The diameter of the generating graph of a finite soluble group

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by  Andrea Lucchini

THE DIAMETER OF THE GENERATING GRAPH
OF A FINITE SOLUBLE GROUP
7
1
ANDREALUCCHINI
0
2
n Abstract. Let G be a finite 2-generated soluble group and suppose that
a ha1,b1i=ha2,b2i=G. IfeitherG′isofoddorderorG′isnilpotent,thenthere
J exists b ∈ G with ha1,bi = ha2,bi = G. We construct a soluble 2-generated
groupGoforder210·32 forwhichthepreviousresultdoesnothold. However
2
1 aweakerresultistrueforeveryfinitesolublegroup: ifha1,b1i=ha2,b2i=G,
thenthereexistc1,c2 suchthatha1,c1i=hc1,c2i=hc2,a2i=G.

]
R
G
. 1. Introduction
h
t Let G be a finite group. The generatinggraphfor G, writtenΓ(G), is the graph
a
m where the vertices are the nonidentity elements of G and there is an edge between
g and g if G is generated by g and g . If G is not 2-generated, then there will
[ 1 2 1 2
be no edges in this graph. Thus, it is natural to assume that G is 2-generated

1 when looking at this graph. There could be many isolated vertices in this graph.
v
For example, all of the elements in the Frattini subgroup will be isolated vertices.
6
We can also find isolated vertices outside the Frattini subgroup (for example the
4
3 nontrivial elements of the Klein subgroup are isolated vertices in Γ(Sym(4)). Let
3 ∆(G) be the subgraph of Γ(G) that is induced by all of the vertices that are not
0 isolated. In [4] it is proved that if G is a 2-generated soluble group, then ∆(G) is
.
1 connected. In this paper we investigate the diameter diam(∆(G)) of this graph.
0
7 Theorem 1. If Gis a2-generated finitesoluble group, then ∆(G) is connectedand
1 diam(∆(G))≤3.
:
v The situation is completely different if the solubility assumption is dropped. It
i
X is an open problem whether or not ∆(G) is connected, but even when ∆(G) is
connected, its diameter can be arbitrarily large. For example if G is the largest
r
a 2-generated direct power of SL(2,2p) and p is a sufficiently large odd prime, then
∆(G) is connected but diam(∆(G))≥2p−2−1 (see [2, Theorem 5.4]).
For soluble groups, the bound diam(∆(G)) ≤ 3 given in Theorem 1 is best
possible. In Section 3 we construct a soluble 2-generated group G of order 210·32
with diam(∆(G)) = 3. However we prove that diam(∆(G)) ≤ 2 in some relevant
cases.
Theorem 2. Suppose that a finite 2-generated soluble group G has property that
|End (V)| > 2 for every nontrivial irreducible G-module which is G-isomorphic
G
to a complemented chief factor of G. Then diam(∆(G)) ≤ 2, i.e. if ha ,b i =
1 1
ha ,b i=G, then there exists b∈G with ha ,bi=ha ,bi=G.
2 2 1 2
1991 Mathematics Subject Classification. 20P05,20D10,20E18.
1
2 ANDREALUCCHINI
Corollary 3. Let G be a 2-generated finite group. If the derived subgroup of G has
odd order, then diam(∆(G))≤2.
Corollary 4. Let G be a 2-generated finite group. If the derived subgroup of G is
nilpotent, then diam(∆(G))≤2.
2. Proof of Theorem 2
We could prove Theorem 2 with the same approach that will be used in the
proofofTheorem1. Howeverweprefer togiveinthis particularcaseaneasierand
shorter proof. Before doing that, we briefly recall some necessary definitions and
results. GivenasubsetX ofafinitegroupG,wewilldenotebyd (G)thesmallest
X
cardinality of a set of elements of G generating G together with the elements of
X. The following generalizes a result originally obtained by W. Gaschu¨tz [7] for
X =∅.
