PRELIMINARIES 18

Preliminaries

CHAPTER 1. PRELIMINARIES 18

Theorem 1.3.20 Let G be a group acting transitively on a setD. LetQED, H = GQ

and X( GIH) be the permutation character of this action. Then

Proof. We have that

G 1 '""" ^-1 1 '"""

(IH) (g) =

THT

^LJ IH(xgx) =

THT

^{LJ 1}

xEG,xgx-1EH xEG,xgx-1EH

Now ifxgx-¹ EH, then xg E Hx. Thus Hxg = Hx and hence Hx is fixed by9E G.

However the summation is taken over all x E G such that xgx-¹ E H. Hence the summation is taken over all x E G for which the coset Hx is fixed by 9 E G. But

\jyE Hx, Hx = Hy and thus we obtain that

L

¹

⁼

^IHII{Hx

^I

^Hxg

⁼

^Hx}1

xEG,xgx-1EH

and hence we obtain that

(IH)G(g) = IHI1H11{Hx1

I

Hxg = Hx}1 = I{Hx

I

Hxg = Hx}1 = X(GIH)(g). D

Let G be a group, H ~ G and X= X(GIH). The following are some properties of permutation characters (see Theorem 2.5.6 in [65]).

(i) deg(x) divides

IGI.

(ii) (X,'ljJ) ~ deg('ljJ) for all'ljJE Irr(G).

(iii) (X, le) = 1.

(iv) X(g) E IN U {O} for all 9 E G.

(v) X(g) ~ X(gm) for all9 E G and m E IN U {O}.

(vi) x(g) = 0 ifo(g) does not divide IGI/deg(x)·

(vii) X(g)dl~t~) is an integer for all 9 E G.

CHAPTER 1. PRELIMINARIES

1.4 Clifford Theory

19

In this section we discuss the technique of Fischer-Clifford matrices. Later we will use this technique to construct the character tables of generalized symmetric groups and some associated groups. For the theory on Fischer-Clifford matrices, we follow the works of Ali [1], Almestady [5], Mpono [47J and Whitely [65J. However for the actual construction of the Fischer-Clifford matrices ofB(rn,n) and associated groups we will follow the works of Almestady [5J and List [35J.

Definition 1.4.1 Let G be a group, H :::; G and () be a character of H. Then for 9 E G, we define ()9: gHg- 1---+ C by ()9(t)

=

()(gtg- 1) for all t E gHg-1. Then ()9

is said to be a G-conjugate of (). If H is a normal subgroup ofG and ()9 = () for all 9 E G, then () is said to be G-invariant.

It is clear that ()9 is a character of gHg-l.

Theorem 1.4.2 (28}(Clifford's Theorem) Let G be a group, H a normal subgroup ofG and X E Irr(G). Let () be an irreducible constituent ofxH and ()1,()2"",()n be distinct conjugates of () in G such that ^()l = (). Then

n

XH = e

L()i,

where e= (XH,()).

i=l

Proof. For hEH we have

()G(h) =

I~I L

^{()O(xhx-1) =}

_I~I L

^()X(h)

xEG xEG

Thus we obtain that

(()G)H = _1

L

^()X

IHI

^xEG

Let </JEIrr(H) such that </J

rt.

{Oi

I

¹

:s;

i

:s;

n}. Then we obtain that

and hence ((()G)H,</J) = O. However by the Frobenius reciprocity theorem, we obtain that (XH, 0)

=

(X,()G). Hence X is an irreducible constituent of()G. Since ((()G) H ,</J)

=

CHAPTER 1. PRELIMINARIES 20 0, then (XH,c/J) = O. Thus c/J is not an irreducible constituent of XH. Hence all the irreducible constituents of XH are among the ()i and thus we obtain that

n n n n

XH

= L.

^(XH,()i)()i

⁼ L.

^(XH,())()i

⁼

^{(XH, ())}

L.

⁽⁾ⁱ

⁼

^e

L.

^{()i ,}

i=l i=l i=l i=l

where e= (XH,()). 0

Definition 1.4.3 Letc/J be a representation ofG anda an automorphism ofG. Then c/JO: is a representation ofG given by

for x,y E G. If the representationc/J affords a character X ofG, then the representation c/JO: affords a character xO: of G which is given by XO:(x) = x(xO:) for x E G. Then the representation </>0: and the character xO: are called the algebraic conjugates of c/J and X respectively induced by the automorphism a.

