Chapter 5 Projective Character Tables

.i(q-l). (ii) Assume that 9 E< b > has order> 2, say 9 = Then 9 : x f-t 'i'oX and (q + 1) t 2i. If'i'o E Irq, then >.i(q-l)2 = 1 and so q + 1 divides i(q - 1). If q is odd, then (q - 1,q + 1) = 2 and so (q + 1)12i, which is impossible. If q is even, then (q - 1,q + 1) = 1 and so (q + 1)li, which is again impossible. Thus 'i'o tf. Irq and so {1, 'i'o} is an Irq-basis of Ir q2 . Let y E GL2(q) satisfy vsv:' = gt. Then yg = s'v and so

v;

Y(/08) = yg(8) = (gt y)(8) = 'i'by(8)

for all

8 E Ir q2

(1)

It follows that for any f3 E Irq,

y(g(f3)8)

y(/0f38) f3y( 'i'08) f3'i'bY( 8)

= Let 1/J : Ir q2

---+

l(f3)y(8)

IF'q2 be defined by

1/'( a

+ 'i'0(3) = a + 'i'b(3

(by (1»

(2)

Projective Character Tables

264

Then 7/J(a and

+ g(,8)) = a + gt(j3).

If l' = a

y(f6) =

+ 1'013,6 E f q2, then l' = a + g(j3)

=

y((a+g(j3))6) (a+g t(j3))y(6)

=

7/J(f)y(6)

(by (2)) (3)

, , , 1Il' h . Hence, given l' ,1' E Jr q2, we ave

7/J(f'1''')y( 1) =

y(f'1''')(by (3) with l' = 1"1''', 6 = 1)

=

7/J( i')y( 1''')

=

7/J( 1")7/J(f")y( 1)

(by (3))

(by (3) with l' = 1''', 6 = 1)

Thus 7/J( 1"1''') = 7/J(f')7/J(1'''), so 7/J is an automorphism of f q2 which is trivial on fq. Hence 7/J E Gal(f q2/fq) and so 7/J = 1 or 7/J(x) = x q for all x E f q2. 1 It follows that 7/J(fo) E hO,1'6}. But 1'6+ = 1, so 1'6 = 1'0 1 and 7/J(fo) E 1 ho, 1'0 } . Thus l is either the map l' ........ 1'01' or the map l' ........ 1'0 11', which shows that gt E {g,g-l}. Finally, assume that ygy-1 = 9 for some y E G. Then t = 1, so by the definition of 7/J, 7/J(f) = l' for all l' E f q2. Applying (3) for 6 = 1, it follows that

y(f) = 1'y(l) This shows that y E <

t.p

>, where Y E<

t.p

t.p :

l' ........ A1'. Thus

> nSL 2 (q) =< b >

proving (ii). (iii) Since pq-1 = Aq2_1 = 1, we see that =< p >. It is now obvious that a = diag(p,p-l) is an element of order q - 1 in SL 2 (q). (iv) Let 9 E < a > be of order > 2. Then 9 is a diagonal nonscalar matrix. Hence GG(g) consists of diagonal matrices of determinant 1 and so GG(g) =< a>.

f;

8 Projective character tables for PSL2(q)

265

Let H consist of all h E G with h- 1gh diagonal. Then H is the set of matrices of the form

{[~

~]}

It follows that

Nc( < 9

» ==

H == (Gc(g), [

~

and so (H : Gc(g)) == 2. Because

[~ -~ r[~

1~-1] [~

we see that 9 is conjugate to g-l in G. But then 9 and powers of 9 conjugate to 9 in G, as required. •

«:'

are the only

Theorem 8.2. Let q be an odd prime power, let {l be a generator of the cyclic group IF; and let the element b of order q + 1 in S L 2( q) be defined as in Lemma 8.1 (i). Put 1 == [

~

o1 ] ,z == [ -10 d ==

[

~

~]

-01] ' e == [ 11

~ ] , a == [ ~

~-1]

For any 9 E SL 2(q), let [gl be the conjugacy class of SL 2(q) containing g. Then SL2(q) has precisely q + 4 conjugacy classes:

[a], [a 2 ],

[1]' [z], [e], [dl, [zel, [zdl . . . , [a(1/2j(Q-3)1, [bl, [b 21, ... , [b(1/2)(Q-1)1

(if q == 3, then the terms [ail should be deleted). Moreover, the cardinalities of these conjugacy classes are :

1[111 == l[zll==l

I[ell == I [d] I == I[zell

== I [zdll == (1/2)(q2 - 1)

I[anll

== q(q+1)

(1~n~(1/2)(q-3))

I[ bml l

==

(1 ::;

q(q - 1)

m::; (1/2)(q -

1))

Projective Character Tables

266

Proof. Note that if c and d are conjugate in G = S L 2 (q), then Jl is a square in IF;, which is impossible. Write q = pn for some prime p and some n ~ 1. Then e and d are of order p, while ze and zd are of order 2p. Thus the conjugacy classes of e, d, ze and zd are all distinct. In addition, we have obviously another two conjugacy classes, namely [I] and [zJ. One easily verifies that

Co(e) Hence

ICo(e)1 =

= Co(d) = {[ ;

ICo(d)1 = 2q and therefore

\[eJI = I[d]1 = lQl = q(q + l)(q 2q

2q

1)

= ~(q2 _

1)

2

It is clear that [zcJ and [zd] have the same cardinality above. By Lemma 8.1, [amJ, ibm], 1 ~ n ~ (1/2)(q- 3), 1 ~ m ~ (1/2)(q-1) are distinct conjugacy classes of G which have prescribed cardinalities. Adding up the cardinalities of the listed conjugacy classes of G, we obtain: 1 + 1 + 4· Since

IGI =

~(q2 - 1) + ~(q - 3)q(q + 1) + ~(q - l)q(q - 1) = q3 - q 2 2 2

q3 - q, the result follows. •

In what follows, we keep the notation of Theorem 8.2.

Let q be an odd prime power. Then (i) a-I is conjugate to a and b- I is conjugate to b. (ii) zan is conjugate to a(I!2)(q-I)-n, 1 ~ n ~ (1/2)(q - 3). (iii) zb" is conjugate to b(1!2)(q+I)-m, 1 ~ m ~ (1/2)(q - 1).

Corollary 8.3.

