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CAPABLE GROUPS OF PRIME EXPONENT AND CLASS TWO
4
0
0 ARTUROMAGIDIN
2
n Abstract. A group is called capable ifitis a central factor group. We con-
a sider the capability of finite groups of class two and exponent p, p an odd
J
prime. Werestate theproblemofcapabilityas aproblemabout lineartrans-
9 formations, which may be checked explicitly for any specific instance of the
2 problem. We use this restatement to derive some known results, and prove
new ones. Among them, we reduce the general problem to an oft-considered
] special case, and prove that a3-generated group ofclass 2 andexponent pis
R
eithercyclicorcapable.
G
.
h
t
a 1. Introduction
m
Inhislandmarkpaper[9]ontheclassificationoffinitep-groups,P.Hallremarked:
[
The question of what conditions a group G must fulfill in order
1 that it may be the central quotient group of another group H,
v
G = H/Z(H), is an interesting one. But while it is easy to write
3 ∼
down a number of necessaryconditions, it is not so easy to be sure
2
4 that they are sufficient.
1 Following M. Hall and Senior [8], we make the following definition:
0
4 Definition 1.1. A groupG is saidto be capable if andonly if there exists a group
0 H such that G=H/Z(H); equivalently, if and only if G is isomorphic to the inner
∼
h/ automorphism group of a group H.
t
a Capability of groups was first studied by R. Baer in [2], where, as a corollary of
m deeper investigations, he characterisedthe capable groups that are a direct sum of
: cyclic groups. Capability of groups has received renewed attention in recent years,
v
thanksto resultsofBeyl, Felgner,andSchmid[3]characterisingthe capability ofa
i
X groupin terms of its epicenter; and more recently to work of Graham Ellis [5] that
r describes the epicenter in terms of the nonabelian tensor square of the group.
a
We will consider here the specialcase of nilpotent groupsof class two andexpo-
nent an odd prime p. This case was studied in [12], and also addressed elsewhere
(e.g., Prop. 9 in [5]). As noted in the final paragraphs of [1], currently available
techniques seem insufficient for a characterization of the capable finite p-groups of
class2,butacharacterizationofthecapablefinitegroupsofclass2andexponentp
seems like a more modest and possibly attainable goal. The present work is a
contribution towards achieving that goal.
We will use some of the techniques used in [16]. We prove that the problem of
capability is equivalent to a linear algebra problem, and then exploit this equiva-
lence. Intherestoftheintroductionwewillrecalltherelevantdefinitionsandbasic
results. In Section 2 we construct a natural candidate to witness the capability of
2000 Mathematics Subject Classification. Primary20D15,Secondary20F12.
1
2 ARTUROMAGIDIN
a given group G. Then we show that the properties of this natural candidate may
be codified using linear algebrain Section 3, anddeduce the promisedequivalence.
In Section 4 we exploit this equivalence and prove several results on capability,
among them reducing the general case to the study of groups G which satisfy
[G,G] = Z(G), an oft-considered special case. In Section 5 we present a geomet-
ric argument which establishes some further cases of capability by using algebraic
geometry.
Throughout the paper, p will be an odd prime. All groups will be written
multiplicatively, and the identity element will be denoted by e; if there is danger
of ambiguity or confusion, we will use e to denote the identity of the group G.
G
The center of G is denoted by Z(G). Recall that if G is a group, and x,y G,
∈
we let the commutator of x and y be [x,y] = x−1y−1xy. We write commutators
left-normed, so that [x,y,z] = [[x,y],z]. Given subsets A and B of G we define
[A,B] to be the subgroup of G generated by all elements of the form [a,b] with
a A, b B. The terms of the lower central series of G are defined recursively
∈ ∈
by letting G = G, and G =[G ,G]. A group is nilpotent of class at most k if
1 n+1 n
and only if G = e , if and only if G Z(G). We usually drop the “at most”
k+1 k
{ } ⊂
clause, it being understood. The class of all nilpotent groups of class at most k is
denoted by N .
k
The following commutator identities are well known, and may be verified by
direct calculation:
Proposition 1.2. Let G be any group. Then for all x,y,z G,
∈
(a) [xy,z]=[x,z][x,z,y][y,z].
