Irreducible characters of Sylow p-subgroups of the Steinberg triality groups 3 D 4 ( p 3m )
aa r X i v : . [ m a t h . R T ] S e p IRREDUCIBLE CHARACTERS OF SYLOW p -SUBGROUPS OFTHE STEINBERG TRIALITY GROUPS D ( p m ) TUNG LE
Dedicated to Professor Geoffrey Robinson on his 60th birthday
Abstract.
Here we construct and count all ordinary irreducible characters ofSylow p -subgroups of the Steinberg triality groups D ( p m ). Introduction
Let F q be a finite field of order q where q is a power of some prime p . A longstanding conjecture by Higman [2] on the number k ( U n ( q )) of conjugacy classes ofthe unitriangular group U n ( q ) of degree n over F q is that k ( U n ( q )) is a polynomialin q with integral coefficients. The unitriangular group U n ( q ) is also known as amaximal unipotent subgroup of the special linear group SL n ( q ). A generalizationof Higman’s Conjecture on the maximal unipotent subgroups U ( q ) of other finitegroups G ( q ) of Lie type is that k ( U ( q )) is a polynomial in q with integral coefficients.Here, we answer this for the Steinberg triality groups D ( q ). Denote by U a Sylow p -subgroup of the Steinberg triality group D ( q ) . Let F ×− := F − − { } . We provethe following result. Theorem 1.1.
The irreducible characters of U are classified into five families aslisted in Table 1. Table 1.
Irreducible characters of Sylow p -subgroups U of D ( q )Family Notation Parameter set Number Degree F χ a,b ,q F × q × F q ( q − q q F χ a ,a ,b ,q F × q × F q × F q ( q − q q F odd χ b,a ,q F × q × F q ( q − q q F even χ b ,q F × q q − q χ b,a,c ,c , q F × q × F × q × F × F q − q − q / F χ b ,b ,q F × q × ( F q / F q ) ( q − q q F lin χ b,alin F q × F q q Date : September 18, 2018.2010
Mathematics Subject Classification.
Primary 20C15, 20D20. Secondary 20C33, 20D15.
Key words and phrases.
Sylow subgroup, root system, irreducible character, Steinberg triality.
It is well-known that the primes 2 and 3 are bad for the Chevalley groups oftype G . Since the Steinberg triality groups D ( q ) also has the Dynkin diagramof type G , we are curious that if the primes 2 and 3 show up as the bad primes of D ( q ) in terms of the representation theory of the Sylow p -subgroup. Theorem1.1 points out that only prime 2 affects on the structure of U and k ( U ), which iscompatible to the global computation on character degrees of D ( q ), see [3]. Corollary 1.2. If q is odd then k ( U ) = 2 q + 2 q − q − q − q . Otherwise, if q iseven, then k ( U ) = 2 q + 5 q − q − q − q + 3 . We approach the Sylow p -subgroup U of D ( q ) by its root system. The methodto construct all irreducible characters of U is quite elementary, mainly using Cliffordtheory. By studying the action of F × q on its F q -hyperplane set and its F p -hyperplaneset, we obtain the structures of U and the factor groups U/Z where Z is normal in U and generated by some root subgroups.Here are some explanations for the information in Table 1. There are 5 families ofirreducible characters of U and each row of Table 1 represents one of these families.The first column gives notation for these families of characters. Notice that thefamily F odd exists only for odd q , while F even exists only if q is even. The indexof this notation describes the positive root j -th of maximal height such that Y j is not contained in the kernels of the irreducible characters in the family, wherethe positive root set of D presented in Subsection 2.2. The family F lin containsall linear characters of U . The second column of Table 1 gives the notation ofirreducible characters in each family. The upper indices are the parameters todecide the uniqueness of each member in their family where a, a i ∈ F q , b, b i ∈ F q and c i ∈ F . These parameters take values from the parameter set in the thirdcolumn. The lower indices show their family and degree. The fourth column liststhe number of distinct irreducible characters in each family and the last columngives their degrees.In this paper, we first present some notations of character theory, introduce theroot system as well as Sylow p -subgroups of the Steinberg triality groups D ( q ).Next, we establish some finite field properties which are used latter for the proof ofTheorem 1.1 in Section 3.2. Basic Setup and Notations
In this section we introduce some fundamental notations of character theory,Sylow p -subgroups of the Steinberg triality D ( q ), and some finite field properties.2.1. Character theory.