Proof. (i) Since l[aJj = l[a-IJj and I[bJI = I[b- I ]I, it follows from Theorem 8.1 that a-I is conjugate to an and b- I is conjugate to b" for some 1 ~ n ~ (1/2)(q - 3), 1 ~ m ~ (1/2)(q - 1). Hence, by Lemma 8.1(iv), an E {a-l,a} which forces an = a. Similarly, by Lemma 8.1(ii), b'" = bas required. (ii) By 0), an is conjugate to a- n and so zan is conjugate to za- n. Since z = a(I!2)(q-I), the required assertion follows. (iii) By (i), bm is conjugate to b- m and so zb": is conjugate to zb"'", Since z = b(1!2)(q+l), the required assertion follows. •

8 Projective character tables for P 5 £2 (q)

267

Assume that q is an odd prime power. Then SL 2(q) is a double cover of PSL 2(q) : Hence we may apply the terminology pertaining to double covers. For convenience, let us recall the following piece of information. For any conjugacy class C of S L 2 ( q), the set zC = {zxlx E C} is another conjugacy class of SL 2(q). We refer to zC as the associate of C. We say that C is selfassociate if C = zC. If C is not self-associate, then {C, zC} is called an associate pair of conjugacy classes of SL 2(q). A conjugacy class of PSL 2(q) splits in SL2(q) if its inverse image in SL2(q) is a union of two conjugacy classes of SL 2(q). Thus, if C is a conjugacy class of SL 2(q) whose image in PSL 2(q) is D, then D splits in SL 2(q) if and only if C is not self-associate. Finally, recall that a group G is ambivalent if each element of G is conjugate to its inverse. Our next task is to identify associate and self-associate conjugacy classes of SL 2(q). It turns out that the answer depends on whether q == 1(mod4) or q == -1(mod4) (since q is odd, these are the only possibilities). Using the notation of Theorem 8.2, we now prove

Theorem 8.4. Let q be an odd prime power and let q _ 1( mod 4). Then (i) S L 2(q) is ambivalent. (ii) [a(I/4)(q-l)] is the only self-associate conjugacy class of SL 2(q). (iii) [1], [c], [d], [a], [a 2], ... , [a(1/4)(q-5)J, [b], [b 2], ... , [b(1/4)(q-l)] are all distinct representatives of associate pairs of conjugacy classes of S L 2 ( q) (by convention, if q = 5, then all terms [ail should be deleted). (iv) The group PSL 2(q) has precisely (1/2)(q + 5) conjugacy classes, namely the images in PSL 2(q) of all conjugacy classes of SL 2(q) described in (ii) and (iii). (v) If the cohomology class of 0' E Z2(PSL 2(q),C") is of order two, then PSL 2(q) has precisely 1/2(q + 3) a-regular conjugacy classes, namely the images in PSL 2(q) of all conjugacy classes of SL 2(q) described in (iii). Proof. (i) Since q == 1(mod4), we see that 41IIr;l. Hence the element -1 of order two in F~ is a square in F~, say -1 = Therefore

,2.

o ] -1

-,

[~

268

Projective Character Tables

o ] -1 1

'

proving that c is conjugate to c- 1 and d is conjugate to a:', Hence zc is conjugate to (zc)-1 and zd is conjugate to (zd)-I. This proves (i), by applying Corollary 8.3 and Theorem 8.2. (ii) That [a(I/4)(q-l)] is a self-associate conjugacy class follows from Corollary 8.3(ii) by taking n = (1/4)(q - 1). The uniqueness is a consequence of property (iii) proved below. (iii) Since q == 1(mod4), we see that (1/2)(q - 3) is odd and (1/2)(q - 1) is even. Hence the desired assertion follows by applying Corollary 8.3(ii), (iii) and Theorem 8.2. (iv) This is a direct consequence of (ii) and (iii). (v) By Theorem 16.1.7 in Vo1.2, SL 2(q) is perfect (since q::J 2,3). Thus we have z E [SL2(q), SL 2(q)J. By Theorem 16.3.2 in Vo1.2, M(PSL 2(q)) has exactly one element of order two. Since a-regularity depends only on the cohomology class of a, we may assume that a is as in Theorem 4.5. Hence the desired assertion follows by applying (iii) and Theorem 4.5(i) . •

Theorem 8.5. Let q be an odd prime power and let q == -1(mod4). Then the following properties hold (i) SL 2(q) is not ambivalent. In fact, SL 2(q) has precisely four irreducible C-characters which are not 'R.-valued. (ii) [b(1/4)(q+l)] is the only self-associate conjugacy class of SL2(q). (iii) [1J, [c], [d], [a], [a 2], ... , [a(1/4)(q-3)], [b], [b 2], . . . , [b(I/4)(q-3)] are all distinct representatives of associate pairs of conjugacy classes of S L 2(q) (by convention, if q = 3, then all terms [ail, [fJJ] should be deleted). (iv) The group PSL2(q) has precisely (1/2)(q + 5) conjugacy classes, namely the images in PSL 2(q) of all conjugacy classes of SL 2(q) described in (ii) and (iii). (v) If the cohomology class of a E Z2(PSL 2(q),C*) is of order two, then PSL 2(q) has precisely (1/2)(q+3) a-regular conjugacy classes, namely the images in PSL 2(q) of all conjugacy classes of SL 2(q) described in (iii). Proof. But if

(i) Since q == -1(mod4), we see that -1 is not a square in

r;.

8 Projective character tables for P S L 2 (q)

269

for some x, y, z, u E r q with xu-yz = 1, then y = 0, u = -x, a contradiction. Thus c is not conjugate to c- I . Since q == -1(mod4), we see that (1/2)(q - 3) is even. Since 4 f Ir;1 and hence J.l2 is a square in we deduce that J.l(1/2)(q-3) = x 2 for some x E Hence

r;,

0][ -11

X [

1

x-I

r;.

O][x 1 1

r

[~ ~ ]

which shows that c l is conjugate to d. Hence (zc)-I is conjugate to zd, d- I is conjugate to c and (Zd)-I is conjugate to zc. Applying Theorem 8.2 and Corollary 8.3(i), we deduce that there are precisely four conjugacy classes C of SL 2(q) with C::j:. C- I . Hence, by Corollary 17.5.2 in Vol.I, SL 2(q) has precisely four irreducible (.>characters which are not R-valued. (ii) That [b(1/4)(q+I)] is a self-associate conjugacy class follows from Corollary 8.3(iii) by taking m = (1/4)(q+ 1). The uniqueness is a consequence of property (iii) proved below. (iii) Since q == -1(mod 4), we see that (1/2)(q-3) is even and (1/2)(q-l) is odd. Hence the desired assertion follows by applying Corollary 8.3(ii), (iii) and Theorem 8.2. (iv) This is a direct consequence of (ii) and (iii). (v) By the proof of Theorem 8.4(v), we need only show that

for q = 3. But, for q = 3, SL 2(q) is a covering group of PSL2(q) (see Theorem 16.3.2 in Vo1.2), as required. • Turning to the case where q is even, we now prove our final result.

Theorem 8.6. Let q = 2t for some t 2: 1, let J.l be a generator of the cyclic group and let the element b of order q + 1 in S L2(q) be defined as in Lemma 8.1 (i). Put

r;

1= [

~

For any g E SL 2(q), let [g] be the conjugacy class of SL 2(q) containing g. Then S L 2 ( q) has precisely q + 1 conjugacy classes: 1 ~ n ~ (1/2)(q-2),

1 ~ m ~ q/2

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270

Moreover, the cardinalities of these conjugacy classes are:

1[1]1 n

I[a

]

=

I =

1,I[c]l=q2-1 q(q + 1) (1 ::; n ::; (1/2)(q - 2)}

(1 ::; m ::; q/2)

Proof.

