IRREDUCIBLE CHARACTERS
OF SMALL DEGREE OF THE
UNITRIANGULAR GROUP
Martin Marjoram
We let q be a power of some prime p and for all n 2 N , we let Un (q)
be the set of n n matries over Fq whih are lower-triangular
and have every diagonal entry equal to 1. Un(q) is alled the
unitriangular group of degree n over Fq . Un (q ) is a Sylow p{
subgroup of GLn (q), the general linear group of invertible n n
matries over Fq . For any nite group G, we will write Irr(G)
for the set of irreduible omplex haraters
of G. Isaas [2℄ has
shown that every element of Irr Un (q) has degree a power of q.
Using this result of Isaas, a theorem of Huppert [1℄ for the ase
where q = p an be extended to prove that
n
(1) :
o
2 Irr Un (q)
=
qa
: 0a
n(n 1)
:
2
In [4℄, the author enumerated the irreduible omplex haraters
of U2m (q) having eah of the three highest degrees, as well as the
irreduible omplex haraters of U2m 1 (q) of highest
degree. In
this paper, we onsider the elements of Irr Un (q) having eah of
the three lowest degrees, namely 1, q, and q2 . As the underlying
eld will remain xed throughout, we will write Un in plae of
Un (q).
It is easy to see that the ommutator subgroup Un0 of Un is
given by
U 0 = f(aij ) 2 Un : ai;i 1 = 0 for 2 i ng:
n
We let Vn;1 = Un=Un0 . Then the elements of Irr(Un) of degree 1
are obtained by extending the irreduible haraters of Vn;1 to Un.
21
22
IMS Bulletin 42, 1999
As Vn;1 is an (elementary) abelian group of order qn 1 , this means
that there are preisely qn 1 irreduible haraters of Un of degree
1. We will show that the irreduible haraters of Un of degrees
q and q2 may be obtained by extending haraters of analogous
fator groups and we will ount the number having eah degree.
For 0 t n 1, we dene
Wn;t = f(aij ) 2 Un : aij = 0 if 1 < i j tg:
Thus, for example, Wn;0 = Un , Wn;1 = Un0 , Wn;3 = Un00 , and
Wn;n 1 = I . Eah Wn;t is a normal subgroup of Un . For 0 t n 1 we dene
Vn;t = Un=Wn;t :
Thus the elements of Vn;t may be identied with lower-triangular
n n matries over Fq with 1's along the diagonal and for whih we
only onsider the entries on the main diagonal and on the t diagonals below the main diagonal. We will write A~ for the elements
AWn;t of Vn;t .
For 1 s n, we dene
Nn;s = f(aij ) 2 Un : aij = 0 for i > j if either i s or j > sg
I
0
s
=
C In s : C 2 Mn s;s (Fq )
and
Hn;s = f(aij ) 2 Un : aij = 0 for i > j where i > s and j sg
A 0 : A2U ;B2U
=
:
s
n
s
0 B
For all s, Nn;s is a normal abelian subgroup of Un and Hn;s is a
subgroup of Un whih omplements Nn;s , i.e. Nn;s \ Hn;s = I and
U = Nn;s Hn;s . Usually
we write the elements of Hn;s as (A; B )
instead of A0 B0 .
The unitriangular group
23
We will make extensive use of the following lemma. It follows partly from a theorem of Gallagher, while the rest an be
dedued from standard Cliord Theory. (See Corollary 6.17 and
Problem 6.18 of [3℄.) The proof is straightforward and we omit it.
Lemma 1. We let G be a nite group, N a normal abelian subgroup of G, and H a subgroup of G whih omplements N . We
let be an element of Irr(N ) and we let TG () be the stabilizer of
in G, i.e.
TG() = fg 2 G : g = g;
where g 2 Irr(N ), g (n) = (g 1 ng ) for all n 2 N . We let
SG () = TG() \ H = TG()=N . Note that TG() = SG ()N . We
dene a harater of TG () by (hn) = (n) for all h in SG ()
and all n in N . Then for eah irreduible harater of SG (),
G is an element of Irr(G), distint for distint 's, and
G =
X
2Irr(S ())
(1) G :
G
Starting instead with an irreduible harater of G, if lies over
2 Irr(N ), then there exists an irreduible harater of SG ()
suh that ours with multipliity (1) in jN and
(1) = (1)
jH j
jSG()j :
For 1 s n and for every s (n s) matrix D over Fq ,
we dene D : Nn;s ! C by
D
Is 0
C In s
!