Let G be a group. Denote G × := G − { } , Irr( G ) theset of all complex irreducible characters of G, and Irr( G ) × := Irr( G ) − { G } . If χ is a character of G and λ a character of a subgroup H of G , we write λ G forthe character induced by λ and χ | H for the restriction of χ to H . We defineIrr( G, λ ) := { χ ∈ Irr( G ) : ( χ, λ G ) > } the irreducible constituent set of λ G .Denote the kernel of χ by ker( χ ) := { g ∈ G : χ ( g ) = χ (1) } . Furthermore, for N E G, let Irr( G/N ) denote the set of all irreducible characters of G with N intheir kernels. For the others, our notations will be quite standard.2.2. Root systems of D and D through the graph automorphism. Let α , α , α , α be fundamental roots of the root system Φ of Lie type D . Here isthe Dynkin diagram of Φ, see Carter [1, Chapter 3]. HARACTERS OF SYLOW p -SUBGROUPS OF D ( q ) 3 t t tt α α α α The positive roots are those roots which can be written as linear combinationsof the simple roots α , α , α , α with nonnegative coefficients and we write Φ + forthe set of positive roots. We use the notation
11 2 1 for the root 2 α + α + α + α and we use a similar notation for the remaining positive roots. The 12 positiveroots of Φ are given in Table 2. Table 2.
Positive roots of the root system Φ of type D .Height Roots5 α :=
11 2 1 α :=
11 1 1 α :=
11 1 0 α :=
10 1 1 α :=
01 1 1 α :=
01 1 0 α :=
10 1 0 α :=
00 1 1 α :=
00 1 0 α :=
01 0 0 α :=
10 0 0 α :=
00 0 1
Let L be a simple complex Lie algebra with root system Φ. We choose a Chevalleybasis { h r | r ∈ ∆ } ∪ { e r | r ∈ Φ } of L such that the structure constants N rs in[ e r , e s ] = N rs e r + s on extraspecial pairs of roots ( r, s ) ∈ Φ × Φ are chosen in such away that they are invariant under the triality automorphism τ , see [1, Section 4.2],i.e. N rs = N τ ( r ) τ ( s ) . Fix a power p f and let D ( p f ) be the Chevalley group oftype D over the field F p f constructed from L , see [1, Section 4.4]. The group D ( p f ) ∼ = P Ω +8 ( p f ) is simple and is generated by the root elements x α ( t ) for all α ∈ Φ and t ∈ F p f . Let X α := h x α ( t ) | t ∈ F p f i be the root subgroup correspondingto α ∈ Φ. For positive roots, we use the abbreviation x i ( t ) := x α i ( t ), i ∈ [1 , x i ( t ) , x j ( u )] = x i ( t ) − x j ( u ) − x i ( t ) x j ( u ) are given in Table 3.All [ x i ( t ) , x j ( u )] not listed in this table are equal to 1. Table 3.
Commutator relations for type D .[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ),[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ),[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ),[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ),[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ),[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ),[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ),[ x ( t ) , x ( u )] = x ( tu ), [ x ( t ) , x ( u )] = x ( tu ).Let UD ( p f ) be the subgroup of D ( p f ) generated by the elements x i ( t ) for i ∈ [1 ,
12] and t ∈ F p f . So UD ( p f ) is a maximal unipotent subgroup and a Sylow TUNG LE p -subgroup of D ( p f ). Recall that the map τ : UD ( p f ) → UD ( p f ) defined by τ ( x i ( t )) = x iγ ( t ) where γ is the permutation (2 , , , , , ,
10) induces anautomorphism of UD ( p f ) of order 3. The signs in Table 3 are chosen to satisfythis choice of τ . Notice that τ is the restriction of a triality automorphism of theChevalley group D ( p f ) to UD ( p f ), see [1, Proposition 12.2.3].Under the action of the permutation γ on the positive roots of Φ, there are sixorbits of roots as follows. S := { α =
00 1 0 } , S := { α =
01 0 0 , α =
10 0 0 , α =
00 0 1 } , S := { α =
01 1 0 , α =
10 1 0 , α =
00 1 1 } , S := { α =
11 1 0 , α =
10 1 1 , α =
01 1 1 } , S := { α =
11 1 1 } , S := { α =
