Limits of Abelian Subgroups of Finite -groups

37 Pages·1997·0.27 MB·English

Limits of Abelian Subgroups of Finite p-groups
(cid:3) y
J.L. Alperin G. Glauberman
University of Chicago University of Chicago
[email protected] [email protected]
1 Introduction
Abelian subgroups play a key role in the theory and applications of (cid:12)nite
p-groups. Our purpose is to establish some very general results motivated by
special results that have been of use. In particular, it is known [KJ] that if
a (cid:12)nite p-group, for odd p, has an elementary abelian subgroup of order pn,
n (cid:20) 5, then it has a normal elementary abelian subgroup of the same order.
Our (cid:12)rst main result is a general one of this type.
Theorem A
If A is an elementary abelian subgroup of order pn in a p-group P, then
there is a normal elementary abelian subgroup B of the same order contained
in the normal closure of A in P, provided that p is at least n4/4.
The additional information about the location of B is essential for the
inductive argument we shall employ. The strategy is to (cid:12)rst reduce this
theorem to another, namely the following.
Theorem B
If A is an abelian subgroup of order pn in a p-group of exponent p and
class at most p +1, then there is a normal abelian subgroup B of the same
order contained in the normal closure of A in P.
The second theorem is now reducible to a question of Lie algebras via
Lazard’s functor connecting p-groups and Lie rings; our hypothesis allows
∗Supported in part by NSF
†Supported in part by NSA
1
use of this functor even though the bound on the class is not p − 1, which
is the usual situation. It is now a consequence of the following result. Fix a
(cid:12)eld F of prime characteristic p.
Theorem C
If A is an abelian subalgebra of the Lie algebra L over F and L is nilpotent
of class at most p, then there is an abelian ideal B of L, in the ideal closure
of A in L, of the same dimension as A.
Our proof of this result is direct but the motivation is entirely from alge-
braicgroups. Theanalogousresult forLiealgebrasover thecomplexnumbers
is due to Kostant [Kos, p.155]. Let L(cid:3) be the extension of L to the algebraic
closure F(cid:3) of F. The idea is that in characteristic p one can use the algebraic
group corresponding to L(cid:3) and apply the (cid:12)xed-point theorem of Borel to the
Grassmannian of all subspaces of the Lie algebra L(cid:3) which areof the same di-
mension as A. In particular, one could use the form of Borel’s theorem which
gives rational (cid:12)xed points. The authors would like to thank W. Fulton for
his helpful comments.
We need to introduce some standard notation. For elements or subsets X
andY ofagroupP welet(Y,X;k),fork non-negative,bede(cid:12)nedinductively,
as usual, by setting
(Y,X;0) = Y; then (Y,X;1) = (Y,X)
isthesubgroupgeneratedbyallcommutatorsoftheform(y,x)asxrunsover
X and y runs over Y; and (Y,X;i+1) = ((Y,X;i),X). We now turn to our
(cid:12)nal main result, which shows that if we are willing to relax the conclusion
on normality in Theorem A then we can achieve striking improvements in
the bounds on the prime p.
Theorem D
If p is at least 5 and A is an elementary abelian subgroup of order pn of
the p-group P, then there is an elementary abelian subgroup B, also of order
pn, satisfying the following conditions:
(a) B is normal in its normal closure in P;
(b) (P,B;3) = 1;
(c) B is normalized by any element x of P satisfying
(cid:18) (cid:19)
p+1
B,x; = 1.
2
2
However, let us occupy the rest of this introductory section with a proof
of the reduction of Theorem A to Theorem B.
Proof. The technique is to prove Theorem A by induction and argue until
the situation of Theorem B arises. In fact, we shall do a double induction,
(cid:12)rst on n and then on the order of P. The case n = 1 is trivial so we can
proceedwiththe induction. AgainifP hasorder pn thereisnothing toprove.
Let Q be a maximal subgroup of P containing A. Thus, there is a normal
elementary abelian subgroup A of Q, contained in the normal closure of A
1
in Q, also of order pn.