(b) [x,yz]=[x,z][z,[y,x]][x,y].
(c) [x,y,z][y,z,x][z,x,y] e (mod G ).
4
≡
(d) [xr,ys] [x,y]rs[x,y,x]s(r2)[x,y,y]r(s2) (mod G4).
≡
(e) [yr,xs] [x,y]−rs[x,y,x]−r(s2)[x,y,y]−s(r2) (mod G4).
≡
Here, n = n(n−1) for all integers n.
2 2
As i(cid:0)n(cid:1)[16],our maintoolwillbe the nilpotentproductofgroups,specifically the
2-nilpotent and 3-nilpotent products. We restrict Golovin’s original definition [7]
to the situation we will consider:
Definition 1.3. Let A ,...,A be cyclic groups. The k-nilpotent product of
1 n
A1,...,An,denotedbyA1 Nk NkAn,isdefinedtobethegroupG=F/Fk+1,
∐ ···∐
where F is the free product of the A , F =A A , and F is the (k+1)-st
i 1 n k+1
∗···∗
term of the lower central series of F.
Note that if G is the k-nilpotent product of the A , then G N , and G/G is
i k k
∈
the (k 1)-nilpotent product of the A .
i
−
Theorem 1.4 (R.R. Struik; Theorem3 in [18]). Let A ...,A be cyclic groups of
1 t
order α ,...,α , respectively; if A is infinite cyclic, let α =0. Let x generate A ,
1 t i i i i
and let F be their free product
F =A A .
1 t
∗···∗
Assume that all primes appearing in the factorizations of the α are greater than
i
or equal to 3. Let α = gcd(α ,α ), α = gcd(α ,α ,α ) for all 1 i,j,k t.
ij i j ijk i j k
≤ ≤
CAPABLE GROUPS OF PRIME EXPONENT AND CLASS TWO 3
Then every g F/F can be uniquely expressed as
4
∈
g = xa11···xatt [xj,xi]aji [xj,xi,xi]ajii[xj,xi,xj]ajij
1≤i<j≤t 1≤i<j≤t
Y Y
[xj,xi,xk]ajik[xk,xi,xj]akij
1≤i<j<k≤t
Y
where the a are taken modulo α , the a modulo α , and the a modulo α .
i i ji ji ℓmn ℓmn
A collectionprocess maybe usedto obtaina multiplicationtable for this group,
althoughthiswillnotbenecessaryforourpurposes. WedirectthereadertoStruik’s
original paper [18] (where she uses a slightly different choice of normal form), or
to[16](wherethemultiplicationformulasagreewiththenormalformgivenabove).
To obtaina descriptionofthe 2-nilpotentproduct, alreadygivenby Golovin[6],
we simply take the quotient modulo the third term of the lower central series; this
is the subgroup of all elements of the form
[xj,xi,xi]ajii[xj,xi,xj]ajij [xj,xi,xk]ajik[xk,xi,xj]akij
1≤i<j≤t 1≤i<j<k≤t
Y Y
for integers a , a , a , and a .
jii jij kij jik
From the definition, it is clear that the k-nilpotent product is the coproduct in
the variety N , so it will have the usual universal property. When we take the
k
k-nilpotent product of infinite cyclic groups we obtain the relatively free group in
the variety N .
k
Finally,whenwesaythatagroupisk-generated wemeanthatitcanbegenerated
byk elements,butmayinfactneedless. Ifwewanttosaythatitcanbe generated
by k elements, but not by m elements for some m < k, we will say that it is
minimally k-generated, or minimally generated by k elements.
2. A natural candidate
Using the 2- and 3-nilpotent product of cyclic groups,we can produce a natural
“candidateforwitness”tothecapabilityofagivenfinitenilpotentgroupsofclass2
and exponent p, which we can then investigate directly. This will be the theme
throughout this article. We begin with an easy observation:
Lemma 2.1 (cf. Lemma 2.1 in [13]). Let G be a capable group, generated by
g ,...,g , n > 1. Then there exists a group H, and elements h ,...,h H
1 n 1 n
∈
such that H is generated by h1,...,hn, and H/Z(H)∼=G, where the isomorphism
is induced by the map sending h to the corresponding g .