Setting G = SL 2(q), we have Ca(c)

= =

{ [

{[

Xy

~ ] lx, y E Irq, x2 = 1}

~ ~ ] Iy E Irq}

Hence we have

ICc(e)1

=q

which implies that l[e]1 = q2 - l. By Lemma 8.1, [an], [b m], 1 :S n :S (l/2)(q- 2), 1 :S m :S q/2, are distinct conjugacy classes of G which have prescribed cardinalities. Adding up the cardinalities of the listed conjugacy classes of G, we obtain

1·1+ 1· q2 + (1/2)(q - 2)(q + 1)2 + (1/2)q(q - 1)2 = q(q2 - 1) =

IGI

as desired . •

c.

The character table for SL2(q)

All characters below are assumed to be C-characters. We begin by providing the character table for SL 2 (q), where q is an odd prime power. All conjugacy classes of SL 2 (q) are given by Theorem 8.2. Note that for any irreducible character X of SL 2 (q), we have

x(ze)

x(z) X(z) = -( )x(e), X(zd) = -()X(d) Xl· X1

for this reason, the columns for the conjugacy classes [zc] and [zd] of SL 2(q) are missing in the character table below.

8 Projective character tables for P S L 2 (q)

271

Theorem 8.7. (Schur (1907), Jordan (1907)). Let q be an odd prime power, let v = (_1)(1/2)(q-1), let c be a primitive (q - l)-th root of 1 in C and let 6 be a primitive (q + 1)-th root of 1 in C. Then the character table for G = SL 2(q) is

[lJ Ie

A (Yi

{3j Xl

X2 /1

/2

1 q q' q" q'/2 q'/2 q"/2 q"/2

where [in

[anJ 1 1 [in

[bmJ 1 -1

[cJ 1

[dJ 1

0

0

0

0

s.;

( _1)n ( _l)n

0 0

1 -1 (1/2)(1 + JVCj) (1/2)(1 - Jvq) (1/2)(-1 + JVCj) (1/2)(-1 - JVCj)

1 -1 (1/2)(1 - JVq) (1/2)(1 + y'i/q) (1/2)(-1 - JVq) (1/2)(-1 + ViJq)

1

(-lr+ (-1r+ 1

0 0

= [in + e:'",

6jm

= _(6 jm + 6- jm),

q'

= q + 1, q" = q -

1 ~ i ~ (1/2)(q - 3),

1 ~ j ~ (1/2)(q - 1)

1 ~ n ~ (1/2)(q - 3),

1 ~ m ~ (1/2)(q - 1)

z 1 q (_l)l q' (-1)1 q" ~vq'

~vq' - ~vq" - ~vq"

1

Proof. For the sake of clarity, we divide the proof into a number of steps. Step 1. Here we demonstrate that the functions A, (Yi, 1 ~ i ~ (1/2)( q - 3) are irreducible characters of G = SL2(q). We denote by S the set of all nonzero squares in IF'q. Then lSI = (1/2)(q - 1). Now put

P={[~ and H

={[ ;

~]IXElF'q}

~-1] Ix E IF'q, y E IF'; }

By hypothesis, q = pk for some odd prime p and some k ~ 1. It is clear that P is a Sylow p-subgroup of G. Moreover, we also have P
< a >, Pn < a > = 1

and

IHI

=

IPII < a> I =

q(q - 1)

Projective Character Tables

272

Next we note that

[c] n

H=

{[

~

~]

Is

E

S}

nH

=

{[

~s

~]

[zc]nH

=

{[

=~

_~] ISES}

{zdJnH

= {[

[d]

Is E S}

-n

_~~

IsE

s}

All of the above sets have cardinality (1/2)(q - 1). If 1 ::; n ::; (1/2)(q - 3), then

~-n ], [~-n

~n]

Ix E Y q }

which is a set of cardinality 2q. If 1 ::; m ::; (1/2)(q - 1), then the linear group < b": > is irreducible. Thus we must have [b m ] n H = 0. It is clear that 1, z E H. Given any character X of H, we now compute XG by using the above information. Let i E {a, 1, ... ,q - 1} and let Ai be the linear character of H given by

0]

fL- t

Then P ~

J( er Ai

and Aq-l

.

-- cIt

= Ao = 1H.

(x E lFq

0::; t ::; q - 2)

The values of the Ay are:

AY(1) AY(ZC)

= =

q + 1, AY(Z) = (_1)i(q + 1), AY(C) = 1 (_1)i, Ay(d) = 1, Ay(zd) = (_l)i

Ay(a n )

=

cin

+ c:- i n , Ay(bm ) = 0

If 1 ~ i ::; (1/2)(q - 3), then O'i = Ay satisfies < O'i,O'i >= 1, so irreducible character of G. Moreover, < Ag >= 2 and

Ag,

O'i

is an

Thus Ag = 1a + A for some irreducible character A of G. This verifies the second and the third row of the character table. Of course, the first row is

8 Projective character tables for PSL 2 (q)

273

obvious, since Ie is the principal character of G. Step 2. We now claim that G has irreducible characters Xl and X2 such that and

Xl(1) ± X2(1)

Xl(ZC) ± X2(ZC) Xl(a n ) ± X2(a n )

+ 1, Xt(z) ± X2(Z) = lI(q + 1), Xt(c) ± Xz(c) = 1, 11, Xl(d) ± X2(d) = 1, Xt(zd) ± X2(zd) = 11 2( -It, X1(b m ) ± X2(bm ) = 0

q

Indeed, put / = A(q-l)/2 (in the notation of Step 1). Then < /,/ >= 2 and so / = Xl ± X2 for some irreducible characters Xl and Xz of G. Since /(z) = 11/(1) implies Xi(Z) = IIXi(l), the claim follows. Step 3. It will be shown here that the functions {3j, 1 :s; j :s; (1/2)(q - 1) listed in the character table are irreducible characters of G. Consider the linear character J-Lj of < b > given by

Then we have

J-Ly(l) m) J-Ly(b

q(q - 1), J-LY(z) 8Jm + 8- j m

= (-l)jq(q -

1)

and J-LY vanishes on the remaining conjugacy classes of G. For each j E {I, ... , (1/2)( q + I)}, let {3j = AAy- Ay - J-LY, where Aj is as in Step 1. Then the values of the generalized character (3j are as follows:

(3j(l) f3j(zc) f3j(an)

q -1, (3j(z) = (-l)j(q - 1), (3j(c)

= -1,

(-l)j+1, f3j(d) = -1, f3j(zd) = (-l)j+I 0, (3j(b m) = _(8 j m + 8- j m)