11 2 1 } .For the construction of D , it requires an automorphism σ = τ ρ of D ( p f )where ρ is a field automorphism of F p f such that ρ = 1, see [1, Section 13.4]. Theexistence of an order 3 field automorphism ρ of F p f forces f = 3 m for some m ∈ N .So from now on we consider the field F q . For all t ∈ F q , denote ¯ t := ρ ( t ) = t q and¯¯ t := ρ (¯ t ). By [1, Proposition 13.6.3], the σ -fixed points of UD ( q ) correspondingto each root orbit are as follows. Y := { y ( t ) := x ( t ) : t = ¯ t ∈ F q } = { y ( t ) : t ∈ F q } ,Y := { y ( t ) := x ( t ) x (¯ t ) x (¯¯ t ) : t ∈ F q } ,Y := { y ( t ) := x ( t ) x (¯ t ) x (¯¯ t ) : t ∈ F q } ,Y := { y ( t ) := x ( t ) x (¯ t ) x (¯¯ t ) : t ∈ F q } ,Y := { y ( t ) := x ( t ) : t = ¯ t ∈ F q } = { y ( t ) : t ∈ F q } ,Y := { y ( t ) := x ( t ) : t = ¯ t ∈ F q } = { y ( t ) : t ∈ F q } . Let { p i } i be Euclidian coordinates of the roots { α i } i . By [1, Section 13.3.4], weset P := p , P := ( p + p + p ), P := ( p + p + p ), P := ( p + p + p ), P := p , P := p . It is easy to check that P = P + P , P = P + 2 P , P = P + 3 P , P = 2 P + 3 P . So { P i : i ∈ [1 , } is the positive root set of D of type G , where P is short and P is long.We call each Y i a root subgroup in the σ -fixed-point subgroup D ( q ). It is clearthat each Y i is abelian, the subgroup generated by all Y i ’s is a maximal unipotentsubgroup and a Sylow p -subgroup of D ( q ), denoted it by U . Since Y , Y , Y have order q and Y , Y , Y have order q , the Sylow p -subgroup U of D ( q ) hasorder q · = q . Using Table 3 we obtain the commutator relations among rootsubgroups Y i by direct computation. All [ y i ( t ) , y j ( u )] not listed bellow are equal to1. For all t, u in the appropriate fields to root subgroups Y i ’s, we have[ y ( t ) , y ( u )] = y ( tu ) y ( − tu ¯ u ) y ( tu ¯ u ¯¯ u ) y ( t u ¯ u ¯¯ u ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ tu ) y ( − t ¯ t ¯¯ u − ¯ t ¯¯ tu − ¯¯ tt ¯ u ) y ( − t ¯ u ¯¯ u − ¯ t ¯¯ uu − ¯¯ tu ¯ u ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ t ¯¯ u + ¯¯ tu ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ t ¯¯ u + ¯¯ tu ),[ y ( t ) , y ( u )] = y ( tu ). HARACTERS OF SYLOW p -SUBGROUPS OF D ( q ) 5 From these commutator relations, it is clear that Z ( U ) = Y , Z ( U/Y ) = Y Y ,Z ( U/Y Y ) = Y Y Y , Z ( U/Y Y Y ) = Y Y Y Y , and U/Y Y Y Y is abelian oforder q . To classify all irreducible characters of U , we come up with the definitionof almost faithful irreducible characters. Definition 2.1.
Let χ be an irreducible character of a group G . χ is said to be almost faithful if Z ( G ) ker( χ ) . Due to the center series of U and the inflation of irreducible characters fromquotient groups, the irreducible characters Irr( U ) are classified as follows. F := { χ ∈ Irr( U ) : Y ker( χ ) } , the set of all almost faithful irreduciblecharacters of U . F := { χ ∈ Irr( U ) : Y ker( χ ) and Y ≤ ker( χ ) } , the set of all almost faithfulirreducible characters of U/Y . F := { χ ∈ Irr( U ) : Y ker( χ ) and Y Y ≤ ker( χ ) } , the set of all almostfaithful irreducible characters of U/Y Y . F := { χ ∈ Irr( U ) : Y ker( χ ) and Y Y Y ≤ ker( χ ) } , the set of all almostfaithful irreducible characters of U/Y Y Y . F lin := { χ ∈ Irr( U ) : Y Y Y Y ≤ ker( χ ) } , the linear character set of U . Remark . For i ≥
3, each χ ∈ F i satisfies that Y i ≤ Z ( χ ) and Y i ker( χ ) where Z ( χ ) = { u ∈ U : | χ ( u ) | = χ (1) } . Thus χ | Y i = χ (1) λ for some λ ∈ Irr( Y i ) × .2.3. Some fundamental field results.
Through out this paper, for each primepower q , we consider the field extension F q / F q . Fix nontrivial linear characters φ : F q → C × , and ϕ : F q → C × . For each a ∈ F q , b ∈ F q , define φ a and ϕ b by φ a ( x ) := φ ( ax ) for all x ∈ F q , and ϕ b ( y ) := ϕ ( by ) for all y ∈ F q . Hence,Irr( F q ) = { φ a : a ∈ F q } and Irr( F q ) = { ϕ b : b ∈ F q } . Recall the Frobenius map ρ : F q → F q , t t q , the notations ¯ t := ρ ( t ) and ¯¯ t := ρ (¯ t ) for all t ∈ F q . Definition 2.3.
For each a ∈ F q and t ∈ F q , we define (i) T a := { x p − a p − x : x ∈ F q } , (ii) A t := { t ¯ u + ¯ t ¯¯ u + ¯¯ tu : u ∈ F q } , (iii) B t := { t q u + tu q : u ∈ F q } . Notice that T = F q , A = { } = B . Now we observe a few important propertiesof T a , A t and B t . Proposition 2.4.