i i
Proof. Let K be any group such that K/Z(K) ∼= G. Let hi ∈ K be any element
mapping to g . Let H = h ,...,h . We then have HZ(K) = K, so it follows
i 1 n
thatZ(H)=Z(K) H anhdhenceH/iZ(H)=H/(Z(K) H)=K/Z(K)=G. (cid:3)
∩ ∼ ∩ ∼ ∼
Let G be a finite nilpotent group of class at most 2 and exponent p, minimally
generated by g ,...,g , n>1. Then G is a quotient of the 2-nilpotent product of
1 n
n cyclic p-groups,
G∼= hx1i∐N2 ···∐N2 hxni /N
where N is the kernelof the map induced by mapping x to g (using the universal
(cid:0) (cid:1) i i
property of the coproduct), and is contained in the commutator subgroup of the
4 ARTUROMAGIDIN
2-nilpotent product. Let y ,...,y generate infinite cyclic groups, let K be the
1 n
3-nilpotent product of the cyclic groups,
K= y N3 N3 y ,
1 n
h i∐ ···∐ h i
and let = K/[K,K]p. By abuse of notation, we denote the images of the y in
i
K K
by y as well.
i
Notethatthecommutatorsubgroupof x N2 N2 x isisomorphictothe
1 r
h i∐ ···∐ h i
subgroupof[ , ]generatedbythecommutatorsoftheform[y ,y ],1 i<j n,
j i
K K ≤ ≤
by mapping [x ,x ] to [y ,y ]. Let be the subgroup of that corresponds to N
j i j i
N K
under this identification.
Let = [ , ]. Finally, let K = / . Again, we also denote the images of
M N K K M
the y in K by y .
i i
Theorem 2.2. Let G, K, , and K be as in the previous three paragraphs. Then
K
G is capable if and only if G=K/Z(K).
∼
Proof. First, note that K/ Kp, ,K is isomorphic to G. Since each of Kp,
3
h NM i
, and K are central in K, it is always be the case that K/Z(K) is a quotient
3
NM
of G.
Assume G is capable, and let H N be a group with H/Z(H) = G. We may
∈ 3 ∼
assume H is generated by elements h ,...,h mapping to g ,...,g , respectively.
1 n 1 n
Then Hp Z(H). We claim this implies that Hp = e .
Indeed,⊂first note that Hp = e . For if c 2H ,{h} H, then using Proposi-
3 { } ∈ 2 ∈
tion1.2we havethat [c,h]p =[c,hp]=e. Since H is abelian,and generatedby all
3
elements [c,h] as above, this shows that H is of exponent p. Now let a,b H; we
3
∈
have (once again using Proposition 1.2):
[a,b]p = [ap,b][a,b,a]−(p2)[a,b,b]−p(12)
= [ap,b][a,b,a]−p(p−21)
= e.
Since H is abelian and generated by all such [a,b], which are of exponent p, we
2
conclude that Hp = e , as claimed.
The natural s2urje{cti}on K G, mapping y to g therefore factors through H
i i
→
and , so we have the following exact diagram:
K
K
↓
1 Z(H) H G 1.
→ → → →
↓
1
If y , then its image in H must map to the trivialelement in G, so its image is
∈N
in the center of H. It follows that =[ , ] is in the kernel of the map from
M N K K
to H, so the map factors through K giving a diagramwith exact row and column:
K
↓
1 Z(H) H G 1.
→ → → →
↓
1
CAPABLE GROUPS OF PRIME EXPONENT AND CLASS TWO 5
Since the center of K maps into the center of H, it follows that K/Z(K) has G as
a quotient. Since we already know that K/Z(K) is a quotient of G, it follows that
K/Z(K)=G, as desired. (cid:3)
∼
Next, note that Z(K/Kp)=Z(K)/Kp, so we have:
Theorem2.3(cf.Theorem8.6in[16]). LetGbeanilpotentgroupofclassatmost2
and exponent an odd prime p, minimally generated by g ,...,g , with n > 1. Let
1 n
A ,...,A be cyclic groups of order p, generated by x ,...,x , respectively, and
1 n 1 n
let be the 2-nilpotent product of the A , = A N2 N2 A . Let N ⊳ be
i 1 n
G G ∐ ···∐ G
the kernel of the homomorphism G induced by mapping x to g . Let K be
i i
the 3-nilpotent product of the A , KG →=A N3 N3A , and identify N with the
i 1 n
∐ ···∐
corresponding subgroup of K . Then G is capable if and only if
2
G=(K/[N,K])/Z(K/[N,K]).
