Short presentations for three-dimensional
unitary groups
Alexander Hulpke and Ákos Seress∗
Department of Mathematics
The Ohio State University
231 West 18th Avenue
Columbus, OH 43202, USA
Abstract
We give a presentation of length O log2 |G| for the groups G ∼
= PSU3 (q).
This result has applications in recent algorithms to compute the structure of
permutation groups and matrix groups.
1
Introduction
A presentation for a group is a description by generators and relators. The length
of a presentation is the number of symbols required to write the presentation; we
count each generator as a single symbol, and allow the relators to use powers of
the generators. The exponents are written in binary form, and the length of ge is
1 + log2 (e).
In [BGK+ 97], Babai,
Goodman, Kantor, Luks, and Pálfy gave presentations
of length O log2 |G| for all finite simple groups G, except the rank one groups of
Lie type PSU3 (q) = 2 A2 (q), 2 B2 (22m+1 ), and 2 G2 (32m+1 ). This result has applications in fast computations with permutation groups and matrix groups [KS99,
KS01, KS]. In order to extend these computational results to input groups which
have composition factors isomorphic to the aforementioned three types of rank
one groups, it is necessary to give presentations which are of polylogarithmic
∗ Research
partially supported by the NSF.
1
length in the group order in these groups as well. Short presentations for these
groups also would prove a conjecture of Mann [Man98] about the number of
groups with a presentation of given length.
Recently, J. Thompson (personal communication to W. Kantor) observed that
Suzuki’s original paper [Suz62, p. 128] characterizing
the groups 2 B2 (22m+1 )
contains a presentation of length O log2 |G| for these groups. The purpose of
this note is to prove a similar result for the groups PSU3 (q).
Theorem 1.
The finite simple group G = PSU3 (q) has a presentation of length
O log2 |G| .
It was known already that a presentation for the Borel subgroup and q extra
relations suffice to present PSU3 (q). We give in Lemma 13 a presentation of
length O log2 q for the Borel subgroup and show in Lemma 14 and Theorems 15,
16 how to reduce the q extra relations to a constant number.
We remark that the same asymptotic result can be achieved if the powers of the
generators are not allowed in the presentation. For each occurrence of a power xn
in the presentation indicated in Theorem 1, let n = ∑ki=0 εi 2i be the binary decomposition of n with εi ∈ {0, 1}. We add new symbols x0 , x1 , x2 , . . . , xk ,y−1 , y0 , y1 , . . . , yk
to the generating(
set and the relators x = x0 , xi+1 = xi · xi for i = 0, 1, . . . , k − 1 and
yj
if ε j+1 = 0,
for j = −1, 0, . . . , k − 1. Then xn can be
y−1 = 1, y j+1 =
y j · x j+1 if ε j+1 = 1,
replaced by yk in the original relators containing xn .
2
The structure of PSU3(q)
Let q = pl be a prime power and take GU3 (q) the group of all 3 × 3 invertible
matrices over GF(q2 ) that preserve the hermitian form


0 0 1
ω :=  0 1 0  ,
1 0 0
i.e., GU3 (q) = {A ∈ GL3 (q2 ) | AωĀt = ω}. Here Ā denotes the matrix obtained
from A by raising each entry to the qth power. We will describe elements of
G = PSU3 (q) via matrix representatives modulo scalars.
2
Our notation now follows [Suz65]: G has a maximal parabolic subgroup hQ, ki
where Q and k are represented by the matrices




1 0 0


2
q
q+1


y 1 0
Q = u(x, y) =
=0
x, y ∈ GF(q ), x + x + y


x −yq 1
and

ρ
0
0
k = diag(ρ) :=  0 ρq−1 0  ,
0
0
ρ−q

where ρ is a generator of the multiplicative group GF(q2 )∗ of GF(q2 ).
We remark that, similarly to the notation Ā above, it is customary to denote wq
by w̄ for elements of GF(q2 ). However, since we have to work with a variety of
powers besides wq , we shall use the power notation for all powers.
Note that |k| = (q2 − 1)/ gcd(3, q + 1) in G and |Q| = q3 . We observe that
k normalizes Q and that hQ, ki is a solvable group. The projection SU3 (q) →
PSU3 (q) is injective on Q and we can thus identify Q with the set of matrices
u(x, y).
The centre of Q is