The following are true. (i) x p − a p − x = Q c ∈ F p ( x − ca ) . (ii) If a ∈ F × q , then T a is an additive subgroup of F q of index p . (iii) For each a ∈ F × q , there exists b ∈ F × q such that b T a = ker( φ ) . Furthermore, cb T a = ker( φ ) iff c ∈ F × p . (iv) { T a : a ∈ F × q } = { ker( φ a ) : a ∈ F × q } are all subgroups of index p in F q . Proof.
Part (i) is clear by checking their solutions directly. For each a ∈ F × q wedefine h a ( x ) := x p − a p − x for all x ∈ F q . It is clear that T a = im( h a ) and h a isan F p -homomorphism with ker( h a ) = a F p by part (i). Thus part (ii) holds. Sincepart (iv) clearly follows part (iii), it suffices to show part (iii).Let S := { T a : a ∈ F × q } . For all a, y ∈ F × q , we have y p T a = T ay since y p h a ( x ) = ( xy ) p − ( ay ) p − ( xy ) = h ay ( xy ) ∈ im( h ay ) = T ay . TUNG LE So F × q acts transitively on the set S . Since S = ∅ and F × p ≤ Stab F × q ( T a ), we have0 < | S | ≤ q − p − . It is enough to show that Stab F × q ( T a ) = F × p by computing | S | . Since the left multiplication action of F × q = { y p : y ∈ F × q } on the F p -hyperplaneset of F q is transitive and the number of F p -hyperplanes of F q is q − p − which equals | S | , the claim holds. (cid:3) Lemma 2.5.
For each t ∈ F × q , A t = F q .Proof. It is clear that A t ⊂ F q since its elements are ρ -invariant. So | A t | ≤ q . Itsuffices to show that | A t | = q .For each a ∈ F q , the equation ¯¯ tu + tu q + ¯ tu q = a has at most q solutions for u in F q . Therefore, when u runs all over F q , A t has at least | F q | /q = q elements.This completes the proof. (cid:3) For each t ∈ F × q , we define an F q -homomorphism f t : F q → F q given by f t ( u ) := t q u + tu q for all u ∈ F q . By Definition 2.3 (iii), B t = im( f t ). Lemma 2.6.
For all t ∈ F × q , the following hold. (i) vx ∈ B t for all v ∈ B t and x ∈ F q . Hence, B tx = B t = x B t for all x ∈ F × q . (ii) If q is odd, then B t = F q . (iii) If q is even, then ker( f t ) = t F q and B t ≤ ( F q , +) of order q .Proof. Part (i) is clear by the F q -homomorphism property of f t . Thus, it sufficesto show parts (ii) and (iii) by finding the kernel of f t .We have t q u + tu q = t q u (1+( t − u ) q − ). So f t ( u ) = 0 iff u = 0 or ( t − u ) q − = − q odd: Since F × q is cyclic of order q − q − q + q +1) and ( q + q +1) isodd, there is no z ∈ F × q such that z q − has order 2. So the equation ( t − u ) q − = − u ∈ F q . This shows that ker( f t ) = { } and im( f t ) = F q .Case q even: The equation ( t − u ) q − = 1 implies that t − u ∈ F × q , i.e. u ∈ t F × q .It is clear that t F q ⊂ ker( f t ). So ker( f t ) = t F q and | B t | = q . (cid:3) Lemma 2.7. If q is even, the following hold. (i) ker( f ) = F q and F q = ker( f ) ⊕ B . (ii) B t = t q +1 B for all t ∈ F × q . (iii) { B t : t ∈ F × q } is the F q -hyperplane set of F q . So | B t ∩ B r | = q iff t r F q . (iv) There exists uniquely t ∈ F × q , up to a scalar of F × q , such that B t ⊂ ker( ϕ ) .Proof. (i) Recall that B = { u + u q : u ∈ F q } and ker( f ) = F q by Lemma 2.6 (iii).Suppose that ker( f ) ∩ B is nontrivial, i.e. there exist x ∈ F × q and u ∈ F × q suchthat x = u + u q .If u ∈ F q , then x = u + u q = 2 u = 0 since char( F q ) = 2, contrary to x ∈ F × q .Notice that F q F q . If u F q , from u q + u q + u ∈ F q , we have u + u q F q ,contrary to x = u + u q ∈ F q . So ker( f ) ∩ B = { } .By Lemma 2.6 (iii) with | ker( f ) | = q and | B | = q , we have F q = ker( f ) ⊕ B .(ii) From t q u + tu q = t q +1 ( ut − +( ut − ) q ) and { ( ut − )+( ut − ) q : u ∈ F q } = B ,the claim is clear.(iii) Since q is even and q − q − q + q + 1), gcd( q + 1 , q −