∼
What are the elements of Z(K/[N,K])? An element k[N,K] will be central in
K/[N,K] if and only if [k,K] [N,K]. This includes the center of K (which is
⊆
equal to K , see Theorem 5.1 in [16]), as well as N. Since K/ N,K = G, it
3 h 3i ∼
follows that G is capable if and only if
Z(K/[N,K])= K ,N /[N,K].
3
h i
The left hand side always contains the right hand side. However, the problem is
that there could be more elements in Z(K/[N,K]). Let k K, and write k in
∈
normal form:
k = xa11···xann [xj,xi]aji [xj,xi,xi]ajii[xj,xi,xj]ajij
1≤i<j≤n 1≤i<j≤n
Y Y
[xj,xi,xk]ajik[xk,xi,xj]akij.
1≤i<j<k≤n
Y
Suppose that there exists i 1,...,n with a = 0. If i < n, consider [x ,k].
i n
∈ { } 6
Using the formulas in Proposition 1.2, it is easy to verify that the normal form for
[x ,k] will have a nonzero exponent for [x ,x ]; since N K , [N,K] K , so
n n i 2 3
⊂ ⊂
[x ,k] [N,K]. If i=n, then consider [k,x ], and once again we find that this
n n−1
6∈
elementhasnonzeroexponentin[x ,x ],so[k,x ] / [N,K]. Thatis,wehave
n n−1 n−1
∈
shown that if k K satisfies [k,K] [N,K], then k K . With this observation,
2
∈ ⊆ ∈
we obtain the following:
Corollary 2.4. Let G, K, and N be as in Theorem 2.3. Then G is capable if and
only if the following condition holds: For every element k K of the form
∈
(2.5) k = [xj,xi]aji
1≤i<j≤n
Y
with 0 a <p, if [k,x ] [N,K] for i=1,...,n, then k N.
ji i
≤ ∈ ∈
Proof. We know that G is capable if and only if
k K [k,K] [N,K] = N,K .
2 3
∈ ⊆ i
If k K2, then [k,K](cid:8)= [k,x1(cid:12)],...,[k,xn] by(cid:9)Prop(cid:10)osition 1.2, so if G is capable
∈ (cid:12)
then all k such that [k,x ] [N,K], i = 1,...,n, must satisfy k N,K . Con-
i 3
(cid:10) ∈ (cid:11) ∈ h i
versely, assume the condition holds and let k K be such that [k,K] [N,K];
2
∈ ⊂
6 ARTUROMAGIDIN
we want to show that k N,K . Multiplying by suitable elements of K , we
3 3
∈ h i
may assume that k is of the form (2.5). Since [k,K] [N,K], it follows that
⊂
[k,x ] [N,K] for each i; so k N, by the condition given. This proves the
i
corollar∈y. ∈ (cid:3)
3. Some Linear Algebra
The main advantage of Corollary 2.4, beyond giving a very precise and explicit
condition to check, is that the condition can also be recast as a statement about
vector spaces and linear transformations. At that point, we have a whole array of
tools that can be brought to bear upon the problem. For example, Theorem 5.6
below uses algebraic geometry to settle several cases.
We will now translate the condition into the promised statement about vector
spaces and linear transformations. The key observation is the simple fact that if
H N , then for each h H the map ϕ : H H given by ϕ (c) = [c,h] is
3 h 2 3 h
∈ ∈ →
an abelian group homomorphism; if H is of exponent p, then the map becomes a
2
linear transformation of spaces over F , the Galois field of p elements.
p
Let n > 1 be a fixed integer. Let K = x N3 N3 x , where x is of
1 n i
orderp, p anoddprime. Then K is a vectorhspiac∐eove·r··F∐, wihth biasisgivenby all
2 p
elements of the form [x ,x ] and [x ,x ,x ], where 1 i < j n, 1 r < s n,
j i s r t
≤ ≤ ≤ ≤
and 1 r t n.