Lemma5([4]Lemma6). LetX beasubsetofGandN anormalsubgroupofGand
suppose that hg ,...,g ,XiN = G. If k ≥ d (G), then there exist n ,...,n ∈ N
1 k X 1 k
so that hg n ,...,g n ,Xi=G.
1 1 k k
It follows from the proof of [4, Lemma 6] that the number, say φ (X,k),
G,N
of k-tuples (g n ,...,g n ) generating G with X is independent of the choice of
1 1 k k
(g ,...,g ). In particular
1 k
φ (X,k)=|N|kP (X,k)
G,N G,N
where P (X,k) is the conditional probability that k elements of G generate G
G,N
with X, given that they generate G with XN.
Proposition 6 ([9] Proposition 16). If N is a normal subgroup of a finite group
G and k is a positive integer, then
µ(H,G)
P (X,k)= .
G,N |G:H|k
X⊆XH≤G
HN=G
where µ is the M¨obius function associated with the subgroup lattice of G.
Corollary 7. Let N be a minimal normal subgroup of a finite group G. Assume
that N is abelian and let q = |End (N)|. For every X ⊆ G, if k ≥ d (G) and
G X
P (X,k)6=0, then P (X,k)≥ q−1.
G,N G,N q
Proof. We may assume that N is not contained in the Frattini subgroup of G
(otherwise P (X,k) = 1). In this case, if H is a proper supplement of N in G,
G,N
then H is a maximal subgroupof G and complements N. Therefore µ(H,G)=−1
and |G:H|=|N|. It follows from Proposition 6 that
c
P (X,k)=1− ,
G,N |N|k
where c is the number of complements of N in G containing X. If c = 0, then
P (X,k) = 1. Assume c 6= 0 and fix a complement H of N in G containing X.
G,N
Let Der (H,N) be the set of derivations δ from H to N with the property that
X
xδ = 1 for every x ∈ X. The complements of N in G containing X are precisely
the subgroups of G of the kind H ={hhδ |h∈H} with δ ∈Der (H,N), hence
δ X
|Der (H,N)|
X
P (X,k)=1− .
G,N |N|k
GENERATING GRAPH 3
Now let F =End (N). Both N and Der (H,N) can be viewed as vector spaces
q G X
over F . Let
q
n=dimFqN, a=dimFqDerX(H,N).
We have
qa
P (X,k)=1− .
G,N qnk
Since P (X,k)6=0, we have a<kq and
G,N
qa 1 q−1
P (X,k)=1− ≥1− = . (cid:3)
G,N qnk q q
Proof of Theorem 2. We prove the theorem by induction on the order of G. We
may assume that G is not cyclic andthat the Frattini subgroupofG is trivial. We
distinguish two cases:
a) All the minimal normal subgroups of G have order 2. In this case G is an
elementary abelian group of order 4 and a and a are nontrivial elements of G. If
1 2
b∈/ {1,a ,a }, then ha ,bi=ha ,bi=G.
1 2 1 2
b) G contains a minimal normal subgroup N with |N| ≥ 3. By assumption q =
|End (N)| ≥ 3. Assume ha ,b i = ha ,b i = G. By induction there exists g ∈ G
G 1 1 2 2
such that ha ,giN =ha ,giN =G. For i∈{1,2}, let
1 2
Ω ={n∈N |ha ,gni=G}.
i i
Since ha ,b i = G, we have d (G) ≤ 1, hence, by Lemma 5, P ({a },1) 6= 0
i i {ai} G,N i
and consequently we deduce from Corollary 7 that
q−1 2|N|
|Ω |=|N|P ({a },1)≥|N| ≥ .
i G,N i
q 3
ButthenΩ ∩Ω 6=∅.Letb=gnwithn∈Ω ∩Ω .ThenG=ha ,bi=ha ,bi. (cid:3)
1 2 1 2 1 2
Proof of Corollary 3. Let G′ be the derived subgroup of G. If |G′| is odd, then
G′ is soluble by the Feit-Thompson Theorem, and consequently G is also soluble.
MoreoverifX andY arenormalsubgroupsofGsuchthatA=X/Y isanontrivial
irreducible G-module then |A| is a power of a prime divisor p of |G′| and F =
End (A) is a finite field of characteristic p. Hence |F|≥ p≥ 3 and we may apply
G
Theorem 2. (cid:3)
Proof of Corollary 4. We may assume Frat(G) = 1. This means that G = M ⋊H
where H is abelian and M = V ×···×V is the direct product of u irreducible
1 u
non trivial H-modules V ,...,V . Let F = End (V ) = End (V ): for each i ∈
