ B E A ,so since A is simple, A ∩ B = 1 or A ∩ B = A . But A ∩ B = A would imply A ≤ B = C G ( A ) which would imply A was abelian. So A ∩ B = 1. Therefore B isa proper subgroup of G and AB ∼ = A × B .Claim: B is finite, and B = 1.Since G is finite and B is a subgroup of G , B is finite, too. The homomorphism θ : G → Aut ( A ) induced by conjugation has kernel B and maps AB onto Inn ( A ),and so G/B ∼ = Im ( θ ) is a subgroup of Aut ( A ) which contains Inn ( A ). Since Aut ( A ) ∼ = Sym (8) and
Inn ( A ) ∼ = Alt (8), it follows that either
G/B ∼ = Alt (8) or
G/B ∼ = Sym (8). Since neither
Alt (8) nor
Sym (8) satisfy hypothesis H , it followsthat we cannot have B = 1.Claim: [ G : AB ] = 2.Since θ maps AB onto Inn ( A ), then by the correspondence theorem we have[ G : AB ] = [ Im ( θ ): Inn ( A )], which divides [ Aut ( A ): Inn ( A )] = [ Sym (8):
Alt (8)] = 2,and so [ G : AB ] divides 2. If [ G : AB ] were to equal 1, we would have G ∼ = A × B ;but since G is a rational group, this would imply that A × B , and therefore both A and B , were rational groups. But A is not a rational group, and so we must have[ G : AB ] = 2.Claim: C G ( B ) ≤ AB .Suppose not. We wish to derive a contradiction. Then since [ G : AB ] = 2, wewould have G = C G ( B ) AB . But since B = C G ( A ), we have A ≤ C G ( B ), and so G = C G ( B ) AB = C G ( B ) B .We will consider two cases: Z ( B ) = 1 and Z ( B ) = 1.Case 1: Assume that Z ( B ) = 1. We have just shown that G = C G ( B ) B . Also, C G ( B ) ∩ B = Z ( B ) = 1. Note that C G ( B ) E N G ( B ) = G and so both C G ( B )and B are normal in G . Thus G ∼ = C G ( B ) × B . In the proof of our first claimabove, we have shown that either G/B ∼ = Alt (8) or
G/B ∼ = Sym (8). But since | G/B | = | G | / | B | = 2 | A || B | / | B | = 2 | A | = 2 | Alt (8) | , we must have G/B ∼ = Sym (8).So C G ( B ) ∼ = Sym (8) and so G ∼ = Sym (8) × B . Sym (8) does not satisfy hypothesis H ; in fact the permutations s = (123456) and s = (123456)(78) have the samecentralizer in Sym (8) but since they have different cycle structures they are notconjugate in
Sym (8), and therefore, since G ∼ = Sym (8) × B , they are not conjugatein G . Therefore, C G ( s ) = C Sym (8) ( s ) × B = C Sym (8) ( s ) × B = C G ( s ) CECIL ANDREW ELLARD so | s G | = [ G : C G ( s )] = [ G : C G ( s )] = | s G | .But since G satisfies hypothesis H , and since the conjugacy classes s G and s G havethe same cardinality, they must be the same conjugacy class, and so s and s mustbe conjugate in G , a contradiction.Case 2: Assume that Z ( B ) = 1. Let z ∈ Z ( B ) with z = 1. So C G ( B ) ≤ C G ( z )and B ≤ C G ( z ), so G = C G ( B ) B ≤ C G ( z ), so z ∈ Z ( G ). But this would give ustwo distinct conjugacy classes z G = { z } and 1 G = { } with the same cardinality,1. But G satisfies hypothesis H , so this is a contradiction.This proves the claim that C G ( B ) ≤ AB .Claim: If b , b ∈ B and if | b B | = | b B | , and if neither C G ( b ) ≤ AB nor C G ( b ) ≤ AB , then b and b are conjugate in B .Assume that b , b ∈ B and that | b B | = | b B | , and that neither C G ( b ) ≤ AB nor C G ( b ) ≤ AB . We wish to show that b and b are conjugate in B . Since[ G : AB ] = 2, AB E G and we have C G ( b )( AB ) = G = C G ( b )( AB ). So by theExtended Frattini Argument, (using AB for N ), we get | b G | / | b AB | = 1 = | b G | / | b AB | and therefore b AB = b G and b