Let≤V b≤e th≤e vector space over F of dimension n , with basis
p 2
(3.1) vji 1 i<j n(cid:0). (cid:1)
≤ ≤
Let W be a vector space over(cid:8)Fp of(cid:12)(cid:12)dimension 2 (cid:9)n2 + n3 with basis
(3.2) wjik 1 i<j n, 1 (cid:0)i(cid:0) (cid:1)k (cid:0) n(cid:1)(cid:1).
≤ ≤ ≤ ≤ ≤
Wewillrefertotheseb(cid:8)asesa(cid:12)sthe“standardbases” forV an(cid:9)dW,andwetalkabout
(cid:12)
V and W corresponding to n to meanthe casewherethe indices ofthe coordinates
range from 1 to n (subject to the conditions listed above). We can clearly identify
K withW,andK withV W. TorestateCorollary2.4intermsofV andW,we
3 2
⊕
justneedtodescribewhathappenswhenwetakethecommutatorof[x ,x ]withx .
j i r
We have two cases: if i r, then [x ,x ,x ] is already in normal form, and there
j i r
≤
is nothing to do. If i > r, then the commutator [x ,x ,x ] must be rewritten in
j i r
normal form. We use Proposition 1.2(c): since [x ,x ,x ][x ,x ,x ][x ,x ,x ]= e,
j i r i r j r j i
we have:
[x ,x ,x ] = [x ,x ,x ]−1[x ,x ,x ]−1
j i r r j i i r j
= [[x ,x ]−1,x ][x ,x ,x ]−1
r j i i r j
= [x ,x ,x ][x ,x ,x ]−1.
j r i i r j
So, for each r = 1,...,n we define a linear transformation ϕ : V W to be
r
→
given by:
w if r i,
(3.3) ϕr(vji)= wjir w if r <≥i.
jri irj
(cid:26) −
Thatis,ϕ codifiesthemapc [c,x ]. Itiseasytoverifythateachϕ isinjective.
r r r
7→
Let G be a non-cyclic group of class 2 and exponent p. Consider the situation
in Corollary2.4,and let X be the subspace of V correspondingto the subgroupN
determined by G (we will refer to such X as the subspace of V corresponding to G
CAPABLE GROUPS OF PRIME EXPONENT AND CLASS TWO 7
ordetermined by G). WedefineY tobethesubspaceofW spannedbytheimages
X
of X; that is:
(3.4) Y = ϕ (X),...,ϕ (X) .
X 1 n
D E
Thus,Y correspondstothesubgroup[N,K]. Finally,letZ bethesubspaceofV
X X
given by:
n
(3.5) Z = ϕ−1(Y ).
X i X
i=1
\
Clearly,X Z , and Z correspondsto the subgroupof all k K , written as in
X X 2
⊆ ∈
(2.5), such that [k,x ] [N,K] for i=1,...,n. We know that G is capable if and
i
∈
only if the subgroup of such k equals N, so we have:
Theorem 3.6. Let G be a finite nilpotent group of class 2 and exponent p, min-
imally generated by g ,...,g , n > 1. Let V and W be the vector spaces over
1 n
F defined in (3.1) and (3.2), let ϕ ,...,ϕ be as in (3.3), and let X be the sub-
p 1 n
space of V determined by G (that is, corresponding to the kernel of the natural map
x N2 N2 x G given by x g , where each x is of order p). Let Y
1 n i i i X
h i∐ ···∐ h i−→ 7→
and Z be defined by (3.4) and (3.5). Then G is capable if and only if X =Z .
X X
Remark 3.7. We will feel free to drop the subscript X from Y and Z when
X X
it is clear from context; also, to avoid multiple subindices, if we have subspaces
X ,...,X , we will denote Y and Z simply by Y and Z , respectively.
1 r Xi Xi i i
4. Some consequences
In this section, we will prove several results regarding the capability of groups
of class 2 and exponent p, based on the restatement of the problem given in The-
orem 3.6. It will be clear that many of the results could be proven by appealing
directly to the normal forms in the groups in question, without having to refer to
linear algebra, but at least the author found that the linear algebra setting was
usually easier to think about (and see also Section 5).
We start with an example of how we can use the result to prove that a given
group is not capable, and which also illustrates how to set up the linear algebra
problem given a specific group.