1 u i H i G i
{1,...,u},V isanabsolutelyirreducibleF H-modulesodim V =1.Nowassume
i i Fi i
that A is a nontrivial irreducible G-module G-isomorphic to a complemented chief
factor of G: it must be A ∼=G Vi for some i, so |EndG(A)| = |Fi| = |Vi| = |A|. It
cannot be |A|=2, otherwise A would be a trivial G-module. Again we may apply
Theorem 2. (cid:3)
Remark 8. Let N be a noncentral and complemented minimal normal subgroup
of a 2-generated soluble group G and assume that F = End (N). It follows from
2 G
[9,Lemma18]and[5,Lemma18]thatifPG,N(X,k)≤1/2,thentherearedimF2N
different complemented factors G-isomorphic to N in every chief series of G. This
means that, with the same arguments used in the proof of Theorem 2, a little bit
strongerresultcanbeproved: Suppose that a finite2-generated soluble groupG has
the following property: if V is a nontrivial irreducible G-module with End (V) =
G
4 ANDREALUCCHINI
F2,thenachiefseriesofGdoesnotcontaindimF2V differentcomplementedfactors
G-isomorphic to V. Then diam(∆(G))≤2.
3. A finite soluble group G with diam(∆(G))>2
Let first recall some results that we will be applied in the discussion of our
example. Let G be a finite soluble group,and let V be a set ofrepresentativesfor
G
theirreducibleG-groupsthatareG-isomorphictoacomplementedchieffactorofG.
For V ∈V let R (V) be the smallest normal subgroup contained in C (V) with
G G G
the property that C (V)/R (V) is G-isomorphic to a direct product of copies of
G G
V andithasacomplementinG/R (V). ThefactorgroupC (V)/R (V)iscalled
G G G
the V-crown of G. The non-negative integer δG(V) defined by CG(V)/RG(V) ∼=G
VδG(V) iscalledtheV-rankofGanditcoincideswiththenumberofcomplemented
factors in any chief series of G that are G-isomorphic to V. If δ (V)6=0, then the
G
V-crownisthesocleofG/R (V). ThenotionofcrownwasintroducedbyGaschu¨tz
G
in [8]. We have (see for example [10, Proposition 2.4]):
Proposition 9. Let G and V be as above. Let x ,...,x be elements of G such
G 1 u
that hx ,...,x ,R (V)i=G for any V ∈V . Then hx ,...,x i=G.
1 u G G 1 u
Now let V be a finite dimensional vector space over a finite field of prime order.
Let K be a d-generated linear soluble group acting irreducibly and faithfully on V
and fix a generating d-tuple (k ,...,k ) of K. For a positive integer u we consider
1 d
the semidirectproduct G =Vu⋊K where K actsin the samewayoneachofthe
u
u directfactors. PutF =End (V). Let n be the dimensionofV overF. We may
K
identify K =hk ,...,k i with a subgroupof the generallinear groupGL(n,F). In
1 d
this identification k becomes an n×n matrix X with coefficients in F; denote
i i
by A the matrix I −X . Let w = (v ,...,v ) ∈ Vu. Then every v can be
i n i i i,1 i,u i,j
viewed as a 1×n matrix. Denote the u×n matrix with rows v ,...,v by B .
i,1 i,u i
The following result is proved in [3, Section 4].
Proposition 10. The group G = Vu⋊K can be generated by d elements if and
u
only if u≤n(d−1). Moreover
(1) rank A ... A =n.
1 d
(cid:0) (cid:1) A ··· A
(2) hk w ,...,k w i=Vu⋊K if and only if rank 1 d =n+u.
1 1 d d (cid:18)B1 ··· Bd(cid:19)
In this section we will use in particular the following corollary of the previous
proposition:
Corollary 11. Let V = F ×F , where F is the field with 2 elements and let
2 2 2
Γ = GL(2,2)⋉V2. Assume that hk ,k i = GL(2,2) and let γ = k (v ,v ), γ =
1 2 1 1 1 2 2
k (v ,v ) in Γ. We have that Γ=hγ ,γ i if and only if
2 3 4 1 2
1−k 1−k
1 2
 v1 v3 6=0.
v v
2 4
 