AB = b G . But | b ( AB )1 | = [ AB : C ( AB ) ( b )] = [ AB : AC B ( b )] and | b ( AB )2 | = [ AB : C ( AB ) ( b )] = [ AB : AC B ( b )]and so since | C B ( b ) | = | C B ( b ) | , it follows that | b ( AB )1 | = | b ( AB )2 | . So | b G | = | b G | .Since G satisfies hypothesis H , we get b G = b G . So b B = b ( AB )1 = b G = b G = b ( AB )2 = b B and so b B = b B , so b and b are conjugate in B . This proves the claim.Claim: B satisfies hypothesis H .We have already shown that B is finite and non-trivial. So we now only need toshow that distinct conjugacy classes of B have distinct cardinalities. Assume to thecontrary that there are elements b and b in B such that | b B | = | b B | but b and b are not conjugate in B . We wish to derive a contradiction.Note that | b B | = | b B | implies that [ B : C B ( b )] = [ B : C B ( b )] and therefore impliesthat | C B ( b ) | = | C B ( b ) | . HE SYM(3) CONJECTURE AND ALT(8) 7
Since b and b are not conjugate in B , then by the previous claim, we must haveeither C G ( b ) ≤ AB or C G ( b ) ≤ AB . Without loss of generality, assume that C G ( b ) ≤ AB . Let a be a 7-cycle of Alt (8) in A and let a = a − . Then a and a are not conjugate in Alt (8), but they generate the same cyclic subgroup of G of order7; since G is a rational group, a and a are therefore conjugate in G . Therefore, a A is a proper subset of a G . Also, | C A ( a ) | = | C A ( a ) | . Since B = C G ( A ), a A = a AB .So by the Extended Frattini Argument, 1 < | a G | / | a AB | = [ G : C G ( a )( AB )] ≤
2. So[ G : C G ( a )( AB )] = 2, so G G ( a ) ≤ AB . Similarly, G G ( a ) ≤ AB . Therefore (using A E G , B E G , and AB ∼ = A × B ), we get: C G ( a b ) = C G ( a ) ∩ C G ( b ) = C AB ( a ) ∩ C AB ( b ) = C A ( a ) × C B ( b ).Similarly, C G ( a b ) = C G ( a ) ∩ C G ( b ) = C AB ( a ) ∩ C AB ( b ) = C A ( a ) × C B ( b ).Thus | C G ( a b ) | = | C A ( a ) | × | C B ( b ) | = | C A ( a ) | × | C B ( b ) | = | C G ( a b ) | .So | ( a b ) G | = | ( a b ) G | . Since G satisfies hypothesis H , it follows that a b isconjugate to a b in G . Choose g ∈ G such that ( a b ) g = a b . Then a g b g = a b .Therefore (again using A E G , B E G , and AB ∼ = A × B ) we get a g = a and b g = b . This implies that g ∈ C G ( a ) and so g ∈ AB . But no element ab of AB can conjugate b to b , because since A centralizes B , this would imply that b conjugates b to b , a contradiction to our assumption that b and b are notconjugate in B . This proves the claim that B satisfies hypothesis H .Claim: B ∼ = Sym (3). B is a finite non-trivial proper subgroup of G which satisfies the hypothesis H ofthe Sym (3) conjecture. Since we are assuming that G is a counterexample of leastcardinality to the Sym (3) conjecture, this implies that B ∼ = Sym (3). This provesthe claim.Therefore G has the normal subgroup AB ∼ = Alt (8) × Sym (3) of index 2. Since B ∼ = Sym (3), let σ be an involution of B and let δ be an element of order 3 of B . Alt (8) has a unique conjugacy class of cardinality 1 ,
680 which is the class of anelement s of Alt (8) of order 6. Also,
Alt (8) has a unique conjugacy class of cardi-nality 2 ,
520 which is the class of an element f of Alt (8) of order 4.Since
Sym (3) ∼ = B E G , σ G ⊆ B and so σ G = σ B and so by the Extended FrattiniArgument, we have G = C G ( σ ) B . So C G ( σ ) cannot be contained in AB . Similarly,we know that C G ( δ ) cannot be contained in AB . Also, we have s A ⊆ s G and s G isa union of A -conjugacy classes of elements of A of order 6 and thus has cardinality1 ,