 1 0 0

Z(Q) =  0 1 0  x ∈ GF(q2 ), x + xq = 0


x 0 1
and is therefore elementary abelian of order q. The factor group Q/Z(Q) is elementary abelian of order q2 . Hence hQ, ki has a generator set {u1 , . . . , u3l , k}
where Z(Q) = hu1 , . . . , ul i and it has a presentation F = hû1 , . . . , û3l , k̂ | R i such
that the map ψ : k̂ 7→ k, ûi 7→ ui (1 ≤ i ≤ 3l) is an isomorphism between F and
hQ, ki. We shall prove in Lemma 13 that u1, . . . , u3l can be chosen so that F =
hû1 , . . . , û3l , k̂ | R i has length O log2 |hQ, ki| = O log2 |G| .
Every u ∈ hQ, ki can be written uniquely in the form u = ∏3l
uαi i kα with
i=1
αi α
0 ≤ α < |k|, 0 ≤ αi < p. We call the corresponding word dec u = ∏3l
i=1 ûi k̂
the decomposition of u (with respect to the generating system (u1 , . . . , u3l , k)).
The action of G on the cosets of hQ, ki is doubly transitive. The group G is
therefore generated by hQ, ki and one extra element, for which we choose


0 0 1
t =  0 −1 0  .
1 0 0
3
We also note that the subgroup Q acts regularly on the cosets different from hQ, ki.
This means that every element of G is either in hQ, ki, or can be written in the form
a · t · b with a ∈ hQ, ki and b ∈ Q. Furthermore kt = k−q , thus t normalizes hki.
Therefore for every u ∈ Q \ {1} there are g(u), f (u) ∈ Q and h(u) ∈ hki such that
ut = g(u) h(u) t f (u).
(2)
If we take an element of Q \ {1}


1 0 0
u = u(x, y) =  y 1 0 
x −yq 1
in matrix form, we can compute g(u), h(u), f (u) and uk explicitly as