Nowwearereadytostarttheconstructionofafinite2-generatedsolubleGwith
diam(∆(G)) > 2. Let H = GL(2,2)×GL(2,2) and let W = V ×V ×V ×V be
1 2 3 4
the direct product of four 2-dimensional vector spaces over the field F with two
2
elements. We define an action of H on W by setting
(v ,v ,v ,v )(x,y) =(vx,vx,vy,vy)
1 2 3 4 1 2 3 4
GENERATING GRAPH 5
and we consider the semidirect product
G=H ⋉W.
Let
N :=C (V )=C (V )={(k,1)|k ∈GL(2,2)},
1 G 3 G 4
N :=C (V )=C (V )={(1,k)|k ∈GL(2,2)}.
2 G 1 G 2
A set of representatives for the G-isomorphism classes of the complemented chief
factor of G contains precisely 5 elements:
• Z, a central G-module of order 2, with R (Z)=G′ =SL(2,2)2⋉W.
G
• U , a non central G-module of order 3, with R (U )=N ⋉W.
1 G 1 2
• U , a non central G-module of order 3, with R (U )=N ⋉W.
2 G 2 1
• V , with R (V )=V ×V ×N .
1 G 1 3 4 2
• V , with R (V )=V ×V ×N .
3 G 3 1 2 1
Let
(x ,y )(v ,v ,v ,v )=g , (x ,y )(v ,v ,v ,v )=g .
1 1 11 12 13 14 1 2 2 21 22 23 24 2
WewanttoapplyProposition9tocheckwhetherhg ,g i=G. Thethreeconditions
1 2
hg ,g iR (Z)=G,hg ,g iR (U )=G,hg ,g iR (U )=G
1 2 G 1 2 G 1 1 2 G 2
are equivalent to hg ,g iW =G, i.e. to h(x ,y ),(x ,y )i=H. Moreover
1 2 1 1 2 2
hg ,g iR (V )=G if and only if hx (v ,v ),x (v ,v )i=(V ×V )⋊GL(2,2),
1 2 G 1 1 11 12 2 21 22 1 2
hg ,g iR (V )=G if and only if hy (v ,v ),y (v ,v )i=(V ×V )⋊GL(2,2).
1 2 G 3 1 31 32 2 41 42 3 4
Applying Corollary 11 we conclude that
hg ,g i=G
1 2
if and only if the following conditions are satisfied:
(1) h(x ,y ),(x ,y )i=H =GL(2,2)×GL(2,2),
1 1 2 2
1−x 1−x
1 2
(2) det v11 v21 6=0,
v v
12 22
 