Example 4.1. A group of class two and exponent p, which is not capable.
Let G= x ,x ,x ,x be the nilpotent group of class two presented by:
1 2 3 4
h i
[x ,x ][x ,x ]−1 =[x ,x ][x ,x ]−1 = e,
3 1 3 2 3 1 4 1
(4.2) G= x ,x ,x ,x [x ,x ]=[x ,x ]=[x ,x ] = e,
* 1 2 3 4(cid:12)(cid:12) 4 2 xp =4 xp3=xp2=x1p = e, +
(cid:12) 1 2 3 4
(cid:12)
(we assume the condition [G(cid:12),G] Z(G) is given, so this is a presentation as an
N -group). This is an extra-s(cid:12)peci⊂al group of order p5; it is mentioned in Section 8
2
of [16], where the fact that it is not capable is deduced by invoking a theorem of
Beyl, Felgner, and Schmid in [3]. We prove that fact here using our set-up.
Let V and W be the vector spaces defined in (3.1) and (3.2), corresponding to
n=4; then X is defined by looking at the identities defining G, that is:
X = v v ,v v ,v ,v ,v .
31 32 31 41 42 43 21
− −
(cid:10) (cid:11)
8 ARTUROMAGIDIN
We have that dim(V) = 4 = 6, and dim(X) = 5. So X = Z if and only if
2 6 X
Z =V. And for that, it is enough to show that v Z .
X 41 X
(cid:0) (cid:1) ∈
To show that v Z , we need to show that each of v , v , v , and v
41 X 411 412 413 414
∈
are in Y = ϕ (X),ϕ (X),ϕ (X),ϕ (X) . Indeed:
X 1 2 3 4
v4(cid:10)11 = ϕ1 v42+(v31 v32) (cid:11) (v31 v41)
− − −
(cid:0) +ϕ2 v31 v41 ϕ3 v21 +ϕ(cid:1)4 v21 .
− −
v412 = ϕ1 v42 +(cid:0)ϕ4 v21 .(cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
v413 = ϕ1(cid:0)v43(cid:1) ϕ2(cid:0)v43(cid:1)+ϕ3 v42 +ϕ4 v31 v32 .
− −
v414 = ϕ3(cid:0)v42(cid:1) ϕ2(cid:0)v43(cid:1)+ϕ4(cid:0)(v31(cid:1) v32(cid:0)) (v31 (cid:1)v41) .
− − − −
Therefore, G is not ca(cid:0)pab(cid:1)le. (cid:0) (cid:1) (cid:0) (cid:1) (cid:3)
Theextremecasesareeasytohandle,sincetheϕ areinjective,andtheirimages
i
span W:
Lemma 4.3. Let n>1, and let V, W, and ϕ ,...,ϕ be given as in (3.1)–(3.3).
1 n
If X = 0 , then Z =X; if X =V, then Z =X.
X X
{ }
Theorem 4.4. Let n>1, p an odd prime. If G is a direct sum of n cyclic groups
of order p, then G is capable. If G is the 2-nilpotent product of n cyclic groups of
order p, then G is capable.
Proof. The direct sum correspondsto the case X =V, while the 2-nilpotent prod-
ucttoX = 0 . Ofcourse,these twoconclusionsaresimply specialcasesofBaer’s
{ }
theorem for abelian groups and its generalization to the k-nilpotent product of
cyclic p-groups, k <p (Theorem 6.4 in [16]). (cid:3)
Indiscussingthesituationinthelinearalgebrasetting,wewillreferto“the(j,i)
coordinate” ofvectorsofV,orto“the(j,i,k)coordinate” ofvectorsofW. Werefer,
of course, to the coefficient of v (resp. of w ) when the vector is expressed in
ji jik
terms of the standard bases given.
The followingeasyobservationswillbe repeatedseveraltimes, so we state them
as a lemma:
Lemma 4.5. Let n>1, and let V, W, and ϕ ,...,ϕ be as in (3.1)–(3.3).
1 n
(a) ϕ (v) has nonzero (j,i,i) coordinate, 1 i < j n, if and only if v has
r
≤ ≤
nonzero (j,i) coordinate, and r =i.