The normal closure of A in P is contained in the normal closure of A in
1
P so we can assume, by replacing A by A, that A is normal in a maximal
1
subgroup of P. In particular, A is normal in its normal closure AP in P.
Thus, the series of intersections of A with the upper central series of AP will
be a series without any equalities until the central series reaches AP. Thus,
AP will be of class at most n, so it will be of exponent p, being generated by
elements of order p and having class certainly less than p.
By induction on n there is an elementary abelian subgroup B of AP of
order pn−1 normal in P. If B does not coincide with its centralizer in AP,
then it is of order pn contained in the center of a subgroup C of AP, where
C is normal in P and has order pn. Since AP is of exponent p, we are done.
Hence, we can assume that B is a maximal normal abelian subgroup of AP.
Wenowformanotherp-group,thesemi-directproductofAP byP/C (AP)
P
and call this group P(cid:3). It su(cid:14)ces now to prove that P(cid:3) is of exponent p and
class less than p, as then Theorem B applies to P(cid:3) and its subgroup A and
gives us the desired subgroup of AP. But AP has order at most pn(n−1)/2
since B has order pn−1 and AP/C (B) is isomorphic with a group of auto-
P
morphisms of B and so has order at most p(n−1)(n−2)/2 ([S I, Thm 2.1.17(ii),
p.93]). Now, setting m = n(n−1)/2, P/C (AP) is isomorphic with a group
P
of automorphisms of the group AP of order at most pm and so has order at
most pm(m−1)/2. Hence, P(cid:3) has order at most p(m+1)/m which is, according
to our assumption that p is at least n4/4, at most pp. Moreover, since P(cid:3) is
generated by elements of order p it is of exponent p ([S I, Lemma 4.3.13(ii)
and Thm 4.3.14(ii), pp. 46-7]).
The rest of the paper is organized in the following way. In Section 2
we study a unipotent automorphism of a non-associative algebra and condi-
tions under which the existence of a subalgebra A with A2 = 0 implies the
3
existence of a similar subalgebra of the same dimension which is invariant
under the automorphism. In Section 3 we extend this result to a group of
automorphisms and derive Theorem C. In the next section we use Lazard’s
work to derive Theorems A and B and then in the last section we can apply
some of the second author’s earlier work [GG 1] to obtain Theorem D.
2 Unipotent automorphisms of algebras
In this section we let L be an algebra (not necessarily associative) over our
(cid:12)eld F of prime characteristic p, α a unipotent automorphism of L (so α−1
is nilpotent) and shall be studying the abelian subalgebras A (those A for
which A2 = 0) of L. We also (cid:12)x a sequence of α-invariant subspaces 0 =
L (cid:20) L (cid:20) ... ,(cid:20) L = L such that α induces the identity on each of the
0 1 s
successive quotients. Fix a dimension d and de(cid:12)ne a partial order (cid:30) on the
set of abelian subalgebras of L of dimension d in the following way. We set
B (cid:30) B0 provided dim(B\L ) (cid:20) dim(B0\L ) for all i > 0, and B\L strictly
i i j
contained in B0 \L for some j.
j
The main result is the following.
Theorem 2.8. If B is maximal under the partial order (cid:30), then B is
α−invariant provided the following conditions hold:
1. (α−1)p = 0;
2. ((α−1)iu)((α−1)jv) = 0 for all u and v in B and for all i,j (cid:21) 1 with
i+j (cid:21) p.
It is worth mentioning that the result holds in characteristic zero without
the two conditions but we do not need this for any application.
To obtain this result, we (cid:12)rst obtain a \limit" B(cid:3) of images of B (Propo-
sition 2.2). This is basically a result from algebraic geometry that requires
only that L be a vector space, and does not require Condition 2. Then we
use Condition 2 to show, in e(cid:11)ect, that taking limits induces an isomorphism
of B onto B(cid:3) (Proposition 2.4(b) and Theorem 2.6). Finally, we show that,
if B 6= B(cid:3), then B < B(cid:3), a contradiction.