1−y 1−y
1 2
(3) det v13 v23 6=0.
v v
14 24
 
Consider the following elements of GL(2,2):
1 0 1 1 1 1
x:= , y := , z := ,
(cid:18)1 1(cid:19) (cid:18)1 0(cid:19) (cid:18)0 1(cid:19)
and the following elements of F2:
2
0=(0,0), e =(1,0), e =(0,1).
1 2
Let
a :=(x,x)(0,e ,0,e ), a :=(x,x)(e ,e ,e ,e )
1 2 2 2 1 2 1 2
b :=(y,z)(e ,0,e ,0), b :=(y,z)(0,0,e ,0).
1 1 1 2 1
It can be easily checked that
h(x,x),(y,z)i=H.
6 ANDREALUCCHINI
Moreover
0 0 0 1
1−x 1−y
1 0 1 1
det 0 e1 =det 0 0 1 0 =1,
 e2 0  0 1 0 0
 
0 0 0 1
1−x 1−z
1 0 0 0
det 0 e1 =det 0 0 1 0 =1,
 e2 0  0 1 0 0
 
0 0 0 1
1−x 1−y
1 0 1 1
det e1 0 =det 1 0 0 0 =1,
 e2 0  0 1 0 0
 
0 0 0 1
1−x 1−z
1 0 0 0
det e1 e1 =det 1 0 1 0 =1,
 e2 0  0 1 0 0
 
so either a ,b as a ,b satisfy the three conditions (1), (2) (3) and therefore
1 1 2 2
ha ,b i=ha ,b i=G.
1 1 2 2
Now we want to prove that there is no b ∈ G with ha ,bi = ha ,bi = G. Let
1 2
b = (h ,h )(v ,v ,v ,v ), and assume by contradiction that ha ,bi = ha ,bi = G.
1 2 1 2 3 4 1 2
We must have in particular that condition (1) holds, i.e. h(x,x),(h ,h )i = H.
1 2
Since (x,x) has order 2 and H cannot be generated by two involutions (otherwise
itwouldbeadihedralgroup)atleastoneofthetwoelementsh ,h musthaveorder
1 2
3: it is not restrictive to assume h = y. Let v = (α,β), v = (γ,δ). Conditions
1 1 2
(2) and (3) must be satisfied, hence we must have
1−x 1−y 1−x 1−y
det 0 v1 =det e1 v1 =1.
e v e v
2 2 2 2
   