If 1 :s; j :s; (1/2)(q - 1), then < f3j,{3j >= 1, (3j(1) > 0 and so {3j is an irred uci ble character of G. This verifies the fourth row of the character table. Step 4. Our aim here is to demonstrate that the irreducible characters of G are Ie, A, al,' .. , a(q-3)/2, {31,"" {3(q-l)/2, Xl, X2, /1, /2

Projective Character Tables

274

where Xl, X2 are as in Step 2 and ,1,,2 are two additional irreducible characters satisfying :

11(Z)

+ 12(1) 11(ZC) + 12(ZC) n 11(an ) + 12(a ) II (1)

-V,1(1), 12(Z) = -v,2(1) q - 1, 11(Z) + 12(Z) = -v(q - 1), 11(C) + 12(C) v, 11(d) + 12(d) = -1, 11(zd) + 12(zd) = v 0, 11(bm) + 12(bm) = 2( _1)m+1

= -1

Moreover, we show that

n(l) Put T

= X2(1) = (1/2)(q + 1), 11(1) = 12(1) = (1/2)(q -

= (3(q+1)/2

(in the notation of Step 3). Then < T, T r = II

1)

>= 2 and therefore

± 12

for some irreducible characters II and 12 of G. A direct verification shows that all inner products among lc,>..,ai,(3j",T are zero. Hence all the irreducible characters listed in Step 4 are distinct, except possibly Xi and "tiBut T(Z) = -vT(I), forcing 11(Z) = -V,1(1), 12(Z) = -v'2(1). Hence, by Step 2, the Xi and Ij are distinct. We next show that T is II + 12, not II - 12, and verify the degrees of Xi and Ij, which will establish the assertion of Step 4 (by applying Step 3). Now u = ± 1 and we treat the cases u = 1 and v = -1 separately. First assume that v = 1. Then 11(Z) = -,1(1), 12(Z) = -,2(1), so J( eq1, J( eq2 do not contain < Z ». On the other hand, by Step 2, J( erX1, J( erX2 do contain < Z >. It follows that the irreducible characters of PSL 2 (q) are:

The sum of the squares of their degrees is IPSL 2(q)1= (1/2)q(q2 -1). Thus we must have 1 + q2

q-5

q-l

+ -4-(q + 1)2 + -4-(q q(q2 _ 1) = --'-----'2

which forces

1)2 + X1(1)2

+ X2(1)2

8 Projective character tables for PSL 2 (q)

275

But X1(1) and X2(1) are positive integers and X1(1) ± X2(1) = q + 1 by Step 2. Hence X1(1) = X2(1) = (1/2)(q + 1). Because the sum of squares of the degrees of all irreducible characters of G is equal to IGI, we deduce that 2 2 1 2 /1 (1) + /2 (1) = 2(q - 1) Now we know that

q - 1 = T(1)

= /1(1) ± /2(1)

and that /1(1) and /2(1) are positive integers. It therefore follows that T(l) = /1(1)+/2(1), so T = /1 +/2, /1(1) = /2(1) = (1/2)(q-1), as desired. Next assume that v = -1. In this case, /1 and /2 are characters of PSL 2(q) and Xl, X2 are not. The desired conclusion can now be obtained by applying the argument similar to the case v = 1. Step 5. Here we prove the theorem for the case u = 1. The assumption v = 1 means that q == l(mod 4). Hence, by Theorem 8.4(i), all irreducible characters of G are lR-valued. It x E G - [a(1/4)(Q-1)j, then by Theorem 8.4(ii), we have 1 ICa/(x < z »1 = 2ICa(x)1 while for

x

E

[a(1/4)(Q-1)j,

ICa/(x < z

»1 = ICa(x)1

Recall also that Xl, X2 are the only two irreducible characters of G/ < z > not fully known. Applying the orthogonality relation in the group G/ < z >, we find the values

Because we also know X1(X) + X2(X) by Step 2, and < Xi,la >= 0, we complete the table of values of Xl and X2. By the foregoing, /1 and /2 are the only irreducible characters of G not fully known. Orthogonality relations in G for ICa ( x)1 give the values of

Because we also know /l(X)

+ /2(X)

by Step 4, and

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276

we easily compute the character table of G. Step 6. In this last step, we prove the theorem for the remaining case u = -1. Now u = -1 means q == -1(mod4). Hence, by Theorem 8.5(i), G has precisely four irreducible characters which are not R-valued. Note that 10, A and the (ti and ;3j are all R-valued. Hence Xl, X2, /1, 1'2 are not R-valued, which forces

Xl

= X2

and

"h

= /2

Note also that, by Theorem 8.5(ii), the only x's in G with xz E [x] are the x's in [b(1/4)(q+l)]. Hence again we know ICo/(x < Z > )1. Now /1 and 1'2 are the only irreducible characters of G/ < Z > not fully known. Applying the orthogonality relations in the group G/ < z > gives us the values of

Because we also know

by Step 4, and < Ii, 10 > = 0, we complete the table of values of 11,12. By the foregoing, Xl, X2 =X1 are the only irreducible characters of G not fully known. Orthogonality relations for ICo( x)1 give the values of

Taking into account that, by Step 2, we also know

X1(X) + X2(X) = X1(a:)

+ X1(X)

and Xi(ZX) = -Xi(X), we easily complete the character table of G. This concludes the proof of the theorem. • We now turn to the case when q is a power of 2. The conjugacy classes of S L 2 ( q) in this case are given by Theorem 8.6 and we use the notation of that theorem. Finally, we remind the reader that if q is a power of 2, then

With this information at our disposal, we now proceed to prove the following result.

8 Projective character tables for PSL 2 (q)

277

Theorem 8.8. (Schur (1907), Jordan (1907)). Let q = 2t for some t ~ 1, let E be a primitive (q - 1)-th root of 1 in C and let fJ be a primitive (q + 1)-th root of 1 in C. Then the character table for G = SL 2(q) is

1G A (Xi

j3j

[1] [cI 1 1 q 0 1 q+1 q - 1 -1

[an] 1 1 Em

[b m ] 1 -1

+ E- m 0

0

-( t5

Jm

+ fJ

Jm)

where

1 ~ i ~ (1/2)(q - 2) , 1 ~ j ~ q/2

1 ~ n ~ (1/2)(q - 2) , 1 ~ m ~ q/2 Proof. For the sake of clarity, we divide the proof into two steps. Step 1. Here we demonstrate that the functions A, (Xi, 1 ~ i ~ (1/2)(q-2), listed in the table, are irreducible characters of G = SL2(q). As a point of departure, we put

and H = { [

~

~-1]

Ix E r q , y E r;}

Then P is a Sylow 2-subgroup of G, P
= P < a >, Pn < a > = 1

and

IHI

=

IPII < a> I =

q(q - 1)

Next observe that [c] n H has order q - 1. If 1 ~ n

~

which is a set of cardinality 2q. Note also that

[b m ] n H = 0

(1 ~ m ~ q/2)

(1/2)(q - 2), then

Projective Character Tables

278

Applying the information above, we can now compute XG , for any character X of H. Given i E {O, 1, ... ,q - I}, let Ai be the linear character of H given by