= !T (tr (CD)) ;
where ! is a primitive pth root of unity in C , T : Fq ! Fp is
the usual trae mapping from an extension eld into the ground
eld, tr denotes the trae of a square matrix, and we identify the
elements of Fp with the integers 0, 1,: : : , p 1.
24
IMS Bulletin 42, 1999
Lemma 2. For 1 s n, Irr(Nn;s ) = fD : D 2 Ms;n s g and
D1 = D2 if and only if D1 = D2 .
Proof: It is easy to hek that D is a degree 1 representation,
and so an irreduible harater, of Nn;s . Nn;s is abelian with
qs(n s) elements, so the number of D 's is equal to the number of
irreduible haraters of Nn;s . The proof will be omplete if we
show that D1 = D2 implies that D1 = D2 . Now,
D1
=
D2
) ! (tr 1 ) = ! (tr 2 ) for all C 2 M (F )
) T tr (C (D D )) = 0 for all C 2 M (F )
) tr (C (D D )) = 0 for all C 2 M (F )
T
(CD )
T
1
1
(CD )
n
2
2
n
n
s;s
s;s
s;s
q
q
q
where the last impliation follows from the well-known fat that
T : Fq ! Fp is non-zero.
But the last ondition above implies that D1 D2 = 0.
For suppose the (i; j ){entry of D1 D2 is non{zero. Then we
an let C = Eji , the (n s) s matrix with 1 in the (j; i){
position and zeros
elsewhere. It is lear that for this hoie of C ,
tr C (D1 D2 ) is equal to the (i; j ){entry of D1 D2 , and so is
not equal to zero, a ontradition. Thus D1 = D2 implies that
D1 = D2 and the proof is omplete.
For 1 s n, we an now apply the result from Lemma 1
to the normal abelian subgroup Nn;s of Un and its omplement
Hn;s . For all elements D of Ms;n s , we let S (D) = SU (D ).
n
Lemma 3. Let D be an element of Irr (Nn;s ) for some value of
s. Then
S (D) = f(A; B ) 2 Hn;s : ADB 1 = Dg:
25
The unitriangular group
Proof: Let (A; B ) be any element of Hn;s . Then
(DA; B)
!
I 0
C I
= D
= D
A 0
0 B
1
I
0
B 1 CA I
= !T (tr (B CAD))
1
= !T (tr (CADB ))
I 0
C I
!
A 0
0 B
!
1
= ADB
1
I 0
C I
!
for all elements CI I0 of Nn;s . Thus (DA; B) = ADB
1
and
o
n
S (D) = (A; B ) 2 Hn;s : (DA; B) = D
n
= (A; B ) 2 Hn;s : ADB
n
1
= D
o
o
= (A; B ) 2 Hn;s : ADB 1 = D :
It follows from the proof above that Hn;s ats on Irr(Nn;s)
via the ation (A; B ) : D 7! ADB 1 . We let O(D ) be the
orbit of D under this ation. Of ourse every element D0 of
O(D ) lies under preisely the same irreduible haraters of Un
as D while S (D0 ) = S (D).
Lemma 4. Let 2 Irr(Un ) and hoose t maximal suh that
Wn;t+1 ker ( ) but Wn;t 6 ker ( ). Then there exists Y 2 Wn;t
suh that Y 62 ker ( ) and Y has preisely one non-zero entry
below the main diagonal.
Proof: Let X = (xij ) be an element of Wn;t
Z = In
n
X
l=t+2
ker ( ). We dene
xl;l t 1 El;l t 1 :
26
IMS Bulletin 42, 1999
We laim that Z 62 ker ( ). For if f is the representation of U
whih aords , then f (ZX ) = I beause ZX 2 Wn;t+1 . Thus
f (Z ) = f (X ) 1 6= I . We note that
Z = (I
(I
xt+2;1 Et+2;1 )(I xt+3;2 Et+3;2 ) : : :
xn;n t 1 En;n t 1 ):
If every term in this produt were an element of ker (f ), then Z
would also be an element of ker (f ). Thus for at least one hoie
of k satisfying t + 2 k n,
xk;k t 1 Ek;k t 1 62 ker ( );
I
as required.
Theorem 5. Take 2 Irr(Un ) and hoose t maximal suh that
Wn;t+1 ker ( ) but Wn;t 6 ker ( ). Then (1) qt .