(b) ϕ (v) has nonzero (j,i,j) coordinate, 1 i < j n, if and only if v has
r
≤ ≤
nonzero (j,i) coordinate, and r =j.
(c) ϕ (v) has nonzero(k,i,j)coordinate, 1 i<j n, i<k n, ifand only
r
≤ ≤ ≤
if either r=j and v has nonzero (k,i) coordinate, or else r =i, and either
k >j and v has nonzero (k,j)coordinate orj >k and v has nonzero (j,k)
coordinate. In the case where r=i, the (k,i,j) coordinate of ϕ (v) is equal
i
to minus the (j,i,k) coordinate.
From these, we obtain:
Lemma 4.6. Let n > 1, and let V, W, and ϕ ,...,ϕ : V W be as in (3.1)–
1 n
→
(3.3). Let X be a subspace of V, and let i and j be fixed integers, 1 i<j n. If
≤ ≤
all vectors in X have zero (j,i) coordinate, then all vectors in Z have zero (j,i)
X
coordinate.
CAPABLE GROUPS OF PRIME EXPONENT AND CLASS TWO 9
Proof. Let v V be a vector with nonzero (j,i) coordinate. It suffices to show
that v / ϕ−1∈(Y ). Indeed, since all vectors in X have zero (j,i) coordinate, it
∈ i X
followsthat allvectorsin Y havezero (j,i,i) coordinate(applying part(a) of the
X
lemma above). Since ϕ (v) has nonzero(j,i,i) coordinate,it cannotlie inY , and
i X
therefore v / ϕ−1(Y ). (cid:3)
∈ i X
We will state most of our results as lemmas about the linear algebra situation,
and deduce the corresponding conclusions for groups of class two and exponent p
as theorems. Since some of our lemmas will appear contrived without the benefit
of knowing the group theory result we are after, we will usually state the theorem
first, and then the corresponding lemma about linear algebra.
A theorem of G. Ellis. The originalimpetus behind this work was my desire to
obtain an alternative proof of the following result:
Theorem 4.7(G.Ellis,Proposition9of[5]). Let Gbea finitelygeneratedgroup of
nilpotencyclassexactlytwoandofprimeexponent. Let x ,...,x beasubsetofG
1 k
{ }
corresponding to a basis of the vector space G/Z(G), and suppose that those non-
trivial commutators of the form [x ,x ], 1 i<j k are distinct and constitute a
j i
≤ ≤
basis for the vector space [G,G]. Then G is capable.
Proof. Let x ,...,x be elements of Z(G) that, together with x ,...,x , con-
k+1 n 1 k
stitute a minimal generating set of G. The conditions given in the statement of
the theorem imply that the subspace X of V corresponding to G will have a basis
consisting of a subset of the standard basis of V; therefore, Ellis’s Theorem will
follow from Lemma 4.8 below. (cid:3)
Lemma 4.8. Let n>1, and let V, W, ϕ ,...,ϕ be given as in (3.1)–(3.3). If X
1 n
has abasis consistingofasubsetof v 1 i<j n (thatis, X isa“coordinate
ji
{ | ≤ ≤ }
subspace” of V), then Z =X.
X
Proof. Assume that X has a basis consisting of a subset S v 1 i<j n .
ji
⊂{ | ≤ ≤ }
If v / S, then all vectors in X have zero (r,s) coordinate, and therefore by
rs
∈
Lemma 4.6, so do all vectors in Z . This means that Z S = X, giving
X X
equality. ⊂ h i (cid:3)
Capability of coproducts. Recall that we define the 2-nilpotent product of two
N groups (not necessarily cyclic) A and B as
2
A N2 B =(A B)/(A B) ,
2
∐ ∗ ∗
where A B is their free product. A theorem of T. MacHenry in [14] shows that
any eleme∗nt of A N2 B can be written uniquely as abc, where a A, b B, and
∐ ∈ ∈
c [B,A], the ‘cartesian’, and that the cartesian is isomorphic to Bab Aab, via
∈ ⊗
the map sending [b,a] to b a.
⊗
Lemma 4.9. Let G be a finite nontrivial group of class two and exponent p, p an
odd prime, and let C be the cyclic group of order p. Then G N2 C is capable.
p p
∐
Proof. The result follows from Theorem 4.4 if G is cyclic, so assume that G is
minimally generatedby g ,...,g , n>1, anddenote the generatorof C by x .