However
0 0 0 1
1−x 1−y
1 0 1 1
det 0 v1 =det 0 0 α β =α,
 e2 v2  0 1 γ δ
 
0 0 0 1
1−x 1−y
1 0 1 1
det e1 v1 =det 1 0 α β =α+1.
 e2 v2  0 1 γ δ
 
However, since α ∈ F either α = 0 or α + 1 = 0, so there is no b ∈ G with
2
ha ,bi=ha ,bi=G.
1 2
GENERATING GRAPH 7
4. A problem in linear algebra
BeforetoproveTheorem1,weneedtocollectaseriesofresultsinlinearalgebra.
Denote by M (F) the set of the r×s matrices with coefficients over the field F.
r×s
Lemma 12. [4, Lemma3] Let V be a finite dimensional vector space over the field
F. If W and W are subspaces of V with dimW = dimW , then V contains a
1 2 1 2
subspace U such that V =W ⊕U =W ⊕U.
1 2
Lemma 13. Let v ,...,v ,w ,...w ∈ Fn, where F is a finite field and either
1 n 1 n
|F|>2 or n6=1. There exist z ,...,z ∈Fn so that the two sequences
1 n
v +z ,...,v +z ,
1 1 n n
w +z ,...,w +z
1 1 n n
are both basis of Fn.
Proof. Equivalently, we want to prove that for every pair of matrices A,B ∈
M (F),thereexistsC ∈M (F),suchthatdet(A+C)6=0anddet(B+C)6=0.
n×n n×n
Since either |F| > 2 or n 6= 1, every element of M (F) can be expressed as the
n×n
sumoftwounits [11]. InparticularA−B =U−V with U,V ∈GL(n,F). We may
take C =U −A=B−V. (cid:3)
Lemma 14. Let F be a finite field and assume r ≤ n. Given R ∈ M (F) and
r×n
S ∈ M (F) consider the matrix R S ∈ M . Assume rank R S = r
r×r r×(n+r)
and let πR,S be the probability that(cid:0)a mat(cid:1)rix Z ∈ Mr×n(F) satisfies th(cid:0)e cond(cid:1)ition
rank(R+SZ)=r. Then
qr
π >1− .
R,S qn(q−1)
Proof. There exist m≤r, X ∈GL(r,F) and Y ∈GL(r,F) such that
I 0
XSY = m ,
(cid:18) 0 0(cid:19)
where I is the identity element in M (F). Since
m m×m
I 0
r=rank R S =rank X R S n =rank XR XSY
(cid:18) (cid:18)0 Y(cid:19)(cid:19)
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
and
rank(R+SZ)=rank(X(R+SZ))=rank(XR+XSZ)
=rank(XR+XSY(Y−1Z)),
it is not restrictive (replacing R by XR, S by XSY and Z by Y−1Z) to assume
I 0
S = m .
(cid:18) 0 0(cid:19)
Denote by v ,...,v the rows of R and by z ,...,z the rows of Z. The fact that
1 r 1 r
the rows of (R S) are linearly independent implies that v ,...,v are linearly
m+1 r
independent vectors of Fn. The condition rank(R+SZ) = r is equivalent to ask
that
v +z ,...,v +z ,v ,...,v
1 1 m m m+1 r
are linearly independent. The probability that z ,...,z satisfy this condition is
1 m
qr−m qr−m+1 qr−m+(m−1)
1− 1− ··· 1− .
(cid:18) qn (cid:19)(cid:18) qn (cid:19) (cid:18) qn (cid:19)
8 ANDREALUCCHINI
Hence
qr−m qr−m+1 qr−m+(m−1)
π = 1− 1− ··· 1−
R,S (cid:18) qn (cid:19)(cid:18) qn (cid:19) (cid:18) qn (cid:19)
qr−m(1+q+···+qm−1)
≥1−
qn
qr−m(qm−1) qr
=1− >1− . (cid:3)
qn(q−1) qn(q−1)
Lemma 15. Assume that F is a finite field and that A, B , B , D and D are
1 2 1 2
elements of M (F) with the property that
n×n
rank A B =rank A B =n,
1 2
(cid:0) (cid:1) (cid:0) (cid:1)
B B
rank 1 =rank 2 =n.
(cid:18)D1(cid:19) (cid:18)D2(cid:19)
Moreover assume that at least one of the following conditions holds:
(1) |F|>2;
(2) detA=0;
(3) n≥2 and (detB ,detB )6=(0,0).
1 2
Then there exists C ∈M (F) such that
n×n
A B A B
det 1 6=0 and det 2 6=0.
(cid:18)C D1(cid:19) (cid:18)C D2(cid:19)
Proof. Let r =rank(A). There exist X,Y ∈GL(n,F) such that
I 0
XAY = r
(cid:18)0 0(cid:19)
whereI istheidentityelementinM (F).LetB ,B ∈M (F)andB ,B ∈
r r×r 11 21 r×n 12 22
M (F) such that
(n−r)×n
B B
XB = 11 , XB = 21 .
1 (cid:18)B12(cid:19) 2 (cid:18)B22(cid:19)
For i∈{1,2}, since
Y 0 I 0 B
n=rank A B =rank X A B =rank r i1 ,
i (cid:18) i (cid:18)0 In(cid:19)(cid:19) (cid:18)0 0 Bi2(cid:19)
(cid:0) (cid:1) (cid:0) (cid:1)
it must be rank(B )=n−r. In particular there exists Z ∈GL(n,F) such that
i2 i
B B∗ B∗
XB Z = i1 Z = i1 i2
i i (cid:18)Bi2(cid:19) i (cid:18) 0 In−r(cid:19)
with B∗ ∈M (F), B∗ ∈M (F). Notice that
i1 r×r i2 r×(n−r)
XAY XB Z X 0 A B Y 0
det i i =det i
(cid:18) CY DiZi (cid:19) (cid:18)(cid:18)0 In(cid:19)(cid:18)C Di(cid:19)(cid:18)0 Zi(cid:19)(cid:19)
A B
=det(X)det(Y)det(Z )det i .
i (cid:18)C Di(cid:19)
This means that it is not restrictive to assume
I 0 B∗ B∗
A= r , B = i1 i2
(cid:18)0 0(cid:19) i (cid:18) 0 In−r(cid:19)
GENERATING GRAPH 9
with B∗ ∈ M (F), B∗ ∈ M (F). Let C ,D ∈ M (F) and C ,D ∈
i1 r×r i2 r×(n−r) 1 i1 n×r 2 i2
M (F) such that C C =C and D D =D. Notice that
n×(n−r) 1 2 i1 i2
(cid:0) (cid:1) (cid:0) (cid:1)
I 0 B∗ B∗
A B r i1 i2
det(cid:18)C Dii(cid:19)=detC0 C0 D0 IDn−r
1 2 i1 i2
 