~-k] ~ Eik

x, : [ ~k

Then P ~ J( er Ai and Aq-l = AO = 1H. Next we note that the values of the

q + 1, E

in

A7 are as follows: A7(c) = 1

+ E- in , A7(bm ) =

°

A7

Assume that i E {I, 2, ... , (1/2)(q - 2)}. Then O'i = satisfies < O'i, O'i >= 1 and therefore O'j is an irred ucible character of G. This verifies the third row of the character table. Now observe that < >= 2 and

Ag, Ag

Ag

Hence == 1G + A for some irreducible character A of G. It is clear that A takes the prescribed values. This verifies the second row of the character table. Step 2. We now show that the functions 1 :S j :S q/2, as listed in the character table, are irreducible characters of G. By Step 1, this will complete the proof of the theorem. Consider the linear character flj of < b > given by

e;

Then flY has the following values:

q(q - 1), flY(C) ==

°

0, fly(b m ) = ajm + a- jm

=

Given j E {1,2, ... ,q/2}, let f3j AAr - Ar - flr, where Aj as in Step 1. Then the values of the generalized character (Jj are as follows: q - 1, /3j(c) = -1 0, /3j(b m) = _(ajm

+ a- jm)

8 Projective character tables for PSL 2 (q)

219

Moreover, < (3j,(3j >= 1 and (3j(l) > O. Thus (3j is an irreducible character of G which takes the prescribed values. Finally, the sum of the squares of the degrees of la, A, ai and (3j is

thus completing the proof. •

D. Projective character tables for P S L 2 (q) We can now easily achieve our aim, which is to provide projective character tables for PSL2(q), q a prime power. The only case which will be omitted is the a-character table for PSL 2(9) ~ A 6 , where the cohomology class of a E Z2(PSL2(9),C*) is of order 3 or 6. In what follows, all characters are assumed to be C-characters. First, we quickly dispose of the case where q is even. In that case

Moreover, by Theorem 16.3.2 in Vol.2, the Schur multiplier of SL 2 (q) is trivial unless q = 4, in which case the Schur multiplier is of order 2. Thus, if q i- 4, then all projective character tables of SL 2 (q) are determined by the ordinary character table of SL 2 (q) given by Theorem 8.8. In case q == 4, we have SL 2(4) ~ As. Hence all projective character tables of SL 2(4) are determined by the character table of SL 2(4) (given by Theorem 8.8) and the spin character table of a covering group of As ~ S L 2 ( 4) (given by Theorem 6.6). Hence all projective character tables for PSL2(q), where q is even, have been already determined. By the above, we can concentrate on the case where q is an odd prime power. We remind the reader that if q i- 9, then the Schur multiplier of P S L 2 ( q) is of order two, while for q = 9 the order of the Schur multiplier of P S L 2 ( q) is 6. Thus, by Theorem 4.5, the ordinary character table for PSL 2(q) together with the spin character table of the double cover SL 2(q) of PSL 2(q) determine all projective character tables for PSL 2(q), with the exception of the a-character table of PSL 2(9), where 0' E Z2(PSL 2(9),C*) is such that the cohomology class of 0' is of order 3 or 6. We now proceed to determine the character table for PSL 2(q), q being odd. For any x E SL 2(q), let x denote the image of x in PSL 2(q). Then the conjugacy class [x] of x in PSL 2(q) is the image of the conjugacy class [x]

As

Projective Character Tables

280

of SL 2(q). Assume that q == 1(mod4). Then, by Theorem 8.4(iv),

[IJ, [eJ, [dJ, raj, [a 2j,

, [a(1/4)(q-5)j, [a(I/4)(q-I)J

[bJ, WJ,

, [b(1/4)(q-I)J

are all distinct conjugacy classes of PSL 2 (q).

Let q be an odd prime power and let q == 1(mod4). Denote by e a primitive (q - 1)-th root of 1 in C and by h a primitive (q + 1)th root of 1 in C. Then the character table for G = PSL2(q) is Theorem 8.9.

1e

[1J 1

[anJ

[bmJ

1

1 -1

q 1 q+1 Ckn /lk q-1 0 Ot (_1)n Xl (1/2)(q + 1) X2 (1/2)(q + 1) (_1)n

A

0

htm 0 0

[eJ 1

[dJ

0

0

1

1 1 -1 -1 (1/2)(1 + Jq) (1/2)(1 - .Jq) (1/2)(1 - Jq) (1/2)(1 + Jq)

q-1 1 -< n -< - 4- ,

q-1 1
1
q- 1 1
-

-

4

'

Proof. The irreducible characters of PSL 2 (q) can be identified with all irreducible characters X of SL 2 {q) with < z >~ J( erx, which is equivalent to X(z) > O. Since q == 1(mod4), we have v = (_1)(1/2)(q-l) = 1. Setting /lk = Q2k and Ot = f32t, the result follows from Theorem 8.7 . • Turning to the spin character table for S L 2(q), q == l( mod 4), we remark that by Theorem 8.4(iii),

1 ::; n ::; (1/4)(q - 5), 1 ::;

m ::;

(1/4)(q - 1)

are all distinct representatives of associate pairs of conjugacy classes of the group SL 2(q).

8 Projective character tables for PSL 2 (q)

281

Theorem 8.10. Let q be an odd prime power and let q == 1(mod4). Denote by E a primitive (q - I)-the root of 1 in C and by 8 a primitive (q + l)-th root of 1 in C. Then the spin character table for SL 2(q) is

[1] J.lk Ot /1 /2

q+1 q- 1 (1/2)(q - 1) (1/2)(q - 1)

where ei«

[an] Ekn

[b m]

0 0 0

btm (_l)m+l (-lr+ 1

0

= E(2k-l)n + E-(2k-l)n, 1~

ti ~

btm

(1/4)(q - 5),

1 ~ k ~ (1/4)(q - 1),

[c] [d] 1 1 -1 -1 (1/2)(-1 +.Jij) (1/2)(-1 - .Jij) (1/2)(-1 - Jij) (1/2)(-1 + Jij)

= _(b(2t-l)m + b-(2t-l)m) 1 ~ m ~ (1/4)(q - 1)

1~ t

~

(1/4)( q - 1)

Proof. The spin character table for SL 2 (q) is obtained from the character table of SL 2(q) by choosing only those irreducible characters X of SL 2(q) for which X(z) < O. Since q == 1( mod 4), we have v = (_1)(1/2)(q-l) = 1. Hence, setting J.Lk = Q2k-l and Ot = (32t-l, the result follows from Theorem 8.7. • We now turn to the case where q == -1( mod 4). Recall that, by Theorem 8.5(iv),

[I] ,[a] , [a 2 ]

,

[b]' [b 2 ] ,

, [a(1/4)(q-3)] ,[c], [d] , [b(1/4)(q-3)] ,

W 4)(Q+l )] 1/

are all distinct conjugacy classes of PSL2(q).