Let X = (Xi(Xj)) be the character table of G, where Xi E Irr(G), 1 :S ⁱ :S n and Xj, 1 :S j :S n are representatives of the conjugacy classes of elements of G.

Then the automorphism a of G induces a permutation on the conjugacy classes of G and therefore also on the columns ofX. For each Xi E Irr(G), we deduce that

xi

EIrr(G). Hence^a induces a permutation on the irreducible characters Xi of G and therefore also on the rows ofX. Moreover since

xi(Xj)

^{= Xi(xj),} then the matrices obtained from X by these two operations are identical. We have the following result known as Brauer's Theorem.

Theorem 1.4.4 (23j(Brauer's Theorem) Let G be a group and K be a group of automorphisms ofG. Then the number of orbits of K as a group of permutations on the irreducible characters ofG is the same as the number of orbits of K as a group of permutations on the conjugacy classes ofG.

Proof. Let X be the character table of G. Then as a matrix, X is square and nonsingular. Let a be an automorphism of G such that a E K. Then a induces a permutation on the conjugacy classes of G and thus induces a permutation on the columns ofX. Hence K acts on the conjugacy classes of G. Since a E K, then to each character Xof G, we obtain a characterxO: of G such that xO: E Irr(G) whenever

CHAPTER1. PRELIMINARIES 21 XE Irr(G). ForyE G, we obtain that XQ(y) = X(yQ). Thusa induces a permutation on the rows ofX. Hence K acts on the irreducible characters of G. Let X^Q denote the image ofX under a. Then we obtain that

P(a)X = X^Q = XQ(a) ,

where P(a), Q(a) are appropriate permutation matrices which are uniquely deter- mined by a E K. Suppose that a, f3 EK. Then we obtain that XQfJ = (XQ)fJ. Also we have that

P(af3)X

=

XQfJ

=

(XQ)fJ

=

(P(a)X)fJ

=

P(f3)P(a)X

and hence P(af3)

=

P(f3)P(a). We also have that XQfJ

=

XQ(af3) and (XQ)fJ = (XQ(a))fJ

=

XQ(a)Q(f3). SinceXQfJ

=

(XQ)fJ,we obtain that XQ(af3)

=

XQ(a)Q(f3).

The non-singularity of X implies that Q(af3) = Q(a)Q(f3). Define mappings 7fl

and 7f2 on K by 7fl(a) = (P(a))t and 7f2(a) = Q(a), where t denotes the trans-

pose operation on matrices. Then 7fl and 7f2 are permutation representations of K.

Let

fh

and

fh

be the permutation characters afforded by 7fl and 7f2 respectively.

Since X-I P(a)X = Q(a), P(a) and Q(a) are similar and thus have the same trace.

Since trace(P(a))t

=

trace(P(a)), we have that trace(P(a))t

=

trace(Q(a)). Hence

(h

=

e

2 and 7fl and 7f2 are equivalent. Let d_{1 ,}d₂ be the number of orbits ofK on the irreducible characters and on the conjugacy classes of G respectively. Thus we observe thatd1is the number of orbits of7fl(K) in its action as a group of permutations. Also d2 is the number of orbits of 7f2(K) in its action as a group of permutations. Since

e

1 is the permutation character of K acting on the irreducible characters of G, we obtain that (e1,!K)

=

d1. Also for e2, we obtain that (e2,!K)

=

d2. Howevere1

=

e2 and thus (e1,!K)

=

(()2,!K) and hence d1

=

d2. 0

Definition 1.4.5 Let () be a character of a subgroup H of a group G. Let Ic(())

=

^{{g ENc(H)}

I

⁽⁾⁹

= ()}.

Then we call Ic(()) the inertia group of () in G. If H is normal in G, then Ic(e)

=

{g E G

I

e⁹

=

e}.

We observe that Nc(H) acts on the characters of H by 9 : () ~ ()9 for all 9 E Nc(H). Then the inertia group of e is the stabilizer of e in Nc(H). Hence Ic(e) ~ Nc(H) ~ G and it is clear that H is a normal subgroup ofIc(e).