I 0 B∗
=(−1)ndet r i1
(cid:18)C1 C2 Di1(cid:19)
I 0 −B∗
I 0 B∗ r i1
=(−1)ndet(cid:18)Cr1 C2 Dii11(cid:19)00 In0−r I0 
r
  
I 0 0
=(−1)ndet r
(cid:18)C1 C2 Di1−C1Bi∗1(cid:19)
=(−1)ndet C D −C B∗ .
2 i1 1 i1
(cid:0) (cid:1)
Assume that we can find C such that
1
rank(D −C B∗ )=rank(D −C B∗ )=r
11 1 11 21 1 21
andletW ,W bethesubspacesofFn spanned,respectively,bythecolumnsofthe
1 2
twomatricesD −C B∗ andD −C B∗ . ByLemma12,thereexistsasubspace
11 1 11 21 1 21
U of Fn such that Fn =W ⊕U =W ⊕U. If C is a matrix whose columns are a
1 2 2
basis for U, then
det C D −C B∗ 6=0 and det C D −C B∗ 6=0
2 11 1 11 2 21 1 21
(cid:0) (cid:1) (cid:0) (cid:1)
and C =(C C ) is a matrix with the request property. Set
1 2
R =DT, R =DT, S =B∗T, S =B∗T, Z =−CT.
1 11 2 21 1 11 2 21 1
The previous observation implies that a matrix C with the requested properties
exists if and only if there exists Z ∈M (F) such that
r×n
(4.1) rank(R +S Z)=rank(R +S Z)=r.
1 1 2 2
Notice that R ,R ∈M (F), S ,S ∈M (F) have the property that
1 2 r×n 1 2 r×r
rank R S =rank R S =r.
1 1 2 2
(cid:0) (cid:1) (cid:0) (cid:1)
First assume that either |F|=q >2 or r<n: by Lemma 14, we have
1 1
π > and π >
R1,S1 2 R2,S2 2
and this is sufficient to ensure that a matrix Z with the requested property exists.
Thereforewe mayassumer =n (i.e. detA6=0)andq =2.Inthis case,weassume
also that at least one of the two matrices B and B is invertible. Let for example
1 2
detB 6=0. This implies detS 6=0. There exist m≤n and X,Y ∈GL(n,F) such
1 1
that
I 0
XS Y = m .
2 (cid:18) 0 0(cid:19)
NoticethatR +S Z isinvertibleifandonlyifXR Y+XS YY−1ZY isinvertible.
2 2 2 2
Moreover R +S Z is invertible if and only if R Y +S YY−1ZY is invertible, if
1 1 1 1
10 ANDREALUCCHINI
andonly if (S Y)−1R Y +Y−1ZY is invertible. This means that(replacing R by
1 1 1
(S Y)−1R Y, R by XR Y and Z by Y−1ZY) we may assume
1 1 2 2
I 0
S =I and S = m .
1 n 2 (cid:18) 0 0(cid:19)
Letv ,...,v be the rowsofR ,w ,...,w the rowsofR andz ,...,z the rows
1 n 1 1 n 2 1 n
of Z. Our request on Z is equivalent to ask that the sequences
v +z ,...,v +z ,v +z ,...,v +z ,
1 1 m m m+1 m+1 n n
w +z ,...,w +z ,w ,...,w ,
1 1 m m m+1 n
are both linearly independent. Notice that the condition rank(R S ) = n implies
2 2
inparticularthatw ,...,w arelinearlyindependent. Firstassumem6=1. For
m+1 n
j >m, letz =v +w sothatz +v =w andletW =hw ,...,w i. We then
j j j j j j m+1 n
workinthe vectorspaceFn/W ofdimensionm andourrequestis thatthe vectors
v +z +W,...,v +z +W and the vectors w +z +W,...,w +z +W are
1 1 m m 1 1 m m
linearly independent: Lemma 13 ensures that this request is fulfilled for a suitable
choiceofz ,...,z .Finallyassumem=1. Asbeforeforj >2,letz =v +w and
1 m j j j
let W =hw ,...,w i. We want to find z and z so that the two vectors v +z +
3 n 1 2 1 1
W,v +z +W andthetwovectorsw +z +W,w +W arelinearlyindependent. This
2 2 1 1 2
isalwayspossible. Firstchoosez sothathw +z +Wi∈/ hw +Wiandv +z ∈/ W.
1 1 1 2 1 1
Once z has been fixed, choose z so that hv +z +Wi∈/ hv +z +Wi. (cid:3)
1 2 2 2 1 1
Remark 16. Notice that when |F| = 2 and detA 6= 0, we cannot drop the as-
sumption (detB ,detB )6=(0,0). Consider for example
1 2
0 1 0 0 0 0 1 0
A= , B =B = , D = , D = .
(cid:18)1 1(cid:19) 1 2 (cid:18)1 0(cid:19) 1 (cid:18)0 1(cid:19) 2 (cid:18)0 1(cid:19)
Then, as we noticed at the end of Section 3, there is no C ∈M (F) with
2×2
A B A B
det 1 6=0 and det 2 6=0.
(cid:18)C D1(cid:19) (cid:18)C D2(cid:19)
This restriction in the statement of Lemma 15 is indeed the reasonwhy we cannot
have diam(∆(G))=2 for every 2-generatedfinite soluble group G.
Proposition 17. Let F be a finite field and let n be a positive integer. Assume
that either n≥2 or |F|>2. Assume that A , A , A , A , B and B are elements
0 1 2 3 0 3
of M (F) with the property that
n×n
rank(A A )=rank(A A )=rank(A A )
0 1 1 2 2 3
and
A A
rank 0 =rank 3 =n.
(cid:18)B0(cid:19) (cid:18)B3(cid:19)
Then there exist B ,B ∈M (F) such that
1 2 n×n
A A A A A A
det 0 1 6=0, det 1 2 6=0, det 2 3 6=0.
(cid:18)B0 B1(cid:19) (cid:18)B1 B2(cid:19) (cid:18)B2 B3(cid:19)
Proof. Set i=1 if either |F|>2 or |F|=2 and detA =0, i=2 otherwise.
1
a) Assume i=1. First choose B so that
2
A A
det 2 3 6=0.
(cid:18)B2 B3(cid:19)