Note that every unipotent linear transformation τ of V is invertible be-
cause, if (τ −1)n = 0,
1−(τ −1)+(τ −1)2 +...+(−1)n−1(τ −1)n−1 = (1+(τ −1))−1 = τ−1.
4
Moreover, if n (cid:20) p−1, we can de(cid:12)ne
(cid:88)p−1 (τ −1)i
Logτ = (−1)i+1 .
i
i=1
Clearly, Log τ is nilpotent because (Logτ)p = 0.
Similarly, suppose ν is a nilpotent linear transformation of V over F and
νp = 0. We de(cid:12)ne
(cid:88)p−1 νi
Exp ν = .
i!
i=0
Then ((Exp ν)−1)p = 0, and Exp ν is unipotent.
The following result appears to be well known, but we have not found a
complete published proof. Most of it is proved by (cid:12)nite methods in [Greg].
We give a shorter proof that requires in(cid:12)nite methods.
Proposition 2.1. Suppose V is a vectorspace overF, τ is a unipotentlinear
transformation of V over F, and ν is a nilpotent linear transformation of V
over F. Assume
(2.1) (τ −1)p = νp = 0.
Then
(a) Exp (Log τ) and Log (Exp ν) are well defined and
Exp (Log τ) = τ, Log (Exp ν) = ν,
and
(b) for every natural number r, Exp rν is well defined and
Exp rν = (Exp ν)r.
Proof. (a) Let Q be the (cid:12)eld of rational numbers and A be the Q-algebra
1
Q[[X]] of all power series in an indeterminate X. In A , de(cid:12)ne
1
(cid:88) (cid:88)
Xi Xi
eX = , e(X) = eX −1, and l(x) = (−1)i−1 .
i! i!
i(cid:21)0 i(cid:21)1
5
De(cid:12)ne eY,e(Y) and l(Y) similarly for every Y in A such that Y has zero
1
constant term. It is easy to see that these give well de(cid:12)ned elements of A
1
and that eX and l(X) are the familiar Maclaurin series for the functions et
and log(t−1). It is shown in [B, p.A.IV.40] that l(e(X)) = e(l(X)) = X.
We take the natural homomorphism Ψ of A onto the algebra
1
A = Q[X]/(Xp).
2
i.e. the quotient of the polynomial algebra Q[X] by the principal ideal gen-
erated by Xp. For X = Ψ(X) and for functions e and l on A de(cid:12)ned like
2
e(X) and l(X) above,
l(e(X)) = e(l(X)) = X.
Let Z[1/(p − 1)!] be the subring of Q obtained by adjoining the number
1/(p−1)! to the ring of integers Z, and let A be the subring of A generated
3 2
by
Z[1/(p−1)!] and X.
Then the functions e and l on A map A into A .
2 3 3
Now we take the natural homomorphism Ψ0 of A into the algebra over F
3
spannedby ν, withX mapping toν. Wede(cid:12)ne thefunctions eandl similarly
on the image of XA under Ψ0. Then, using our previous de(cid:12)nitions, we have
3
ν = l(e(ν)) = l(expν −1) = log(expν).
A similar homomorphism of A with X mapping to τ −1 yields
3
τ −1 = e(l(τ −1)) = e(logτ) = −1+exp(logτ),
and thus τ = exp(logτ).
(b) The proof is similar to that of (a). For commuting indeterminate
X,Y and the algebra of formal power series Q[X,Y], it is proved in [B,
pp.A.IV.39-40] that
eX+Y = eXeY.
By the method of (a), this yields a proof of (b) by induction on r.
Now take an indeterminate t over F, and let F(t) be the (cid:12)eld obtained
by adjoining t to F. For a vector space V over F, let V(t) be the vector
space F(t)⊗ V over F(t).