x
0
0
1
0
0
1
0 ,
g(u) =  −y/x
h(u) =  0 xq−1 0  ,
q
q
1/x y /x 1
0 0 x−q


1
0 0
1 0 ,
f (u) =  −y/xq
q
1/x y /x 1


1
0
0
1
0 .
uk =  yρ2−q
xρq+1 −yq ρ2q−1 1
We want to write a presentation for PSU3 (q) using a generating set {û1 , . . . , û3l , k̂}
corresponding to the generators of hQ, ki and a generator tˆ corresponding to t. Its
−q
relators will contain the relations R for hQ, ki, the relations tˆ2 = 1 and tˆ k̂ tˆ = k̂
tˆ
as well as extra relations of the form dec u = dec g(u)h(u) tˆ dec f (u) for
some elements u ∈ Q. If 36 q + 1 then we will need 3 extra relations and if 3 q + 1
then we will need 7 extra relations.
Definition 3. For a set of elements a1 , . . . , am ∈ Q \ {1}, we define a finitely presented group P(a1 , . . . , am ) as follows.
P(a1 , . . . , am ) :=
k̂, û1 , . . . , û3l , tˆ | R ,
tˆ2 = 1, tˆ k̂ tˆ = k̂
−q
,
tˆ dec ai tˆ = dec g(ai )h(ai ) tˆ dec f (ai ) (1 ≤ i ≤ m)
4
Since the length of the relators of the form
tˆ dec ai tˆ = dec g(ai )h(ai ) tˆ dec f (ai )
is O (log q), Lemma 13 implies that the length of the presentation in Definition 3
is O log2 q + m log q .
For a given finitely presented group P(a1 , . . . , am ) and u ∈ Q\{1}, we consider
the following condition analogous to (2):
There are f (u), g(u) ∈ Q, h(u) ∈ hki such that
tˆ dec u tˆ = dec g(u) dec h(u) tˆ dec f (u)
(4)
holds in P(a1 , . . . , am ).
The following Lemmas 5 and 7 are well known (their statement is found implicitly in [Ste81] and [BGK+ 97]).
Lemma 5. For every set {a1 , . . . , am } ⊂ Q \ {1}, the map k̂ 7→ k, ûi 7→ ui , tˆ 7→ t is
an epimorphism from P(a1 , . . . , am ) onto PSU3 (q).
If condition (4) is fulfilled in P(a1 , . . . , am ) for every u ∈ Q \ {1} then this
epimorphism is an isomorphism and the presentation will define PSU3 (q).
The proof of Lemma 5 is based on the observation that if (4) holds for every
u ∈ Q\{1} then for any u1 , u2 , u3 , u4 ∈ Q and integers e, f , the product dec u1 k̂etˆ dec u2 ·
dec u3 k̂ f tˆ dec u4 can be written in P(a1 , . . . , am ) in the form dec u5 k̂ctˆ dec u6
for suitable u5 , u6 ∈ Q and integer c. The precise argument is similar to the proof
of Lemma 14, which we shall give in full detail.
However, to keep the presentation for PSU3 (q) short, we cannot add relators
of the form (4) for every u ∈ Q. Instead we will use only a few relators of the
form (4) and will have to show that these suffice to prove (4) for every u ∈ Q.
Definition 6. If G is a group, g ∈ G, and U ≤ G, then the U-class of g is the orbit
of g under conjugation by U.
Lemma 7. Suppose that (4) holds for some u ∈ Q in P(a1 , . . . , am ). Then (4) holds
for each element of the hki-class of u.
Again, the proof is similar to the proof of Lemma 14.
Lemma 7 implies that if {a1 , . . . , am } is a set of representatives of the hkiclasses in Q, then P(a1 , . . . , am ) ∼
= PSU3 (q). However, this presentation is still too
long.
5
3
hki-classes
To reduce the number of class representatives
needed, we will investigate
the hki
classes on Q in more detail. Let S := s ∈ GF(q2 ) | s + sq = 0 . Then Z(Q) =
{u(s, 0) | s ∈ S}, S is an additive subgroup of GF(q2 ), and S∗ is a coset of GF(q)∗
in GF(q2 )∗ satisfying (S∗ )2 = GF(q)∗ .
Lemma 8. Let 0 6= ξ ∈ S. Then every nonidentity element of Z(Q) is conjugate
under hki to u(ξ, 0).
Proof. Let 0 6= s ∈ S. Then
q
ξq −ξ ξ
ξ
= q =
= ,
s
s
−s
s
and so ξ/s ∈ GF(q). We can thus find an exponent e such that (ρe )q+1 = s/ξ.
Accordingly, ke will conjugate u(ξ, 0) to u(s, 0).
Lemma 9. Consider elements u(x, y), u(w, z) ∈ Q (thus x + xq + yq+1 = 0, w +
wq + zq+1
= 0) with x, y, w, z 6= 0.
(a) If 36 q + 1 then u(x, y) and u(w, z) are conjugate under hki if and only if w/x ∈
GF(q). (b) If 3 q + 1 then u(x, y) and u(w, z) are conjugate under hki if and only if w/x ∈
GF(q) and z/y = ρ3i for some integer i.
