Fitting Subgroup

Home , Fitting length, Fitting subgroup

Universidade Federal de Minas Gerais Instituto de Ciências Exatas Programa de Pós Graduação em Matemática

Fitting subgroup

Inácio Augusto Rabelo Pinto Júlio César Magalhães Marques Lucas Abraão Mateus de Castro

Course: Grupos e Representações (MAT 889) Professor: Csaba Schneider

November 14, 2019 Contents

Introduction Deﬁnition and ﬁrst results Examples Properties of Fitting subgroup Fitting Series

1/29 Contents

Introduction Deﬁnition and ﬁrst results Examples Properties of Fitting subgroup Fitting Series

2/29 Contents

Introduction Deﬁnition and ﬁrst results Examples Properties of Fitting subgroup Fitting Series

3/29 Proposition T 1. Op(G) = P ; P Syl (G) ∈ p 2. Op(G) char G;

3. Op(G) is the largest normal p-subgroup of G;

Deﬁnition Let G be group and p be a prime.

| Op(G) = N E G N is p-subgroup of G .

4/29 2. Op(G) char G;

3. Op(G) is the largest normal p-subgroup of G;

Deﬁnition Let G be group and p be a prime.

| Op(G) = N E G N is p-subgroup of G .

Proposition T 1. Op(G) = P ; P Syl (G) ∈ p

4/29 3. Op(G) is the largest normal p-subgroup of G;

Deﬁnition Let G be group and p be a prime.

| Op(G) = N E G N is p-subgroup of G .

Proposition T 1. Op(G) = P ; P Syl (G) ∈ p 2. Op(G) char G;

4/29 Deﬁnition Let G be group and p be a prime.

| Op(G) = N E G N is p-subgroup of G .

Proposition T 1. Op(G) = P ; P Syl (G) ∈ p 2. Op(G) char G;

3. Op(G) is the largest normal p-subgroup of G;

4/29 Deﬁnition The Fitting subgroup of a group G is deﬁned as

F(G) = hN E G | N is nilpotenti.

I There are in the literature another deﬁnitions for the Fitting subgroup. I If G is nilpotent, then F(G) = G. In particular, if G is abelian, F(G) = G.

5/29 D | E 2. F(G) = Op(G) p is a prime ;

3. F(G) = p π(G) Op(G) (internal direct product); ∈ 4. F(G) is the largest nilpotent normal subgroup of G.

Proposition Let G a group. The following statements are equivalent: 1. F(G) = hN E G | N is nilpotenti;

6/29 3. F(G) = p π(G) Op(G) (internal direct product); ∈ 4. F(G) is the largest nilpotent normal subgroup of G.

Proposition Let G a group. The following statements are equivalent: 1. F(G) = hN E G | N is nilpotenti;

D | E 2. F(G) = Op(G) p is a prime ;

6/29 4. F(G) is the largest nilpotent normal subgroup of G.

Proposition Let G a group. The following statements are equivalent: 1. F(G) = hN E G | N is nilpotenti;

D | E 2. F(G) = Op(G) p is a prime ;

3. F(G) = p π(G) Op(G) (internal direct product); ∈

6/29 Proposition Let G a group. The following statements are equivalent: 1. F(G) = hN E G | N is nilpotenti;

D | E 2. F(G) = Op(G) p is a prime ;

3. F(G) = p π(G) Op(G) (internal direct product); ∈ 4. F(G) is the largest nilpotent normal subgroup of G.

6/29 Corollary The subgroup F(G) of a group G may be described as the uni- que maximal nilpotent normal subgroup of G. Remark If G is a solvable group, then its Fitting subgroup is nontrivial.

Corollary - Fitting’s Theorem If H and K are nilpotent normal subgroups of a ﬁnite group G, then also HK is a normal nilpotent subgroup of G.

7/29 Corollary - Fitting’s Theorem If H and K are nilpotent normal subgroups of a ﬁnite group G, then also HK is a normal nilpotent subgroup of G. Corollary The subgroup F(G) of a group G may be described as the uni- que maximal nilpotent normal subgroup of G. Remark If G is a solvable group, then its Fitting subgroup is nontrivial.