F
6
Proposition 2.2. Suppose λ is a linear transformation of a vector space V
over F, and λp = 0. For each u 2 V, let
(cid:88)p−1 λi(u)
(2.2) τ (u) = ‘ ti (in V(t)).
t
i!
i=1
Suppose U is a subspace of V. Let s = dimU, and take a basis u ,... ,u
1 s
of U. On the exterior product space (cid:3)s(V(t)), let
s((cid:88)p−1)
(2.3) c(t) = τ (u )^τ (u )^...^τ (u ) = c tk,
t 1 t 2 t s k
k=0
where c 2 (cid:3)s(V) for each k. Then
k
c = u ^u ^...^u .
0 1 2 s
Take the largest integer e for which c 6= 0. Then the following conditions
e
are satisfied:
(a) Up to scalar multiples, c(t) is independent of the choice of the basis
u ,... ,u .
1 s
(b) For some linearly independent vectors u(cid:3),u(cid:3),... ,u(cid:3) in V,
1 2 s
u(cid:3) ^u(cid:3) ^...^u(cid:3) = c .
1 2 s e
(c) The subspace U(cid:3) = hu(cid:3),u(cid:3),... ,u(cid:3)i of V has dimension s and is invari-
1 2 s
ant under λ.
(d) For each λ-invariant subspace X of V,
dimU(cid:3) \X (cid:21) dimU \X.
(e) Suppose 0 = V (cid:20) V (cid:20) ...V = V is an increasing series of subspaces
0 1 m
of V such that
(2.4) λV (cid:20) V , for i = 1,2,... ,m.
i i−1
If U is not invariant under λ, then
U(cid:3) \V > U \V for some i.
i i
7
Remark. By (c) and (d), U(cid:3) is contained in the smallest λ-invariant sub-
space of V containing U, i.e., in
U +λ(U)+λ2(U)+...+λn−1(U).
Proof. (a) Since
τ (u+v) = τ (u)+τ (v), and τ (au) = aτ (u)
t t t t t
for every u,v 2 V and a 2 F, this follows from the standard properties of
the exterior product.
(b)Leth (t),... ,h (t)bepolynomialsintwithcoe(cid:14)cientsfromV chosen
1 s
such that
(2.5) h (t)^h (t)^...^h (t) = c(t)
1 2 s
(cid:80)
and degh is minimal subject to (2.5).
i
For each i,(cid:80)let di be the degree of hi.and u(cid:3)i be the coe(cid:14)cient of tdi in
h (t). Set d = d . Clearly, from (2.5),
i i
(2.6) u(cid:3) ^u(cid:3) ^...^u(cid:3) is the coe(cid:14)cient of td in c(t)
1 2 s
and is equal to c if u(cid:3) ^u(cid:3) ^...^u(cid:3) 6= 0 (in which case d = e).
e 1 2 s
Suppose u(cid:3),... ,u(cid:3) are linearly dependent. Let
1 s
a u(cid:3) +(cid:1)(cid:1)(cid:1)+a u(cid:3) = 0,
1 1 s s
where a 2 F for each i and a 6= 0 for some i. Among all subscripts i for
i i
which a 6= 0, choose one, i for which d has the highest degree. We may
i 0 i
assume that i = 1 and that a = 1. Let
0 1
(cid:88)
h(cid:3)(t) = h (t)+ a td1−dih (t).
1 1 i i
2(cid:20)i(cid:20)s
Then h(cid:3) ^h (t)^h (t)^...^h (t) = c(t), but
1 2 3 s
(cid:88) (cid:88)
(cid:3)
degh + degh < degh + degh ,
1 i 1 i
2(cid:20)i(cid:20)s 2(cid:20)i(cid:20)s
contrary to the choice of h (t),... ,h (t). This contradiction shows that
1 s
u(cid:3),... ,u(cid:3) are linearly independent. Now (b) follows from (2.6).
1 s
8
(c) From (b), U(cid:3) has dimension s. Since λp = 0, we may let
(cid:88)p−1 λi
τ = τ = Exp λ = .