Theorem 8.11. Let q be an odd prime power and let q == -1(mod4). Denote by E a primitive (q -1 )-th root of 1 in C and by b a primitive (q+ 1 )-th root of 1 in
Projective Character Tables

282

--- --------_._--

[an] [c] [b m] [d] 1 1 1 1 1 1a -1 A q 1 0 0 1 1 0 q+1 Ekn Ilk q-1 8 -1 -1 0 Ot tm 11 (1/2)(q - 1) 0 (_l)m+I (1/2)(-1 + R) (1/2)(-1- R) 12 (1/2)(q - 1) 0 (_1)m+T (1/2)(-1 - R) (1/2)( -1 + R)

[1]

1 ~ n ~ (1/4)(q - 3), 1 ~ k ~ (1/4)(q - 3),

1 ~ m ~ (1/4)(q+ 1) 1 ~ t ~ (1/4)(q - 3)

Proof. The irreducible characters of PSL 2 (q) can be identified with all irreducible characters X of SL 2(q) for which X(z) > O. Since q == -1(mod4), we have 11 = (-1 )(1/2)(q-1) = -1. Setting Ilk = D:2k and (}t = f32t, the result follows by virtue of Theorem 8.7. • Finally, we consider the spin character table for SL 2(q), q == -1(mod4). Owing to Theorem 8.5(iii),

(1 ~ n

~

(1/4)(q - 3), 1 ~ m ~ (1/4)(q - 3))

are all distinct representatives of associate pairs of conjugacy classes of SL 2 (q).

Theorem 8.12. Let q be an odd prime power and let q == -1(mod4). Denote by E a primitive (q-1 )-th root of 1 in C and by 8 a primitive (q+ 1 )-th root of 1 in C. Then the spin character table for S L 2 ( q) is

[an] [b m] [1] [c] [d] q+1 0 1 1 Ekn Ilk ()t q-1 -1 0 s.; -1 0 (1/2)(1 + R) (1/2)(1- R) Xl (1/2)(q + 1) (_1)n 0 (1/2)(1- R) (1/2)(1 + R) X2 (1/2)(q+ 1) (-l)n where Ekn = E(2k-l)n + C(2k-1)n, 8tm = _(8(2t-l)m

+ 8-(2t-l)m)

1 ~ n ~ (1/4)(q - 3),

1 ~ m ~ (1/4)(q - 3)

1 ~ k ~ (1/4)(q - 3),

1 ~ t ~ (1/4)(q

+ 1)

9 Nonisomorphlc groups with the same projective characater tables

283

Proof. By definition, the spin character table for S L 2 ( q) is obtained from the character table of SL 2(q) by choosing only those irreducible characters X of SLz(q) for which X(z) < O. Since q == -1(mod4), we have v = (_l)(1/Z)(Q-l) = -1. Hence, setting Ilk = a2k-l and ()t = f32t-l, the result follows from Theorem 8.7. •

9

Nonisomorphic groups with the same projective character tables

A. Introduction All characters below are assumed to be C-characters. Let G be a finite group. Then the character table of G carries a lot of information on the structure of G. In fact, by Theorem 22.1.1 in Vol.I, it determines the degrees of irreducible characters, the orders of centralizers of elements (in particular, the order of G), the number of elements in each conjugacy class of G, Z(G) and G', the lattice of normal subgroups of G and the set of primes which divide the order of any given element g of G. Unfortunately, the character table of G does not determine G up to isomorphism, since dihedral and quaternion groups of order 8 have the same character tables. On the other hand, such classical groups as Sn and An are determined by their character tables (see Nagao (1957) and Oyama (1964)). The proof of the facts concerning Sn and An relies on the property that the class multiplication constants are determined by the character table (see Theorem 21.1.2 in VoLl). The question of whether the character table of a given group determines this group up to isomorphism is especially important in the theory of finite simple groups. For example, in characterizing a known simple group, it may be convenient to know that no other group has the same character table. A typical situation is described in G. Higman's paper (Higman (1971)) where it is shown that Janko's first group is determined by its character table. It is natural to enquire what happens if "the character table of G" is replaced by "the projective character tables of G". In other words, do all the projective character tables of G determine G up to isomorphism? According to Hoffman and Humphreys (1987b), this question was first raised by Morris. Let us first of all give a precise meaning to our question. The following definition takes into account that the projective character tables corresponding to cohomologous cocycles determine each other (see

Projective Character Tables

284

Lemma 1.2.5). Let G 1 and G 2 be two finite groups and let M( Gi) be the Schur multiplier of Gi, i = 1,2. By saying that G 1 and G 2 have the same projective character tables, we mean that there is a bijective correspondence Q:i f-+ 13i between the representatives of cohomology classes in M(Gd and M(G 2 ) , respectively, such that for each i, the Q:i-character table C, of G 1 coincides with the 13i-character table D, of G 2 (upon a suitable permutation of rows and columns of Dd. Thus if G1 and G 2 have the same projective character tables, then M( Gd and M( G 2 ) have the same order. For this reason, the dihedral and quaternion groups of order 8 cannot have the same projective character table, since the orders of their Schur multiplier are different. In case G 1 and G 2 satisfy IM(Gdl = IM(G 2 )1 = 2, our question can be reformulated as follows. Let Gj and Gi be any given covering groups of G 1 and G 2 , respectively. Then all projective character tables of G; are determined by : (i) The character table of Gi. (ii) The spin character table of Gi. (see Theorem 4.5). Thus our question can be reformulated as follows: Assume that the groups G 1 and G 2 have the same character table and that Gj and Gi have the same spin character table. Is it true that G 1 ~ G 2 ? It is the purpose of this section to provide a negative answer. The corresponding example involves groups G 1 and G 2 of order 32.

B. The groups G 1 and G 2 Let E E {O, I} and let G = G( E) be the group generated by 91,92,93,94 and 95 subject to the relations

9i [92,g5]

[9i,9j]

g? = 1, 95

= gl, gg = 93

and

91

[93,94] = gl, [94,95] = g2, 1 for the remaining 1:S i, j

= 9~

:s 5

Let G 1 be the group G(O) and G 2 be the group G(l). Our aim here is to show that G 1 and G 2 are nonisomorphic groups of order 32 with the same character table. We treat both groups simultaneously by examining the structure of G = G(E). In what follows, we put /3(G) = [G',G].

Lemma 9.1.

The group G = G(E) satisfies the following properties:

9 Nonisomorphic groups with the same projective characater tables

(i) G' =< 91,92 > is of order 4. (ii) GIG' ~ 1£4 X 1£2 and,in particular, (iii) /3(G) = Z(G) =< 91 .>.

IGI

285

= 32.