7/29 Contents

Introduction Deﬁnition and ﬁrst results Examples Properties of Fitting subgroup Fitting Series

8/29 Fitting subgroup of Sn

I ' For n = 2: S2 Z2 is nilpotent. Then F(S2) = S2.

≥ Remark: For n 3 we have Z(Sn) = 1, then Sn is non-nilpotent and F(Sn) < Sn.

I For n = 3: I Normal subgroups of S3: 1,A3 and S3; I ' A3 Z3 is nilpotent. Then F(S3) = A3.

9/29 ≥ Remark: For n 4 we have Z(An) = 1, then An is non-nilpotent and F(An) < An.

I For n = 4: I h i Normal subgroups of S4: 1,A4,X = (12)(34),(13)(24) and S4; I ' × X Z2 Z2 is nilpotent. Then F(S4) = X.

I For n ≥ 5: I Normal subgroups of Sn: 1,An and Sn. Then F(Sn) = 1.

10/29 Fitting subgroup of Dn

j ≥ Remark: Dn is nilpotent if and only if n = 2 for some j 0.

I j ≥ F(Dn) = Dn if and only if n = 2 for some j 0.

I Suppose n = 2jm for some j ≥ 0 and m ≥ 3 odd. D E | n 2 1 Dn = a,b a = b = 1,bab = a− D E I d | Normal subgroups of Dn: 1, a for some d n. In ad- D E D E diction, if n is even, G = a2,ab and H = a2,b are also normal. These are all proper and normal subgroups of Dn.

11/29 I h i h i ≤ a E Dn is nilpotent, then a F(Dn).   1    hai I hai ≤ F(D ) =  D E n  2  G = a ,ab , neven  D E  H = a2,b , neven

I h i Then F(Dn) = a if n is not a power of two.

12/29 Contents

Introduction Deﬁnition and ﬁrst results Examples Properties of Fitting subgroup Fitting Series

13/29 1.F(G)char G

Proof: I Suppose α ∈ Aut(G) I Q α(F(G)) = p π(G) α(Op(G)) ∈ I ∈ For each p π(G), Op(G)char G I Q Then p π(G) Op(G) = F(G) ∈

14/29 Proof: I F(N)char N E G, then F(N) E G I Note that F(N) 6 F(G). Thus F(N) ≤ N ∩ F(G) I N ∩ F(G) E G and is nilpotent, because N ∩ F(G) 6 F(G). I Thus N ∩ F(G) ≤ F(N). I Hence N ∩ F(G) = F(N)

2. If N E G, then N ∩ F(G) = F(N)

15/29 2. If N E G, then N ∩ F(G) = F(N)

Proof: I F(N)char N E G, then F(N) E G I Note that F(N) 6 F(G). Thus F(N) ≤ N ∩ F(G) I N ∩ F(G) E G and is nilpotent, because N ∩ F(G) 6 F(G). I Thus N ∩ F(G) ≤ F(N). I Hence N ∩ F(G) = F(N)

15/29 Preliminar result

Theorem Let G be a group, Z ≤ Z(G). Then G/Z is nilpotent if, and only if, G is nilpotent. Corollary Let G be a group, Z ≤ Z(G) and H ≤ G such that Z ≤ H. Then H/Z is nilpotent if, and only if, H is nilpotent.

16/29 Proof: I Write F(G/Z) = H/Z, where Z ≤ H ≤ G.

I By the Correspondece Theorem H E G and Z 6 H.

I So, by the previous Corollary, H is nilpotent.

I So H ≤ F(G) and this implies F(G/Z) = H/Z ≤ F(G)/Z

I Z ≤ F(G), by previous Corollary, F(G)/Z is nilpotent. Note that F(G) E G.

I Hence F(G)/Z E G/Z. Thus F(G)/Z ≤ F(G/Z).

3. If Z ≤ Z(G), then F (G/Z) = F(G)/Z.

17/29 3. If Z ≤ Z(G), then F (G/Z) = F(G)/Z.

Proof: I Write F(G/Z) = H/Z, where Z ≤ H ≤ G.