1) = 1 and { t q +1 : t ∈ F × q } = F × q . By part (ii), F × q acts transitively on the set of all B t ’s. HARACTERS OF SYLOW p -SUBGROUPS OF D ( q ) 7 However, the left multiplication action of F × q is also transitive on the F q -hyperplaneset of F q . This shows that { B t : t ∈ F × q } is the F q -hyperplane set of F q .By part (ii), B t = B r iff ( tr − ) q +1 ∈ Stab F × q ( B ). The transitivity of F × q on the F q -hyperplane set, whose cardinality is q − q − , implies that | Stab F × q ( B ) | = q − F × q ( B ) = F × q . It is clear that ( tr − ) q +1 ∈ F × q iff t ∈ r F × q .By the property of F q -hyperplanes in F q , the rest of the claim is clear.(iv) The uniqueness follows by part (iii) since B t + B s = F q ) ker( ϕ ) for any B t = B s . It is clear that B is contained in ker( ϕ b ) for some ϕ b ∈ Irr( F q ) × . Theexistence follows by the transitivity of the left multiplication action of F × q on the F p -hyperplane set of F q , also known as { ker( ϕ r ) : ϕ r ∈ Irr( F q ) × } . (cid:3) Characters of Sylow p -subgroups of the triality groups D ( q )Let χ ∈ Irr( U ). We shall prove Theorem 1.1 by constructing χ through each fam-ily from F to F , ..., F lin as discussed in Subsection 2.2. We recall the commutatorrelations computed in Subsection 2.2.[ y ( t ) , y ( u )] = y ( tu ) y ( − tu ¯ u ) y ( tu ¯ u ¯¯ u ) y ( t u ¯ u ¯¯ u ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ tu ) y ( − t ¯ t ¯¯ u − ¯ t ¯¯ tu − ¯¯ tt ¯ u ) y ( − t ¯ u ¯¯ u − ¯ t ¯¯ uu − ¯¯ tu ¯ u ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ t ¯¯ u + ¯¯ tu ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ t ¯¯ u + ¯¯ tu ),[ y ( t ) , y ( u )] = y ( tu ).3.1. Family F where χ is almost faithful. Let T := Y Y Y Y Y and V := Y Y Y . It is easy to check that V is abelian, V ⊳ T ⊳ U and Z ( T ) = Y . By Lemma2.5 and Clifford theory with the transversal Y Y of V in T , all λ ∈ Irr( V ) such that λ | Y = 1 Y satisfy λ T ∈ Irr( T ) of degree q . Thus all almost faithful irreduciblecharacters of T have degree q . Since χ | T decomposes into sum of almost faithfulirreducible characters, χ (1) ≥ q . Since λ T | Y Y is the regular character of theabelian group Y Y , χ | V has a linear constituent θ such that θ | Y = 1 Y .Let H := Y Y Y Y . Clearly that H ≤ U and Z ( H ) = Y Y . By the existence of θ , let ξ be an irreducible constituent of χ restricted to H such that ξ | Y = ξ (1)1 Y .Since Y ≤ ker( ξ ), ξ can be considered as a character of H/Y ∼ = Y Y Y , which isabelian. So ξ is linear. We shall show that ξ U ∈ Irr( U ), which implies that χ isthe unique constituent in Irr( U, ξ ), i.e. χ = ξ U . Since H ∩ T = V and U = HT ,by Mackey formula for the double coset H \ U/T we have ξ U | T = ξ | H ∩ T T ∈ Irr( T ).Thus ξ U ∈ Irr( U ) as claimed. So all almost faithful irreducibles have degree q .By the extension of each λ T to U through Y , we obtain ( q − q almost faithfulirreducible characters of degree q , parameterized by ( Y × , Y ) ∼ = ( F × q , F q ) anddenoted by χ a,b ,q .3.2. Family F where Y ⊂ ker( χ ) and Y ker( χ ) . We study the quotientgroup ¯ U := U/Y . Abusing terminology slightly we call the image under the naturalprojection of a root group of U , a root group of ¯ U . Recall that the commutatorrelations in ¯ U are as follows.[ y ( t ) , y ( u )] = y ( tu ) y ( − tu ¯ u ) y ( tu ¯ u ¯¯ u ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ tu ) y ( − t ¯ t ¯¯ u − ¯ t ¯¯ tu − ¯¯ tt ¯ u ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ t ¯¯ u + ¯¯ tu ). TUNG LE