Proof. (i) By definition, [91,92J = 1 and so < 91,92 > is of order 4. Since 91 and 92 are commutators, we have < 91,92 >~ G'. By definition, we also have < 91 >~ Z(G). Since

we see that < 91,92 > is a normal subgroup of G. Since the corresponding factor group is obviously abelian, it follows that the group G' =< 91,92 > is of order 4. (ii) The abelian group GIG' is generated by the elements 94G' and 95G' of order 2 and 4, respectively. Hence GIG' ~ 1£4 X 1£2. (iii) It is clear that 91 = [92,95J E /3(G). But, by (ii), G is a 2-group, so /3(G) is a proper subgroup of /2(G) = G'. Hence /3(G) =< 91 ». Assume that 9 E Z( G) and write 9 = 9t9~9J9~, 0 :S; i,j, k :S; 1,0 :S; t :S; 3. Then 9~9~91 E Z(G) and we show that j = k = t = 0, which will complete the proof. Since 92 commutes with 94, we see that 92 commutes with 91. Hence [92, 95J =11 forces t = O. Since 93 commutes with 92, it follows that 93 commutes with 9~' Hence [93,94J =11 forces k = O. Finally, since [92,95) =11, it follows that j = 0, as required . • In what follows, for any 9 E G, [9) denotes the conjugacy class of G. Lemma 9.2. as follows:

All distinct conju9acy classes of the qroup G

= G(E)

are

[lJ, [91], [92), [93], [9293), [94], [9394) [95), [9395], [9495], [939495J

The cardinalities of these conju9acy classes are

and the remaininq conju9acy classes have cardinality 4. Moreover,

G(O) '1- G(l) Proof. The assertion regarding conjugacy classes of G and their cardinalities follow directly from the defining relations of G. If E = 1, then there

Projective Character Tables

286

are precisely 3 elements of order two, namely 91,92,9192' If E = 0, then there is at least one more such element, e.g. 94. Hence G(O) ~ G(1). • To simplify the notation, let us label the conjugacy classes as follows:

where 1 E C l , 91 E C 2 , 92 E C3 , 93 E C4 , 9293 E C s, 94 E C6, 9394 E C 7 , 95 E C s , 9395 E C g, g4g5 E C lO , 93g495 E Cn. Lemma 9.3.

Xl

X2 X3 X4 XS X6 X7 XS Xg XlO Xu

The group G = G(E) has the [ollounnq character table

Cl

C2

C3

C4

Cs

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 -1 -1 -1 -1

1 1 1 1 -1 -1 -1 -1

2 2 4

C6

1 1 -1 -1 1 1 -1 -1 2 -2 2 -2 0 2 -2 -2 2 0

-4

0

0

0

0

C lO

1 1 1 1 -1 -1 -1 1 1 -1 -1 -1 -1 -t t -1 - t t -t 1 t 1 -t t 0 0 0

1 -1 -1 1

1 -1 -1 1

t

-t

-t

t

0 0

Cs

Cu

Cg

C7

0 0

0 0

-t

t

t

-t

0 0

0

0

0 0

Proof. By Lemma 9.1(ii), GIG' is of order 8. Hence G has precisely 8 linear characters, say Xl, ... ,Xs. Now GIG' =< g4G' > X < g5G' >, where the first factor is of order 2 and the second is of order 4. Hence, if X is a linear character of G, then X(94) = ±1 and X(9s) = ±1, ±i. Since, by Lemma 9.1(i), G' =< 91,92 >, it follows that Xl, ... , Xs have the prescribed values. Since IGI = 32 and G has 11 conjugacy classes, the remaining three irreducible characters of G, say Xg, XlO and Xn satisfy

X9(1) = XlO(1) = 2 and

Xn(1) = 4

If < 91 > ~ J( erXn, then a group of order 16 has an irreducible character of degree 4, which is impossible. Thus Xu (Yl) = -4 and row orthogonality

9 Nonisomorphic groups with the same projective characater tables

287

shows that Xu vanishes on the remaining nine conjugacy classes. By the foregoing, we are left to verify that X9 and XlO have the prescribed values. Applying the column orthogonality for C l and C2 , we see that

Thus X9 and XlO contain 91 in their kernel, so X9 and XlO can be regarded as irreducible characters of G = GjZ(G). Note also that

Moreover, since by Lemma 9.1(i), G' =< 91,92 >, we see that 92 rJ. J(erX9, 92 rJ. J(erxlO. Thus X9(C3) = -2 and XlO(C3) = -2. Since 93Z(G) is a central element of order two in G, we also have

By column orthogonality relation, applied to Cl and C 4 , X9(C4 ) and XlO(C4 ) must have the opposite sign. Hence we may assume that X9(C4) = 2 and XlO(C4 ) = -2, in which case X9(C5 ) = -2 and XlO(C5 ) = 2. The remaining entries for X9 and XlO follow from < X9,X9 >=< XlO, XlO >= 1. • Corollary 9.4. character table.

The qroups G1

= G(O)

and G2

= G(l)

have the same

Proof. The character table for G = G(£), e E {O, I}, does not depend upon the choice of E• •

C. The double covers of G l and G 2 Let e E {O,I} and let G*(c) be the group generated by a central involution Z together with Xl, X2, X3, X4 and X5 subject to the relations:

=

21

Xl

2

'X2

= Z, X32 = ZXl, X52 = X3

an d

= Xl

2"

X4

= ZXI, [X2, X4] = Z, [X3' X4] = Xl, [X4, X5] = X2 [Xi, Xj] = 1 for the remaining 1 ~ i,j ~ 5

[X2, X5]

Then it is easily seen that G*( E) is a double cover of G(c) via the exact sequence: 1

-.<

Z

>-. G*(E) L G(c) -. 1

Projective Character Tables

288

where

f(Xd = gi

for all

1

~

i ~ 5

Since Z = [X2,X4] E [G*(E),G*(E)] n Z(G*(E», we see that

IM(G(E))I

is even

It will be shown later (see Theorem 9.7) that M (G( E)) is of order two. In what follows, for any 9 E G*( E), [g] denotes the conjugacy class of g. Lemma 9.5.