I By the Correspondece Theorem H E G and Z 6 H.

I So, by the previous Corollary, H is nilpotent.

I So H ≤ F(G) and this implies F(G/Z) = H/Z ≤ F(G)/Z

I Z ≤ F(G), by previous Corollary, F(G)/Z is nilpotent. Note that F(G) E G.

I Hence F(G)/Z E G/Z. Thus F(G)/Z ≤ F(G/Z).

17/29   F(PSL (F )) = 1 PSL (F ) is simple =⇒  n n  F F  F(PSLn( )) = PSLn( )

F F SLn( ) F F F(SLn( )) F = PSLn( )=F(PSLn( )) = F Z(SLn( )) Z(SLn( )) F which implies SLn( )nilpotent. F F F Then F(PSLn( )) = 1 and F(SLn( )) = Z(SLn( )).

Fitting subgroup of PSL(n,F ) Let (n,|F |) , (2,2),(2,3) and F is a ﬁnite ﬁeld. F F F PSLn( ) = SLn( )/Z(SLn( ))

I F Note that SLn( ) is perfect, so is non-nilpotent.

18/29 F F(SLn( )) = F Z(SLn( )) F which implies SLn( )nilpotent. F F F Then F(PSLn( )) = 1 and F(SLn( )) = Z(SLn( )).

Fitting subgroup of PSL(n,F ) Let (n,|F |) , (2,2),(2,3) and F is a ﬁnite ﬁeld. F F F PSLn( ) = SLn( )/Z(SLn( ))

I F Note that SLn( ) is perfect, so is non-nilpotent.   F(PSL (F )) = 1 PSL (F ) is simple =⇒  n n  F F  F(PSLn( )) = PSLn( )

F SLn( ) F F F = PSLn( )=F(PSLn( )) Z(SLn( ))

18/29 F which implies SLn( )nilpotent. F F F Then F(PSLn( )) = 1 and F(SLn( )) = Z(SLn( )).

Fitting subgroup of PSL(n,F ) Let (n,|F |) , (2,2),(2,3) and F is a ﬁnite ﬁeld. F F F PSLn( ) = SLn( )/Z(SLn( ))

I F Note that SLn( ) is perfect, so is non-nilpotent.   F(PSL (F )) = 1 PSL (F ) is simple =⇒  n n  F F  F(PSLn( )) = PSLn( )

F F SLn( ) F F F(SLn( )) F = PSLn( )=F(PSLn( )) = F Z(SLn( )) Z(SLn( ))

18/29 F F F Then F(PSLn( )) = 1 and F(SLn( )) = Z(SLn( )).

Fitting subgroup of PSL(n,F ) Let (n,|F |) , (2,2),(2,3) and F is a ﬁnite ﬁeld. F F F PSLn( ) = SLn( )/Z(SLn( ))

I F Note that SLn( ) is perfect, so is non-nilpotent.   F(PSL (F )) = 1 PSL (F ) is simple =⇒  n n  F F  F(PSLn( )) = PSLn( )

F F SLn( ) F F F(SLn( )) F = PSLn( )=F(PSLn( )) = F Z(SLn( )) Z(SLn( )) F which implies SLn( )nilpotent.

18/29 Fitting subgroup of PSL(n,F ) Let (n,|F |) , (2,2),(2,3) and F is a ﬁnite ﬁeld. F F F PSLn( ) = SLn( )/Z(SLn( ))

I F Note that SLn( ) is perfect, so is non-nilpotent.   F(PSL (F )) = 1 PSL (F ) is simple =⇒  n n  F F  F(PSLn( )) = PSLn( )

F F SLn( ) F F F(SLn( )) F = PSLn( )=F(PSLn( )) = F Z(SLn( )) Z(SLn( )) F which implies SLn( )nilpotent. F F F Then F(PSLn( )) = 1 and F(SLn( )) = Z(SLn( )). 18/29 Properties of Fitting subgroup

Theorem Let G be a group and Φ(G) its Frattini subgroup. Then: 1. [F(G),F(G)] ≤ Φ(G) ≤ F(G).