Let λ be a linear character of the abelian normal subgroup H := Y Y Y Y ⊳ ¯ U .By Lemma 2.5, for each y ( t ) ∈ Y × , there exists some y ( u ) ∈ Y such that[ y ( u ) , y ( t )] ker( λ ), i.e. y ( t ) λ ( y ( u )) = λ ( y ( u )). So the inertia group I ¯ U ( λ ) = H . By Clifford theory, λ ¯ U ∈ Irr( ¯ U ).Since q − Y extend to ( q − q linear charactersof H , by the above argument we obtain ( q − q irreducibles of degree q in thisfamily, parameterized by ( Y × , Y , Y ) ∼ = ( F × q , F q , F q ) and denoted by χ a ,a ,b ,q .3.3. Family F where Y Y ⊂ ker( χ ) and Y ker( χ ) . We study the quotientgroup ¯ U := U/Y Y . Recall that the commutator relations in ¯ U are as follows.[ y ( t ) , y ( u )] = y ( tu ) y ( − tu ¯ u ),[ y ( t ) , y ( u )] = y ( t ¯ u + ¯ tu ).Let λ be a linear character of the abelian normal subgroup H := Y Y Y ⊳ ¯ U such that λ | Y = 1 Y . Since Y is a transversal of H in ¯ U , to obtain the inertiagroup I ¯ U ( λ ), we find all y ( t ) ∈ Y such that y ( t ) λ = λ . First we compute thestabilizer Stab Y ( λ | Y ). We have y ( t ) λ ( y ( u )) = λ ( y ( u ) y ( t ) )= λ ( y ( u )) λ ([ y ( u ) , y ( t )])= λ ( y ( u )) λ ( y ( − ( t ¯ u + ¯ tu ))) . By Lemma 2.6 (ii) with q odd, the inertia group I ¯ U ( λ ) = H . By Clifford theory, λ ¯ U ∈ Irr( ¯ U ). Hence there are ( q − q irreducible characters of degree q in thisfamily, parameterized by ( Y × , Y ) ∼ = ( F × q , F q ) and denoted by χ b,a ,q ∈ F odd .Now we observe the case where q is even. Fix λ | Y = ϕ c where c ∈ F × q asdescribed in Subsection 2.3. We shall find the orbit of λ under the action of Y .Recall the function f t ( u ) := t q u + tu q defined right before Lemma 2.6. We have y ( t ) λ ( y ( u )) = λ ( y ( u )) λ ( y (( t ¯ u + ¯ tu ))) = ϕ ( c u ) ϕ c ( f t ( u )) . So y ( t ) λ | Y = λ | Y iff ϕ c ( f t ( u )) = 1 for all u ∈ F q , i.e. im( f t ) ⊂ ker( ϕ c ). ByLemma 2.7 (iv), there exists unique t ∈ F × q , up to a scalar of F × q , such thatim( f t ) ⊂ ker( ϕ c ). By Lemma 2.7 (iii) St := Stab Y ( λ | Y ) = { y ( rt ) : r ∈ F q } .Thus, under the action of Y , Irr( Y ) decomposes into q orbits; each has size q .Let λ | Y = ϕ c , c ∈ F q . Now we compute St := Stab Y ( λ ) = Stab St ( λ | Y ). y ( rt ) λ ( y ( v )) = λ ( y ( v )) λ ([ y ( v ) , y ( rt )])= λ ( y ( v )) λ ( y ( vrt ) y ( − vrt rt ))= λ ( y ( v )) ϕ ( c vrt + c vr t q +10 )= λ ( y ( v )) ϕ c ( vrt ( c c − + rt q )) . So y ( rt ) λ | Y = λ | Y iff ϕ c ( vrt ( c c − + rt q )) = 1 for all v ∈ F q . We find r ∈ F × q with parameter c ∈ F q to obtain that vrt ( c c − + rt q ) ∈ ker( ϕ c ) for all v ∈ F q .Notice that if c c − t − q ∈ F × q then it is a nontrivial solution for r . We shall see that q − c ∈ F × q such that c c − t − q ∈ F × q correspond to q − λ whose Stab Y ( λ ) has order 2, and this is the unique nontrivial solution for r ∈ F × q ineach orbit. Now we suppose that c c − t − q F × q . So c c − + rt q = 0 for all r ∈ F × q .For each r ∈ F × q , let T ( r, c ) := { vrt ( c c − + rt q ) : v ∈ F q } be a one-dimensional F q -subspace of F q . Assume that T ( r, c ) ⊂ ker( ϕ c ) ⊂ F q . HARACTERS OF SYLOW p -SUBGROUPS OF D ( q ) 9 If T ( r, c ) ∩ B t = { } , then F q = T ( r, c ) ⊕ B t ⊂ ker( ϕ c ), a contradiction.Therefore, T ( r, c ) ∩ B t is nontrivial. By Lemma 2.6 (i), we have T ( r, c ) ⊂ B t . So rt ( c c − + rt q ) ∈ B t . By Lemma 2.7 (ii) with B