Proof. We have hki = diag(σ) | σ ∈ GF(q2 )∗ . Assume first that
u(w, z) = u(x, y)diag(σ) = u(x σq+1 , y σ2−q ).
Then w/x = σq+1 ∈ GF(q). If 3 q + 1 then 3 2 − q and so σ2−q = z/y is a third
power.
Conversely, suppose that w/x ∈ GF(q) and, in the case 3 q + 1, z/y = ρ3i
for some integer i. First we show that there is a σ ∈ GF(q2 )∗ with σ2−q = z/y.
Note that gcd(q2 − 1, 2 − q) = gcd(3, 2 − q). If 36 q + 1, we therefore have that
2
gcd(q
guarantees the existence of such a σ.
− 1, 2 − q) = 1 which
3i
If 3 q + 1 and so z/y = ρ then let σ = ρi(q+2) . Then
σ2−q = ρi(q+2)(2−q) = ρi(−q
2 +1+3)
= ρ3i = z/y.
Next we claim that if σ2−q = z/y, then u(x, y)diag(σ) = u(w, z). We have that
u(x, y)diag(σ) = u(x σq+1 , y σ2−q ) = u(x σq+1 , z) ∈ Q.
6
Hence, denoting w/x ∈ GF(q) by a, we get:
(x + xq )σq+1 = xσq+1 + (xσq+1 )q = −zq+1 = w + wq
= ax + (ax)q = (x + xq )a.
As x+xq = −yq+1 6= 0, we deduce that a = σq+1 . Therefore u(x, y)diag(σ) = u(w, z)
as claimed.
Lemma 10. Let x ∈ GF(q2 )\S and let C be a coset of GF(q)∗ in GF(q2 )∗ , different
from S∗ . Then there exists s ∈ S such that x + s ∈ C.
Proof. Let c ∈ C and v ∈ S∗ . As S is a GF(q)-subspace of GF(q2 ) and x 6∈ S, {v, x}
is a GF(q)-basis for GF(q2 ) and so we have c = αx + βv with α, β ∈ GF(q). Here
α 6= 0 because c 6∈ S. Let s := βv/α ∈ S. Then x + s = x + βv/α = c/α ∈ C.
Lemma 11. Assume that 36 q + 1 and u(x, v) ∈ Q \ Z(Q). Then for any hki-class
K ⊆ Q \ Z(Q) there exists u(s, 0) ∈ Z(Q) such that u(x, v) · u(s, 0) ∈ K.
Proof. The condition u(x, v) 6∈ Z(Q) implies that x + xq = −vq+1 6= 0, so x 6∈ S.
By Lemma 9 (a), the class K consists of those u(x0 , y) for which x0 belongs to the
same fixed coset C of GF(q)∗ in GF(q2 )∗ . Since K 6⊆ Z(Q), we have that C 6= S∗
and Lemma 10 implies that x + s ∈ C for some s ∈ S. Then u(x, v) · u(s, 0) =
u(x + s, v) ∈ K.
The analogous statement for the case 3 q + 1 is:
Lemma 12. Assume that 3 q + 1. Let x ∈ GF(q2 ) \ S and let C be a coset of
GF(q)∗ in GF(q2 )∗ different from S∗ . Moreover, for j = 0, 1, 2, let K j = {u(c, y j ) |
q+1
c ∈ C, y j ∈ ρ j hρ3 i, c + cq + y j = 0}. (By the characterization in Lemma 9 (b),
these sets are hki-classes.)
q+1
Then there exist v j ∈ ρ j hρ3 i for j = 0, 1, 2 such that x + xq + v j = 0 and, for
any such v0 , v1 , v2 , there exists u(s, 0) ∈ Z(Q) such that u(x, v j ) · u(s, 0) ∈ K j for
0 ≤ j ≤ 2.
Proof. Let v ∈ GF(q2 ) such that x + xq + vq+1 = 0. Then x + xq + (vρq−1 )q+1 =
0 and x + xq + (vρ2(q−1) )q+1 = 0 as well. Moreover, since 36 q − 1, the three
elements v, vρq−1 , vρ2(q−1) are in different cosets of hρ3 i and so we can choose
them as v0 , v1 , v2 .
The second statement of the lemma follows from the fact that, by Lemma 10,
there is an s ∈ S such that x + s ∈ C. Then u(x, v j ) · u(s, 0) = u(x + s, v j ) ∈ K j .
7
Every solvable group H has a power-commutator presentation of length O log3 |H|
[LNS84]; hence hQ, ki has a presentation of length O log3 q . However, for a
suitable choice of generators, a subset of the commutator
relations suffices and
achieves a presentation for hQ, ki of length O log2 q .
Lemma 13.
The group hQ, ki has a presentation F = hû1 , . . . , û3l , k̂ | R i of length
O log2 q .
Proof. Suppose first that 36 q + 1. We choose generators for a power-commutator
presentation such that u1 , . . . , ul generate Z(Q) and ul+1 , . . . , u3l generate Q/Z(Q);
moreover, u2 , . . . , ul are hki-conjugate to u1 and ul+2 , . . . , u3l are hki-conjugate to
ul+1 . By Lemma 8 and 9(a), we can achieve this by setting ui := u(si , 0) (1 ≤ i ≤ l)
for a generating set {s1 , . . . , sl } of the additive subgroup S of GF(q2 ), and setting
ui = u(xi , ρi−l ) (i > l) for suitable xi .
The subgroups hu1 , . . . , ui i (1 ≤ i ≤ 3l) form a subnormal series of Q. We
write a presentation for hQ, ki in a generator set {û1 , . . . û3l , k̂} corresponding to