680 or 5 , B ≤ C G ( s ) and so AB ≤ C G ( s ) A . Then | s G | / | s A | = [ G : C G ( s ) A ] ≤ [ G : AB ] = 2, so CECIL ANDREW ELLARD | s G | ≤ | s A | = 3 , | s G | = 1 ,
680 = | s A | . So s G = s A . Similarly, we have f A ⊆ f G and f G is a union of A -conjugacy classes of elements of A of order 4 andthus has cardinality 1 ,
260 or 3 , B ≤ C G ( f ) and so AB ≤ C G ( f ) A . Then | f G | / | f A | = [ G : C G ( f ) A ] ≤ [ G : AB ] = 2, so | f G | ≤ | f A | = 2 , | f G | = 1 ,
260 = | f A | . So f G = f A . Since s G = s A and f G = f A , we know from the Extended Frattini Argument that neither C G ( s ) nor C G ( f ) can be contained in AB .Claim: C G ( sσ ) is not contained in AB .Since C G ( s ) is not contained in AB , we can choose g ∈ C G ( s ) such that g / ∈ A × B .Since Sym (3) ∼ = B E G , g permutes the three involutions of B . But g cannotfix all three for otherwise we would have g ∈ C G ( B ) and by the previous claim, C G ( B ) ≤ AB . So either g fixes one involution and transposes the other two, or g permutes the three involutions in a 3-cycle. But g cannot permute the involutionsin a 3-cycle, because in this case, g (which is also not in AB , since [ G : AB ] = 2)would be in C G ( B ), which by a previous claim is a subgroup of AB . So g mustfix one involution and transpose the other two. Without loss of generality, we as-sume that σ is the involution of B fixed by g . Then g ∈ C G ( s ) ∩ G G ( σ ). Thus, g ∈ C G ( sσ ). So g is an element of C G ( sσ ) which is not an element of AB . Thisproves the claim that C G ( sσ ) is not contained in AB .Claim: C G ( f δ ) is not contained in AB .Since C G ( f ) is not contained in AB , we can choose h ∈ C G ( f ) such that h / ∈ AB .Since Sym (3) ∼ = B E G , h permutes the two elements of order 3 of B . If h fixesboth of them, then h ∈ C G ( δ ), and so we can take h as our element in C G ( f δ )which is not in AB . Otherwise, assume that h transposes the two elements of order3 of B . Since C G ( δ ) cannot be contained in AB , there is an element h ′ ∈ G which isnot in AB which centralizes δ . Every element in G which is not AB can be writtenas hab , so write h ′ = hab . Then ( δ ) hab = δ so since ( δ ) h ∈ B , and A centralizes B ,we get ( δ ) ha = ( δ ) h so ( δ ) hab = ( δ ) hb . So hb centralizes δ . But hb also centralizes f (since both h and b do), and so hb is in in C G ( f δ ). And hb is not in AB (since b is, but h isn’t). So hb is an element of C G ( f δ ) which is not an element of AB .This proves the claim that C G ( f δ ) is not contained in AB .Thus, | C G ( sσ ) | = 2 | C A ( s ) × C B ( σ ) | = 2 | C A ( s ) | × | C B ( σ ) | = 2(12)(2) = 48.Also, | C G ( f δ ) | = 2 | C A ( f ) × C B ( δ ) | = 2 | C A ( f ) | × | C B ( δ ) | = 2(8)(3) = 48.Thus, | C G ( sσ ) | = 48 = | C G ( f δ ) | . Therefore the conjugacy classes in G of sσ and f δ have the same cardinality. Since G satisfies hypothesis H , it follows that sσ and f δ are conjugate in G . But sσ has order 6 while f δ has order 12, and so theycannot be conjugate in G . So we have reached the desired contradiction. It followsthat Soc ( G ) ′ is not isomorphic to Alt (8). (cid:3)
HE SYM(3) CONJECTURE AND ALT(8) 9
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