t = t q +10 B , there exists y ∈ B such that rt ( c c − + rt q ) = t q +10 y . Solve this equation for c with parameter r ∈ F × q , we have c ∈ { c t q ( r − y + r ) : r ∈ F × q , y ∈ B } =: I .We claim | I | = ( q − q by proving that if r − y + r = s − z + s for some r, s ∈ F × q and y, z ∈ B then r = s and y = z . Since B is an F q -vector space, r − y + s − z ∈ B . From r − y + r + s − z + s = 0 ∈ B , we have r + s ∈ B . ByLemma 2.7 (i) with F q = F q ⊕ B , we have r + s ∈ F q ∩ B = { } , i.e. r = s .This shows that each c ∈ I determines uniquely r ∈ F × q and y ∈ B suchthat c = c t q ( r − y + r ). Notice that r − y + r ∈ F × q iff y = 0 where r ∈ F × q and y ∈ B . The above argument also confirms our discussion on q − c ∈ F × q such that c c − t − q = r ∈ F × q which is the unique nontrivial solution of r ,and the q -size orbit of λ with this r is given by obtaining all c ∈ I from runningall y ∈ B .These ( q − q linear characters λ of H corresponding to c ∈ I × (counting λ | Y for other q copies) have the stabilizer St = Stab Y ( λ ) = { , y ( r t ) } and I ¯ U ( λ ) = HSt . It is easy to see that [
HSt, HSt ] ≤ ker( λ ). Thus, these ( q − q linearcharacters extend to 2( q − q linear characters (of their inertia groups) and induceirreducibly to ¯ U , which gives 2( q − q / ( q /
2) = 4( q −
1) irreducible charactersof degree q /
2. Together with all c ∈ F × q , we obtain 4( q − q −
1) irreduciblecharacters of degree q / Y × , Y ∗ , Y ∗ , Y ∗ ) ∼ = ( F × q , F × q , F , F )and denoted by χ b,a,c ,c , q ∈ F even , which is proven bellow in details.The other q linear characters λ corresponding to c ∈ F q − I have Stab Y ( λ ) = { } . This shows that Y acts transitively on these q linear characters and theirinertia groups equal H . (Notice that we can choose λ with c = 0.) So we obtain q − U of degree q , parameterized by ( Y × ) ∼ = ( F × q ) anddenoted by χ b ,q ∈ F even .3.3.1. Proof of the parametrization ( Y × , Y ∗ , Y ∗ , Y ∗ ) ∼ = ( F × q , F × q , F , F ) of χ b,a,c ,c ,q / .Fix c ∈ F × q . The parameter Y ∗ ∼ = F × q corresponds to q − Y ) sat-isfying c ∈ I under the action of Y . It suffices to show the parametrization of( Y ∗ , Y ∗ ) ∼ = ( F , F ). Fix c ∈ I , let µ ∈ Irr( Y Y ) such that µ | Y i = ϕ c i , i = 3 , µ is extendable to K := I ¯ U ( λ ) = HSt where H = Y Y Y and St = { , y ( r t ) } . Let η , η be two extensions of η to K . Moreover, µ extendsto K := Y Y St where St = Stab Y ( λ | Y ) = { , y ( rt ) : r ∈ F q } . Let θ be anextension of µ to K and S := Stab Y ( θ ). An easy way to read this is described inthe following diagram. Y Y : µ ւ ց H = Y Y Y : λ θ : Y Y St = K ↓ ↓ K = HSt : η i γ : KS = K ց ւ ¯ U : η ¯ Ui , γ ¯ U ∈ Irr( ¯ U )We shall show the follows. (i) | S | =2 and θ extends to K := I ¯ U ( θ )= KS . Call γ an extension of θ to K .(ii) ( η ¯ Ui , γ ¯ U ) = 1 iff η i | St = γ | St and η i | S = γ | S for i = 1 , η ¯ U , η ¯ U ) = 1 iff η | St = η | St and η | S = η | S . Proof. (i) Notice that µ extends to K because [ K, K ] ≤ ker( µ ). Let T be atransversal of St in Y . It is clear that T Y is a transversal of K in ¯ U . Since Y Y Y is abelian, it is enough to find Stab Y ( θ | St ). For t, r ∈ F q , we have y ( t ) θ ( y ( rt )) = θ ( y ( rt )[ y ( rt ) , y ( t )])= θ ( y ( rt ) y ( trt ) y ( trt rt ))= θ ( y ( rt )) ϕ ( c trt + c tr q +1 t q +10 )= θ ( y ( rt )) ϕ c ( trt ( c c − + rt q )) . To obtain Stab Y ( θ ) = { } we find t ∈ F × q such that trt ( c c − + rt q ) ∈ ker( ϕ c ) forall r ∈ F q . By the uniqueness of ( c , r , y ) ∈ I × F × q × B with c = c t q ( r − y + r ),it requires that trt q +10 ( r − y + r + r ) ∈ ker( ϕ c ) for all r ∈ F q . By Lemma 2.6(i) and Lemma 2.7 (ii) with trr − ∈ F q and t q +10 y ∈ t q +10 B = B t ⊂ ker( ϕ c ), itsuffices to find t ∈ F × q such that tt q +10 r ( r + r ) ∈ ker( ϕ c ) for all r ∈ F q .If tt q +10 r ( r + r ) ∈ B t ⊂ ker( ϕ c ), then by Lemma 2.6 (i) with t, r, r ∈ F × q wehave t q +10 ∈ B t . By Lemma 2.7 (ii) with B t = t q +10 B , it implies that 1 ∈ B whichcontradicts Lemma 2.7 (i). Thus, tt q +10 r ( r + r ) B t for all r ∈ F × q .Let A t := { tr ( r + r ) : r ∈ F q } for t ∈ F × q . Notice that A t is an F -hyperplane of F q due to the F -homomorphism h r : F q → F q , r r r + r . If t q +10 A t ⊂ ker( ϕ c ),using the above argument we have ker( ϕ c ) = B t ⊕ t q +10 A t , an F -hyperplane of F q . By Lemma 2.7 (ii) we have ker( ϕ c ) = t q +10 ( B ⊕ A t ). Notice that F × q actstransitively on the F -hyperplanes of F q and the number of F -hyperplanes of F q is q −
1. So there exists uniquely t ∈ F × q such that ker( ϕ c ) = t q +10 ( B ⊕ A t ), whichshows that S = Stab Y ( θ ) = { , y ( t ) } as claimed. Here, the extension of θ to I ¯ U ( θ ) = KS is clear by [ KS , KS ] ≤ ker( θ ).(ii) Let γ be an extension of θ to K = KS and η an extension of λ to K = HSt .We shall show that ( η ¯ U , γ ¯ U ) = 1 iff η | St = γ | St and η | S = γ | S . Notice that both η ¯ U , γ ¯ U ∈ Irr( ¯ U ) by Clifford theory.Since K ⊳ ¯ U = K K , by Mackey formula for the double coset K \ ¯ U /K andFrobenius reciprocity we have( η ¯ U , γ ¯ U ) = ( η ¯ U | K , γ )= ( η | K ∩ K K , γ )= ( η | K ∩ K , γ | K ∩ K ) . Since K ∩ K = Y Y StS and both η, γ are linear, the claim holds.(iii) Choosing γ ∈ Irr( K ) such that γ | K ∩ K = η | K ∩ K , we have η ¯ U = γ ¯ U bypart (ii). Again by part (ii) we have ( η ¯ U , γ ¯ U ) = 1 iff η | St = γ | St and η | S = γ | S ,which completes the proof. (cid:3) Family F where Y Y Y ⊂ ker( χ ) and Y ker( χ ) . Set ¯ U := U/Y Y Y .Recall that the unique commutator relation in ¯ U is [ y ( t ) , y ( u )] = y ( tu ). Let λ be a linear character of the abelian normal subgroup H := Y Y ⊳ ¯ U such that λ | Y = ϕ c for some c ∈ F × q . Clearly Y is a transversal of H in ¯ U . HARACTERS OF SYLOW p -SUBGROUPS OF D ( q ) 11 For each y ( t ) ∈ Y and all y ( u ) ∈ Y , we have y ( t ) λ ( y ( u )) = λ ( y ( u )) λ ([ y ( u ) , y ( t )])= λ ( y ( u )) λ ( y ( tu ))= λ ( y ( u )) ϕ c ( tu ) . So y ( t ) λ = λ iff ϕ c ( tu ) = 1 for all u ∈ F q , clearly iff t = 0. Thus, the inertiagroup I ¯ U ( λ ) = H . By Clifford theory, λ ¯ U ∈ Irr( ¯ U ) of degree q .So F contains ( q − q irreducible characters of degree q , parameterized by( Y × , Y ∗ ) ∼ = ( F × q , F q / F q ) and denoted by χ b ,b ,q , where Y ∗ is the representative setof q orbits of Irr( Y ) under the action of Y for a fixed c ∈ F × q .3.5. Family F lin where Y Y Y Y ⊂ ker( χ ) . Set ¯ U := U/Y Y Y Y . Since ¯ U isabelian, this family is the set of all linear characters of ¯ U . Here we obtain q linearcharacters parameterized by ( Y , Y ) ∼ = ( F q , F q ) and denoted by χ b,alin . References [1] R.W. Carter, Simple Groups of Lie Type, Pure Appl. Math., Wiley, London, 1972. 2, 3, 4[2] G. Higman, Enumerating p-groups. I: Inequalities, Proc. London Math. Soc. (3), 10 (1960),24-30. 1[3] F. Luebeck, Online Data on Character Degrees of the Steinberg triality groups D ( q ) . 2 Mathematics Department, North-West University, Mafikeng, South Africa
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