the generators {u1 , . . . , u3l , k}. Let Q̂ = hû1 , . . . , û3l i.
p
The relation set R for this presentation consists of the power relations ûi =
2
1 (1 ≤ i ≤ 3l) and kq −1 = 1; commutator relations [û1 , ûi ] = 1(2 ≤ i ≤ 3l),
[ûl+1 , ûi ] = dec [ul+1 , ui ] (l + 2 ≤ i ≤ 3l), and [ûi , k̂] = dec [ui , k]] (1 ≤ i ≤ 3l);
e
e
as well as relations û1 = ûk̂i i (2 ≤ i ≤ l) and ûl+1 = ûk̂i i (l + 2 ≤ i ≤ 3l) that show
that the generators ui lie in two hki-orbits.
−e
The relations imply that û1 ∈ Z(Q̂). In particular, û1 commutes with ûk̂j i for
2 ≤ i ≤ l and 1 ≤ j ≤ 3l; conjugating by k̂ei , we obtain that ûi ∈ Z(Q̂) for 2 ≤ i ≤ l.
Similarly, using the commutator
identity [a,bc] = [a, c][a, b]c , we can deduce the
−e
−e
i
relation [ûl+1 , dec ukj ] = dec [ul+1 , ukj i ] for l + 2 ≤ i ≤ 3l and l + 1 ≤ j ≤ 3l.
Conjugating by k̂ei , we obtain the remaining commutator relations among the ûi .
Thus R implies a full power-commutator presentation.
If 3 q + 1 then u1 , . . . , ul are defined as above, but for l + 1 ≤ i ≤ 3l we
require that ui = u(xi , ρi−l ) is conjugate under hki to one of ul+1 , ul+2 , ul+3 .
In the presentation R , the commutator relators [ûl+2 , ûi ] = dec [ul+2 , ui ] and
[ûl+3 , ûi ] = dec [ul+2 , ui ] are added for l + 1 ≤ i ≤ 3l and the order relation for k̂
2
is replaced by k̂(q −1)/3 = 1. The same proof as in the case 36 q + 1 gives that R
implies all relations in the power-commutator
presentation
of hQ, ki.
Clearly, in both of the cases 36 q + 1 and 3 q + 1 the presentation R consists
of O (l) relators of total length O l 2 log p = O log2 q .
8
4
A presentation for PSU3(q)
In proving that a presentation defines PSU3 (q) we will have to show that (4) holds
for representatives u from all hki-classes.
The key observation for this is the following lemma that permits us to deduce
conjugation relations for new hki-classes.
Lemma 14. Let a, b ∈ Q and assume that in a group P(a1 , . . . . , am ) the relation
(4) holds for the hki-classes of a, b, and ab. Then (4) must hold for the class of
f (a)g(b) as well.
Proof. Using the relations in P(a1 , . . . . , am ) and using that (4) holds for the denoted classes, we get:
dec g(ab) dec h(ab) tˆ dec f (ab)
= tˆ dec ab tˆ = tˆ dec a tˆ tˆ dec b tˆ
= dec g(a)h(a) tˆ dec f (a) dec g(b)h(b) tˆ dec f (b)
tˆ
= dec g(a)h(a) tˆ dec f (a)g(b) tˆ dec h(b) dec f (b)
= dec g(a)h(a) tˆ dec f (a)g(b) tˆ dec h(b)t f (b)
with dec h(b)t ∈ hk̂i. Therefore
tˆ dec f (a)g(b) tˆ = dec (g(a)h(a))−1 g(ab)h(ab) tˆ dec f (ab) f (b)−1 (h(b)t )−1 .
As k normalizes Q and is normalized by t, we have that
f (ab) f (b)−1 (h(b)t )−1 = kn f (u)
for some integer n and f (u) ∈ Q. Hence
tˆ dec f (a)g(b) tˆ = dec (g(a)h(a))−1 g(ab)h(ab) tˆ dec kn dec f (u)
= dec (g(a)h(a))−1 g(ab)h(ab) dec k−nq tˆ dec f (u)
= dec (g(a)h(a))−1 g(ab)h(ab) k−nq tˆ dec f (u)
with (g(a)h(a))−1 g(ab)h(ab) k−nq ∈ hQ, ki and f (u) ∈ Q.
We now fix a field element x ∈ ρ S∗ and then s ∈ S such that x + s ∈ S∗ /ρ. (By
Lemma 10 such s exists.)
9
Theorem 15. Assume that 36 q + 1 and choose v ∈ GF(q2 )∗ such that x + xq +
vq+1 = 0. Let a = u(x, v) and b = u(s, 0). Then
P = P(a, b, ab) ∼
= PSU3 (q).
Proof. By Lemmas 5 and 7, it is enough to prove that (4) holds in P for representatives of all hki-classes in Q \ {1}. Since b ∈ Z(Q) \ {1}, condition (4) holds for
all nonidentity elements of Z(Q).
Let K1 , . . . , Kq be the hki-classes in Q \ Z(Q), where Ki = {u(y, z) ∈ Q | y ∈
∗
S /ρi }. By induction on i = 1, . . . , q, we show that (4) holds in P for a representative of Ki (and so for the whole of Ki ).
Since ab ∈ K1 , this statement is true for i = 1. Suppose now that (4) holds for
the elements of Ki . By Lemma 11, there exists w ∈ S∗ such that u(x, v) u(w, 0) ∈
Ki . Then, by Lemma 14, (4) holds in P for