2. F(G/Φ(G)) = F(G)/Φ(G).

19/29 Contents

Introduction Deﬁnition and ﬁrst results Examples Properties of Fitting subgroup Fitting Series

20/29 Upper Fitting series For a group G, deﬁne: I F0 = 1 I F1 = F(G) I Fi+1 is such that Fi+1/Fi = F(G/Fi) ≤ ≤ ··· ≤ ≤ 1 = F0 F1 Fn 1 Fn = G −

Deﬁnition A Fitting series, or nilpotent series, of a group G is a normal series ≤ ≤ ··· ≤ ≤ 1 = G0 G1 Gn 1 Gn = G − such that each factor Gi+1/Gi is nilpotent.

21/29 Deﬁnition A Fitting series, or nilpotent series, of a group G is a normal series ≤ ≤ ··· ≤ ≤ 1 = G0 G1 Gn 1 Gn = G − such that each factor Gi+1/Gi is nilpotent. Upper Fitting series For a group G, deﬁne: I F0 = 1 I F1 = F(G) I Fi+1 is such that Fi+1/Fi = F(G/Fi) ≤ ≤ ··· ≤ ≤ 1 = F0 F1 Fn 1 Fn = G −

21/29 Fitting Series of G = S4

I h i F0 = 1, F1 = F(S4) = X := (12)(34),(13)(24) . I F2 is such that F2/X = F(S4/X). • Note that S4/X is non-abelian with order 6, then ' S4/X S3. I Then ' F2/F1 = F2/X = F(S4/X) F(S3) = A3.

The only subgroup of S4 with order 12 is A4. Thus F2 = A4. I F3/F2 = F3/A4 = F(S4/A4) = S4/A4. Therefore F3 = S4.

1 < X < A4 < S4

22/29 Fitting series of G = Dn

h n 2 1i Dn = a,b : a = b = 1,bab = a−

I j ∈ N If n = 2 for some j , then F(Dn) = Dn.

1 < Dn

I Suppose n = 2jm for some j ∈ N and m ≥ 3 odd. h i F1 = F(Dn) = a . h i h i h i F2/ a = F(Dn/ a ) = Dn/ a . Then F2 = Dn. h i 1 < a < Dn

23/29 Definition We define the Fitting length as the smallest j for which Fj = G. If G is trivial, G has Fitting length 0. If Fi , G for all i, G has no Fitting length. Theorem A finite group G is solvable if, and only if, has Fitting length.

Fitting Length Theorem Suppose that

≤ ≤ ··· ≤ ≤ 1 = G0 G1 ...Gn 1 Gn = G − ≤ ∀ is a nilpotent series, then Gi Fi i.

24/29 Theorem A ﬁnite group G is solvable if, and only if, has Fitting length.

Fitting Length Theorem Suppose that

≤ ≤ ··· ≤ ≤ 1 = G0 G1 ...Gn 1 Gn = G − ≤ ∀ is a nilpotent series, then Gi Fi i. Deﬁnition We deﬁne the Fitting length as the smallest j for which Fj = G. If G is trivial, G has Fitting length 0. If Fi , G for all i, G has no Fitting length.

24/29 Fitting Length Theorem Suppose that

24/29 I F(G/Fk) = Fk+1/Fk = 1 I G is solvable, then G/Fk also.

I A non-trivial solvable group has non-trivial Fitting subgroup.

I Then G = Fk and G has Fitting length.

Proof

Suppose G is solvable. I Since G is ﬁnite, Fk = Fk+1 for some k.

25/29 I G is solvable, then G/Fk also.

I A non-trivial solvable group has non-trivial Fitting subgroup.

I Then G = Fk and G has Fitting length.

Proof

Suppose G is solvable. I Since G is ﬁnite, Fk = Fk+1 for some k. I F(G/Fk) = Fk+1/Fk = 1

25/29 I A non-trivial solvable group has non-trivial Fitting subgroup.

I Then G = Fk and G has Fitting length.

Proof

Suppose G is solvable. I Since G is ﬁnite, Fk = Fk+1 for some k. I F(G/Fk) = Fk+1/Fk = 1 I G is solvable, then G/Fk also.