Proof: We hoose Y = I + Ek;k t 1 not in ker ( ). We note
that Y is an element of Nn;k 1 . Thus there is at least one element
D of Irr(Nn;k 1 ) for whih Y is not an element of ker (D ), i.e.
tr E1;k t 1 D 6= 0, where D is a (k 1) (n k + 1) matrix over
Fq . If D = (dij ), then it follows that dk t 1;1 6= 0. We let = db1
be the rst non{zero entry in olumn 1 of D. Then b k t 1.
(In fat it must be the ase that b = k t 1, but we do not need
to prove this.) We let
n
X
dl1
E :
lb
l=b+1
A=I
We set D0 = AD. Then D0 is an element of O(D ) and
D0 = AD
n
X
d 1E
= +1
= I
l
= E 1 +
b
l
!
E 1 +
b
lb
i
j
n
d0 E ;
ij
k
l
1
k+1
ij
!
X
2 1
i
j
X
2 d 1E 1 +
l
l=b+1
b
1
n
X
n
d E
ij
k
1
k+1
ij
The unitriangular group
27
for some d0ij 's in Fq . Thus D0 has preisely one non{zero entry in
olumn 1, namely in the (b; 1){position.
Now we have
S (D) = S (D0 ) = f(A; B ) 2 Hn;k 1 : AD0 B 1 = D0 g:
Writing A = (aij ), the equation AD0 = D0 B gives us relations
whih the k 1 b entries ab+1;b , ab+2;b , : : : , ak 1;b must satisfy.
Thus
n;k 1 j
jS (D)j = jS (D0 )j jH
k
q 1 b
qk jH1 n;k(k 1t j 1)
jH j
= n;kt 1 :
q
By Lemma 1, there exists 2 Irr S (D) suh that
jH j
(1) = (1) n;k 1
jS (D)j
jHjHn;k j1=qj t
n;k 1
t
=q :
Denition: For a nite group G and a positive integer l, we
dene C (G; l) to be the number of irreduible haraters of G
having degree l.
Corollary 6. Choose t suh that 1 t n 1. Then for
0 a t 1, we have
C (Un ; qa ) = C (Vn;t ; qa ):
Proof: If is an element of Irr(Un ) suh that (1) q t 1 , then
by Theorem 5 Wn;t ker ( ). Thus is an element of Irr(Vn;t ).
Conversely, given any element of Irr(Vn;t), we an extend to
28
IMS Bulletin 42, 1999
an element of Irr(Un ) by letting (A) = (I ) for all elements A
of Wn;t .
To nd the number of irreduible haraters of Un of degrees
q and q2 , we need only alulate C (Vn;2 ; q) and C (Vn;3 ; q2 ). While
our main interest in Vn;2 is in the degree q haraters, it turns out
that for this group there is a simple losed formula desribing the
number of irreduible haraters having any degree. The formulas
we derive
oeÆients. We adopt the onventions
involve binomial
that 00 = 1 and st = 0 whenever s < t.
Theorem 7. f(1) : 2 Irr(Vn;2 )g = fqa : 0 a n 2 1 g and,
for all values of a in this range,
C (V 2 ; q ) =
n;
a
n
a
a
1
q
n
a
2
(q 1) +
a
n
a
2
a
q
n
2
a
(q 1)
a+1
:
Proof: We proeed by indution on n. V2;2 = U2 has q irreduible
haraters, all of degree 1, as the theorem states. V3;2 = U3 has
q2 haraters of degree 1 while every other harater degree is a
power of q. Reall that the squares of the degrees of the irreduible
haraters of a nite group G add up to the order of G. It follows
that f(1) : 2 Irr(V3;2 )g = f1; qg and that C (V3;2 ; q) = q 1,
whih again is in agreement with the statement of the theorem.
For n > 3, we let
n
o
Jn = (aij ) + Wn;2 2 Vn;2 : an;n 2 = an;n 1 = 0
and
n
o
Kn = (bij ) + Wn;2 2 Vn;2 : bij = 0 for i 6= n :
Now Kn is a normal abelian subgroup of Vn;2 , having Jn as omplement. For all and in Fq , we dene ; : Kn ! C by
; (bij ) + Wn;2 = !b
n;n
2
+b
n;n
1
:
It is easy to prove that Irr(Kn) = f; : ; 2 Fq g and
0 ;0 = ; if and only if 0 = and 0 = . Furthermore, for
29
The unitriangular group
any elements A~ = (aij ) + Wn;2 of Jn and B~ = (bij ) + Wn;2 of Kn,
~ (B~ ) = A~ (b ) + W A;
n;2
; ij
1
= ; A (bij )A + Wn;2
= ; I + (bn;n 2 + bn; n 1 an 1;n 2 )En;n 2 +
bn;n 1En;n 1 + Wn;2
= !T ((b 2+b 1 a 1 2 )+b 1)
= !T (b 2+(+a 1 2 )b 1 )
= ;(+a 1 2) (B~ );
n;n
n;n
n
n;n
n
n
;n
n;n
;n
n;n
;n
~ =
i.e. A;
;( +") , where " = an 1;n 2 .