1 n p n+1
Let V be the vector space of dimension n with basis v , 1 i < j n, and
n 2 ji ≤ ≤
let V be the larger vector space that also includes the basis vectors v ,
n+1 (cid:0) (cid:1) (n+1)k
1 k n. Identify V with the obvious subspace of V . If X is the subspace
n n+1
≤ ≤
10 ARTUROMAGIDIN
of V corresponding to G, then X is also the subspace of V corresponding to
n n+1
G N2 C . This subspace does not contain any vectors with a nonzero (n+1,k)
p
coo∐rdinate, 1 k n. The result will now follow from Lemma 4.10 below. (cid:3)
≤ ≤
Lemma 4.10. Let n>1, let V, W, ϕ ,...,ϕ be as in (3.1)–(3.3), and let X be a
1 n
subspace of V. Suppose that there exists j, 1 j n, such that for all v X and
≤ ≤ ∈
all i, 1 i n, i = j, v has zero (i,j) or zero (j,i) coordinate (whichever makes
≤ ≤ 6
sense). Then X =Z .
X
Proof. We claim that
ϕ (V) ϕ (X),...,ϕ (X),ϕ (X),...,ϕ (X) = 0 .
j 1 j−1 j+1 n
∩
Indeed,everynonzeDrocoordinateofavectorinϕ (V)hasatleaEston(cid:8)e(cid:9)index(either
j
the second or the third) equal to j. But no vector in any of ϕ (X), i = j, has a
i
6
nonzero coordinate involving a j (in any position). Thus, neither do any of their
linear combinations. This means that ϕ−1(Y ) = X (since ϕ is injective), and
j X j
therefore that Z =X. (cid:3)
X
In fact, we can strengthen this result considerably:
Theorem 4.11. Let G and H be any two nontrivial finite groups of class at most
two and exponent p, p an odd prime. Then G N2H is capable.
∐
Proof. If either G or H are cyclic then the result follows from Lemma 4.9. If
they are both noncyclic, let G be minimally generated by g ,...,g , a > 1, and
1 a
let H be minimally generatedby h ,...,h , b>1. Let X be the subspace of
a+1 a+b 1
v 1 i<j a correspondingtoGundertheobviousidentification,andletX
ji 2
h | ≤ ≤ i
be the subspace of v a+1 i<j a+b corresponding to H. Then X X
ji 1 2
is the subspace of Vh =|v 1≤ i<j≤ a+bi that corresponds to G N2 H.⊕The
ji
result will follow from Lhemm| a≤4.12 bel≤ow. i ∐ (cid:3)
WewillneedsomenotationpriortostatingLemma4.12. Leta,b>1. LetV,W,
andϕ ,...,ϕ beasin(3.1)–(3.3),correspondington=a+b. WedecomposeV as
1 n
V =V V V ,whereV isgeneratedbythevectorsv ,1 i<j a(the“small
s m ℓ s ji
⊕ ⊕ ≤ ≤
index”vectors);V is generatedbythevectorsv ,a+1 i<j a+b(the “large
ℓ ji
≤ ≤
index”vectors);andV isgeneratedbythe vectorsv with1 i a<j b(the
m ji
≤ ≤ ≤
“mixed index” vectors). For any vector v V, we write
∈
v=v +v +v ,
s m ℓ
where v V , v V , and v V . The idea here is that the vectors in
s s m m ℓ ℓ
∈ ∈ ∈
the “small index” part of V will correspond to the relations defining G, while the
vectors in the “large index” part of V will correspond to relations defining H.
Similarly, we decompose W as W = W W W W ; W is generated
s ⊕ m1 ⊕ m2 ⊕ ℓ s
by the basis vectors which have all three indices smaller than or equal to a; W
m1
by the basis vectors which have exactly two indices smaller than or equalto a, one
largerthana; W by the basis vectors whichhave exactly one index smaller than
m2
orequalto a,andtwolargerthana;andW by the basisvectorsinwhichallthree
ℓ
indices are larger than a. As above, we can decompose any vector w W as
∈
w=w +w +w +w ,
s m1 m2 ℓ
with w W , etc.
s s
∈