25/29 I Then G = Fk and G has Fitting length.

Proof

Suppose G is solvable. I Since G is ﬁnite, Fk = Fk+1 for some k. I F(G/Fk) = Fk+1/Fk = 1 I G is solvable, then G/Fk also.

I A non-trivial solvable group has non-trivial Fitting subgroup.

25/29 Proof

Suppose G is solvable. I Since G is ﬁnite, Fk = Fk+1 for some k. I F(G/Fk) = Fk+1/Fk = 1 I G is solvable, then G/Fk also.

I A non-trivial solvable group has non-trivial Fitting subgroup.

I Then G = Fk and G has Fitting length.

25/29 I i = 0, F0 = 1. I For i = k, suppose Fk is solvable. I Fk+1/Fk is solvable. Then Fk+1 is also. So the claim is proved.

I There exists j such that Fj = G. Hence G is solvable.

Proof

Suppose that G has Fitting length j. We claim that Fi is solvable for each i. By induction:

26/29 I For i = k, suppose Fk is solvable. I Fk+1/Fk is solvable. Then Fk+1 is also. So the claim is proved.

I There exists j such that Fj = G. Hence G is solvable.

Proof

Suppose that G has Fitting length j. We claim that Fi is solvable for each i. By induction: I i = 0, F0 = 1.

26/29 I Fk+1/Fk is solvable. Then Fk+1 is also. So the claim is proved.

I There exists j such that Fj = G. Hence G is solvable.

Proof

Suppose that G has Fitting length j. We claim that Fi is solvable for each i. By induction: I i = 0, F0 = 1. I For i = k, suppose Fk is solvable.

26/29 I There exists j such that Fj = G. Hence G is solvable.

Proof

Suppose that G has Fitting length j. We claim that Fi is solvable for each i. By induction: I i = 0, F0 = 1. I For i = k, suppose Fk is solvable. I Fk+1/Fk is solvable. Then Fk+1 is also. So the claim is proved.

26/29 Proof

I There exists j such that Fj = G. Hence G is solvable.

26/29 Moral

I Criterion of solubility.

I Not every solvable group is nilpotent. However, in ﬁnite case, Fitting length intuitively measures how far such group is to be nilpotent.

I In general, to study solvable groups by nilpotent factors of its Fitting series.

27/29 ReferênciasI

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[4] K. Conrad, Simplicity of PSLn(F ). Expository papers, available from: https://kconrad.math.uconn.edu/blurbs/grouptheory/PSLnsimple.pdf. [Accessed 29/10/2019] [5] K. Conrad, Subgroups Series II. Expository papers, available from: https://kconrad.math.uconn.edu/blurbs/grouptheory/subgpseries2.pdf. [Accessed 28/10/2019] [6] K. Doerk, T. Hawkes, inite soluble groups. Walter de Gruyter, 1992. [7] D. Gorenstein, Finite Groups. 2nd ed. Chelsea Publishing Company, 1980.

28/29 ReferênciasII

[8] I. M. Isaacs, The Fitting and Frattini subgroups. Notes 3, 2002. https://www.math.wisc.edu/ boston/notes3.pdf [Accessed 03/11/2019] ∼ [9] H. Kurzweil and B. Stellmacher, The theory of ﬁnite groups - An introduction. 1st ed. Universitex - Springer, 2003. [10] J. B. Olsson, Group theory. version 2007, lecture notes. http://web.math.ku.dk/ olsson/manus/Gruppe-2009/gruppe2007en_all.pdf [Accessed 03/11/2019]∼ [11] D. J. S. Robinson, A course in the theory of groups. 2nd ed. Springer, 1996. [12] H. E. Rose, A course on ﬁnite groups. Springer, 2009. [13] J. J. Rotman, An Introduction to the Theory of Groups. 4nd ed. Springer, 1995.

[14] R. Schoof, The symmetric groups Sn. Lecture notes - Algebra II, available from: https://www.mat.uniroma2.it/ eal/S6.pdf. [Accessed 29/10/2019] [15] B. A. F. Wehrfritz, Finite groups: a second course on group theory. World Scientiﬁc Publishing Company, 1999.

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