 
 

1
1
1
v
  0 1
 =  − xvq 1
.
f (u(x, v)) g(u(w, 0)) =  − xq 1
q
q
1
v
1
x+w v
x
x 1
w 0 1
xw
x 1
Since x + w ∈ S∗ /ρi and xw ∈ (ρS∗ )(S∗ ) = ρ GF(q)∗ , we have that (x + w)/xw ∈
S∗ /ρi+1 and thus f (u(x, v)) g(u(w, 0)) ∈ Ki+1 .
Theorem 16. Assume that 3 q + 1 and choose v0 ∈ hρ3 i, v1 ∈ hρ3 iρ, v2 ∈ hρ3 iρ2
q+1
such that x + xq + v j = 0 for j ∈ {0, 1, 2}. (By Lemma 12, such v j exist.) Let
a j = u(x, v j ) and b = u(s, 0). Then
P = P(a1 , a2 , a3 , b, a1 b, a2 b, a3 b) ∼
= PSU3 (q).
Proof. As in the proof of Theorem 15, it is enough to show that (4) holds in P for
a representative of each hki-class in Q \ Z(Q). For 0 ≤ j ≤ 2 and 1 ≤ i ≤ q, let Ki, j
be the hki-class in Q \ Z(Q) given by
Ki, j = u(y, z) ∈ Q | y ∈ S∗ /ρi and z ∈ ρ j hρ3 i .
By induction on i = 1, . . . , q, we show that (4) holds in P for a representative of
Ki, j for 0 ≤ j ≤ 2.
Since a j b ∈ K1, j , this statement is true for i = 1. Suppose that (4) holds for
the elements of Ki, j for 0 ≤ j ≤ 2. By Lemma 12, there exists w ∈ S∗ such that
u(x, v j ) u(w, 0) ∈ Ki, j for 0 ≤ j ≤ 2. Then by Lemma 14, (4) holds in P for


1
 vj

f (u(x, v j )) g(u(w, 0)) =  − xq 1q
.
x+w v j
xw
x 1
10
As in the proof of Theorem 15, (x + w)/xw ∈ S∗ /ρi+1 . Moreover, for j = 0, 1, 2,
the field elements −v j /xq are in different cosets of hρ3 i in GF(q2 )∗ , so (4) holds
in P for all three hki-classes Ki+1,0 , Ki+1,1 , Ki+1,2 .
Acknowledgement
We are indebted to Bill Kantor for many helpful comments.
References
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[KS]
William M. Kantor and Ákos Seress, Algorithms for finite linear
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[KS99]
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