1
i!
i=1
Then τ operates on V and, in an obvious way, on V(t). From (2.2), we may
describe τ (u) as (Exp λt)(u) for each u 2 V. The usual calculations for
t
multiplying exponentials show that, for u 2 V,
(cid:32) (cid:33)
(cid:88)p−1 λi (cid:88)p−1 λj(u) (cid:88)p−1 λk(u)
τ(τ (u)) = tj = (t+1)k = τ (u)
t t+1
i! j! k!
i=1 j=0 k=0
Therefore, from (2.3) and (2.6),
(cid:88)l
τ(c )tk = τ(c(t)) = τ(τ (u )^τ (u )^...^τ (u ))
k t 1 t 2 t s
k=0
= τ (u )^τ (u )^...^τ (u )
t+1 1 t+1 2 t+1 s
(cid:88)e
= c(t+1) = c (t+1)k,
k
k=0
and (by comparison of coe(cid:14)cients of te)
τ(u(cid:3) ^u(cid:3) ^...^u(cid:3)) = τ(c ) = c = u(cid:3) ^u(cid:3) ^...^u(cid:3).
1 2 s e e 1 2 s
Hence, τ(U(cid:3)) = U(cid:3).
Since
(cid:88)p−1 (τ −1)i
λ = Log (Exp λ) = Log τ = (−1)i+1 ,
i
i=1
we have λ(U(cid:3)) (cid:20) U(cid:3).
(d) Let r = (dimU)\X. We may assume that u ,... ,u 2 X.
1 r
Let us apply the process in the proof of (b) to X and U \X in place of
V and U. We obtain polynomials f (t),... ,f (t) in t with coe(cid:14)cients from
1 r
X such that
(2.7) f (t)^f (t)^...^f (t) = τ (u )^τ (u )^...τ (u )
1 2 r t 1 t 2 t r
and the leading coe(cid:11)cients u(cid:3),... ,u(cid:3) of f (t),... ,f (t) are linearly indepen-
1 r 1 r
dent.
9
From (2.3) and (2.7), and the associativity of the exterior algebra of V(t),
f (t)^...f (t)^τ (u )^...τ (u ) = c(t).
1 r t r+1 t s
Nowwe choose polynomials h (t),... ,h (t) in twithcoe(cid:14)cients fromV such
(cid:80) 1 s
that degh is minimal subject to
i
(2.8)
h (t)^...^h (t) = c(t), and the leading coe(cid:14)cients of h (t),... ,h (t)
1 s 1 s
(cid:3) (cid:3)
span a subspace of V that contains u ,... ,u .
1 r
Then h ,... ,h exist, since
1 s
f (t),f (t),... ,f (t),τ (u ),... ,τ (u )
1 2 r t r+1 t s
satisfy (2.8). The proof of (b) can be adopted to this case; note that, in that
proof, the leading coe(cid:14)cient of h is a linear combination of the leading co-
1
e(cid:14)cients of h ,... ,h , so that replacing h by h(cid:3) either preserves or enlarges
2 s 1 1
the subspace of V spanned by the leading coe(cid:14)cients.
Thus,
hu(cid:3),... ,u(cid:3)i (cid:20) U(cid:3) \X,
1 r
and dimU(cid:3) \X (cid:21) r = dimU \X.
(e) Assume U is not invariant under λ. Take i to be maximal subject to
(2.9) U(cid:3) \V = U \V .
i i
Then 0 (cid:20) i < m and
(2.10) U(cid:3) \V 6= U \V ,
i+1 i+1
Suppose U \V (cid:20) U(cid:3)\V . By (2.10), U(cid:3)\V > U \V , as required.
i+1 i+1 i+1 i
Next, suppose U \V is not contained in U(cid:3) \V . Take u 2 U \V
i+1 i+1 i+1
such that u 62 U(cid:3) \ V . Then u 62 U(cid:3) and λp(u) = 0. Take r minimal so
i+1
that λr(u) = 0. Let
X = hu,λ(u),λ2(u),... ,λr−1(u)i.
Then X is invariant under λ. By (d),
dimU(cid:3) \X (cid:21) dimU \X.
10