Xl X2 X3 X4

Xs X6 X7

XS

[1] 2 2 2 2 2 2 2 2

The spin character table for G*(E) is as follows:

[Xl] -2

-2

[X3]

0 0 0 0

-2 -2 2 2i 2 -2i 2 -2i 2 2i

[xs] 1 +i 1- i -1+i -1-i

[X3 XS]

[X4XS]

[X3X4 XS]

-2i 2i 2i -2i

1- i 1 +i -1- i -1+i

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

l+i 1- i -1 + i -1- i

-l+i -1-i 1 +i 1- i

[X2 X3]

Proof. For the sake of clarity, we divide the proof into three steps. Step 1. Here we show that it suffices to construct spin characters Xl, ... , Xs with the prescribed values on the chosen conjugacy classes. So assume that such spin characters have been found. By Theorem 4.4(ii), the sum of the squares of the degrees of spin characters of G*(E) is equal to IG(E)I = 32. Thus XI, ... ,XS are all characters of G*(E). Since no two columns in the table are the same, the chosen conjugacy classes are distinct. Since there are no zero columns, no chosen conjugacy class is self-associate (see Lemma 4.1(iii»). Finally, since no column is obtained from another column by multiplying by -1, we see that no two distinct conjugacy classes are associate (see Lemma 4.1(i). Thus the chosen conjugacy classes form a full set of distinct representatives of associate pairs of conjugacy classes of G*(E). Step 2. Here we construct spin characters Xl, X2, X3 and X4 with the prescribed values. Put H = G*(E) and HI =< Z, Xl, X2, Xs >. Then HI is a subgroup of index 2 in H with H / HI =< X4HI >. It is clear that H~ =< ZXI >

9 Nonisomorphic groups with the same projective characater tables

289

and HdH~

=< xzH~

>

X

< xsH~

>~ 1£4

X

1£4

Let ,X be the linear character of HI with 'x(xz) = 'x(xs) = i. Then 'x(X3) = 'x(xs)Z = -1, 'x(XI) = 'x(z) = 'x(xz)Z = -1. Now put Xl = ,XH. Then XI(Z) = -2 < 0 and Xl is irreducible (since otherwise, by Frobenius reciprocity, ,X extends to a character of H, which is impossible in view of Z E H'). Hence Xl is a spin character. We have:

xI(Xd XI(X3) XI(XZ X3) XI(XS) XI(X3XS)

'x(Xd + 'x(x 4 lxIX4) = 'x(xd + 'x(xd = -2 'x(X3) + 'x(x4lx3X4) = 'x(X3) + 'x(x3xd = 0 'x(XZX3) + 'x(x4lxZX3X4) = 'x(XZX3) + 'x(XZZX3X4) = -2i 'x(xs) + 'x(x4lxSX4) = 'x(xs) + 'x(XSX2 1) = 1 + i 'x(X3XS) + 'x(X4IX3XSX4) = 'x(X3XS) + 'x(X3XIXSX21) = 1- i

Setting Xz = Xl, we see that the spin characters Xl and Xz have the prescribed values. Now let ,X be another linear character of HI, namely 'x(xz) = -i, 'x(xs) = i. Then 'x(X3) = -1, 'x(xd = 'x(z) = -1. Setting X3 = ,XH, it follows that X3 is a spin character of H with X3(X4XS) = X3(X3X4XS) = O. Moreover, we have

-2, X3(X3) = 0, X3(XZX3) = 2i -1 + i, X3(X3XS) = -1 - i Setting X4 = X3, we see that the spin characters X3 and X4 have the prescribed values. Step 3. We now complete the proof by constructing spin characters Xs, X6, X7 and Xs. Put Hz =< Z, Xl, Xz, X3, X4XS >. Then Hz is a subgroup of index 2 in H with H/H z =< X4HZ >. It is clear that H~ =< Xl > and

Let ,X be the linear character of Hz with 'x(xz) = 'x(X4XS) = i. Then 'x(z) = 'x(xz)Z = -1, 'x(xd = 1, 'x(XZX3) = 'x((X4XS)2) = -1, 'x(X3) = i (since

Projective Character Tables

290

(X4X5)2 = x~+lX2X3)' Setting X5 of H. We have

= >..H, it is clear that X5 is a spin character

+ >"(x41XIX4) = 2 >"(X3) + >"(X41x3X4) = >"(X3) + >"(X3Xt) = 2i >"(X2X3) + >"(x41X2X3X4) = >"(X2X3) + >"(X2ZX3Xl) = 0 >"(X4X5) + >"(x41x4X5X4) = i + 1 >"(X3X4X5) + >"(X41x3X4X5X4) = -1 + i

>"(Xd

x5(xd X5(X3) X5(X2 X3) X5(X4 X5) X5(X3 X4 X5)

Since X5, X3X5 rt. H 2, we also have X5(X5) = X5(X3X5) = O. Setting X6 = X5, we see that spin characters X5 and X6 have the prescribed values. Now let>" be another character of H 2, namely >"(X2) = -i, >"(X4X5) = i. Then X7 = >..H is a spin character of H with the prescribed values. Finally, setting XS = X7, the result follows. •

Corollary 9.6. Let G 1 = G(O), G2 = G(1), Gj = G*(O) and Gi G*(1). Then Gi is a double cover of Gi, i = 1,2, and the spin character tables for

Gi

and

Gi are

the same.

Proof. We have already observed that Gi is a double cover of Gi, i =:: 1,2. Moreover, by Lemma 9.5, the spin character table for G*(E) does not depend upon the choice of E E {O, I}. So the corollary is true. • D. The main result We have now accumulated all the necessary information for the proof of the main result below. In view of our discussion in Sec.A, it provides two nonisomorphic groups with the same projective character tables.

Theorem 9.7. (Hoffman and Humphreys (1987b)). Let £ E {O, 1} and let G = G(E) be the qroup generated by 91,92,93,94 and 95 subject to the relations 9r = 9~ = 1, 9~ = 91, 9g = 93 and 9~ = gf [92,95]

= [93,94] = 91, [94,95] = 92,

= 1 for the remainin9 1 S; i,j S; 5 Put G 1 = G(O) and G2 = G(I). Then Gl and G2 are nonisomorphic groups of order 32 with IM(Gdl = IM(G2 )1= 2 such that: [gj,gj]

9 Nonisomorphic groups with the same projective characater tables

291

(i) G 1 and G 2 have the same character table. (ii) A covering group Gi of G 1 and a covering group Gi of G 2 have the same spin character table.

Proof. Let the group G*(E) be defined as in Sec.C, let Gi = G*(O) and let Gi = G*(l). By Corollary 9.4, the groups G 1 and G2 have the same character table, while by Corollary 9.6, Gi is a double cover of Gi, i = 1,2, and the spin character table for Gi and Gi are the same. Note that, since in the construction of G*(E), Z E [G*(E),G*(E)], Gi is a covering group of Gi, i = 1,2, provided we show that M(Gd is of order 2. Since, by Lemmas 9.1 and 9.2, G t and G 2 are nonisomorphic groups of order 32, we are left to verify that for E E {O, I} IM(G(E))I = 2 Setting N =< g5,g2 >, we see that N is a normal subgroup of G = G(E) having index 2 and

g58 = 1,g22 = 1,92-1 9592 = 955 Hence, by Corollary 10.1.26 in Vol.2, M(N) = 1. Setting 9 = 94, we have GIN =< gN >. Denote by CNIN,(g) the subgroup of elements of NIN' fixed under conjugation by g. Since N' =< gl >, a direct calculation shows that CNINI(g) = {N',g2N',g~N',g2g~N'} Let L be the subgroup of CN/NI(g) generated by g2 N' elements of the form

(x E N)

(XN')(g-l xg)N' Then a direct calculation shows that

Hence, by Theorem 11.10.4 in Vo1.2,

/M(G)/ = ICNIN,(g)1 = 2

ILl

as desired. •

= gl N' = N'

and all