We dene S (; ) = SV 2 (; ). Then for all non-zero
elements of Fq and all elements of Fq ,
n;
S (0; ) = Jn = Vn 1;2
and
S (; ) = S (; 0)
n
= (a ) + W
ij
n;2
2J : a
= (a ) + W
n;2
2V
n
ij
=V
n
2;2
n
n;2
n
:a
1;n
n
2
1;n
=0
2
o
=a
n;n
=a
1
o
n;n
2
=0
(F ; +):
q
For all non-zero elements of Fq and all elements of Fq ,
Lemma 1 tells us that
n
=
and
n
2 Irr(Vn;2 ) : jN ; 0;
0; V
n;2
o
N 6= 0
: 2 Irr S (0; )
o
30
IMS Bulletin 42, 1999
n
=
=
n
n
o
2 Irr(Vn;2 ) : jN ; ; N 6= 0
o
2 Irr(Vn;2 ) : jN ; ;0 N 6= 0
;0 V
n;2
: 2 Irr S (; 0)
o
:
Lemma 1 also implies that, with the same notation as in the sets
above,
0; V 2 (1) = (1)
and
;0 V 2 (1) = q(1):
Thus for 0 a n 22 1 + 1,
n;
n;
C (V 2 ; q ) =qC (V 1 2 ; q
=q 1 1 q
q 1 2q
q (q 1) 2
q (q 1) 2
2
=
q
2
q
1
1
=
q
n;
a
n
;
n
a
n
a
a
a
n
a
a
a
n
a
n
a
a
n
a
a
n
a
a
n
a
n
a
a
a
n
a
a
1
+
a
a
n
a
;
a
n
n
n
a
a
n
) + q(q 1)C (V 2 2 ; q 1 )
1
2
(q 1) +
1
2
(q 1) +1 +
(
1) 1
q 2 ( 1) 2 (q 1)
1
(
1) 2
q 2 ( 1) 2 (q 1)
1
3
2
2
(q 1) +
q
(q
3
2
2
(q 1) +
q
(q
1
2
2
2
(q 1) +
q
(q
a
a
n
a
n
a
a
n
a
a
n
a
a
a
n
a
n
a
a
1)
1)
1)
a+1
+
a+1
a+1
:
It seems to be muh more diÆult to nd a formula for
C (Vn;3 ; qa ) whih holds for all relevant values of a. In this ase,
we will onne our attention to the haraters of degree q2 and
state without proof our ndings in this ase.
Theorem 8. We have C (V4; 3 ; q2 ) = q(q 1) and for n 5,
C (V
n; 3
; q2 ) =
(q 1)2 + 2 4 q 4 (q 1)3 +
(n 3)q 2 (q 1) + (n 5)q 2 (q 1)2 :
n
2
3
q
4
n
n
n
n
n
The unitriangular group
31
We an now provide formulae for the number of irreduible
haraters of Un having the degrees 1, q and q2 .
Theorem 9. For n 2, Un has qn 1 irreduible haraters of
degree 1. For n 3, Un has (n 2)q n 3(q 1)+(n 3)q n 3(q 1)2
irreduible haraters of degree q . U4 has q (q 1) irreduible
haraters of degree q 2 ; for n 5, Un has n 2 3 q n 4 (q 1)2 +
n 4q n 4 (q 1)3 + (n 3)q n 2 (q 1) + (n 5)q n 2 (q 1)2
2
irreduible haraters of degree q 2 .
Proof: This follows from Corollary 6, Theorem 7, and Theorem 8.
Referenes
[1℄ B. Huppert, A remark on harater degrees of some p{groups, Arhiv
der Math. 59 (1992), 313-318.
[2℄ I. M. Isaas, Charaters of groups assoiated with nite algebras,
J. Algebra 177 (1995), 708-730.
[3℄ I. M. Isaas, Charater Theory of Finite Groups. Dover: New York,
1994.
[4℄ M. Marjoram, Irreduible haraters of Sylow p{subgroups of lassial
groups, Ph. D. thesis (1997), National University of Ireland, Dublin.
Martin Marjoram
Department of Mathematis
University College
Beleld
Dublin 4