. Thus P 00= P 0by (4), and P 00is conjugate to Upk.
If G = U or if p is odd, then P = P 00and we are done. If p = 2 and G = O or
Sp , then N(P 00)=P 00~=C2x Sp 2k(F2) by (8). Furthermore, NG(n)(P ) NG(n)(P 0*
*0) by
construction of P 00; so P=P 00is 2-stubborn in N(P 00)=P 00, and hence must be*
* pre-
cisely the C2 factor (|P=P 00| 2). It follows that P is conjugate to O2k+1 O *
*(2k+1 )
or Sp2k Sp(2k); and that N(P )=P ~=Sp2k(F2).
Case B Now assume that A ~= {1}. In particular, p = 2 and G = O or Sp .
Then any central extension of the form given in (5) and (6) is isomorphic, eith*
*er to
a central product of copies of D(8), or a central product of one copy of Q(8) a*
*nd
copies of D(8). In either case P 0has a unique effective irreducible representa*
*tion:
of real type in the first case, and of symplectic type in the second case. In *
*other
words, P 0 G(n) is conjugate to one of the pairs O2k O (2k) or Sp2kSp (2k). As
in Case A, any automorphism of P 0which induces the identity on P 0=A is inner;
and so P = P 0by (4).
Finally, since P has a unique effective irreducible representation,
ae O+ (F ) if G = O
NG(2k)(G2k)=(G2k) ~=Out (G2k) ~= 2k- 2
O2k+2 (F2) if G = Sp
(see [12, x4] for the definitions of these groups). By a theorem of Dieudonne *
*[12,
pp. 47-51], for k 3, any nontrivial normal subgroup of O 2k(F2) contains its
commutator subgroup, which is simple. So O 2k(F2) is 2-reduced when k 3;
and the following list shows that O 2k(F2) is 2-reduced in all other cases exce*
*pt for
O +2(F2):
O +2(F2) ~=C2; O -2(F2) ~=3; O +4(F2) ~=3 o 2; and O -4(F2) ~=5:
Thus O2kand Sp2kare 2-stubborn for all k 0, except O2~=D(8).
6
Step 2 We now assume that there exists a subgroup A K P such t*
*hat
KC N(P ), and such that V |K splits as a sum of nonisomorphic irreducible K-
representations. Write V |K~=V1 x . .x.Vr (r > 1), where each Vi is a sum of
isomorphic K-representations, and where for i6=j the irreducible summands of Vi
and Vj are nonisomorphic.
Since V is irreducible an an P -representation, P=K permutes the Vi transiti*
*vely.
In particular, they all have the same dimension: set m = n=r = dim (Vi). So (af*
*ter
conjugating) we may assume that K G(m)r.
Let K0=K P=K be the subgroup of elements which leave each Vi invariant. Let
Ki G(m) be the image of K0 in the i-th factor, so that
K0 K = K1 x . .x.Kr G(n):
Since KC N(P ), the conjugation action of N(P ) permutes the Vi, and so K is
normalized by N(P ). Lemma 1 now applies to show that K P .
Now set L = P=K . Then L permutes the Vi effectively and transitively. Als*
*o,
since OV (g) = 0 for all g2 P rK , the action of any 16=x2 L must permute the Vi
freely. So we can regard L as a free transitive subgroup of r. Then P is conjug*
*ate
to K1 o L, and
NG(n)(P )=P ~= NG(m) (K1)=K1 x Nr (L)=L ~=NG(m) (K1)=K1 x Aut(L): (9)
In particular, Aut (L) is p-reduced, so L must be elementary abelian, and is co*
*n-
jugate to Epk pk (r = pk). Also, K1 is p-stubborn in G(m). So by induction,
we see that P is conjugate to an iterated wreath product o Eq1 o . .o.Eqr, whe*
*re
qi = 2tiand is one of the groups listed in (2). Also, each of the o Eq1 o . .*
*o.Eqi
is p-stubborn in the appropriate G(mi) and normal in N(P ); and so
NG (P )=P ~= NG(m) ()= x GLt1(F2) x . .x.GLtr(F2):
Note in particular that q1 4 if = O (1) (and p = 2) _ since O (1) o 2 is not
2-stubborn in O (2).
Step 3 In order to finish the proof of point (i), it remains to show that al*
*l of the
groups listed there are p-stubborn (and with N(P )=P as given). Fix a subgroup *
*of
the form P = o Eq1 o . .o.Eqr G(n): where qi = 2ti, G(m) is one of the
groups listed in (2), and q1 4 if = O1= O (1). When r = 0, was shown to be
p-stubborn in Step Otherwise,
qr
P 0= o Eq1 o . .o.Eqr-1 C P
is the subgroup generated by elements of nonzero character on V . In particula*
*r,
P 0CN(P ), and so
N(P )=P ~= N(P 0)=P 0x GLtr(Fp):
7
It now follows inductively that
N(P )=P ~= NG(m) ()= x GLt1(Fp) x . .x.GLtr(Fp):
This is p-reduced since N()= is (Step 1); and so P is p-stubborn.
Step 4 Now let P G(n) be an arbitrary p-stubborn subgroup. Assume
that the corresponding P -representation factors as a product V1 x : :x:Vs of i*
*r-
reducible representations. In other words, after conjugating, we may assume th*
*at
P G(n1) x . .x.G(ns), where ni = dim (Vi), and where the image Pi G(ni) of
P in the i-th factor is irreducible for each i.
By Lemma 1, CG(n)(P ) = Z(P ) _ unless possibly p is odd and G = O in
which case [CG(n)(P ) : Z(P )] 2. In this latter case, CG(n)(P ) is a product*
* of
unitary groups and one copy of O (m), where m = dim (V P). Also, m 1, since
CG(n)(P )=Z(P ) O (m) is the number of trivial summands in V .
In either case, CG(n)(P ) is abelian, so the Viare distinct as P -representa*
*tions, and
are permuted by N(P ). In particular, N(P ) N(P1 x . .x.Ps), and so Proposition
2.2 applies to show that P = P1 x . .x.Ps.
After reindexing, we can write P = (P1)m1 x . .x.(Pr)mr , where the Pi G(ni)
are irreducible and pairwise distinct as representations. Then
N(P )=P ~= NG(n1)(P1)=P1 o m1 x . .x. NG(nr)(Pr)=Pr o mr :
In particular, since N(P )=P is p-reduced by assumption, each of the N(Pi)=Pi m*
*ust
be p-reduced, and so each of the Pi must be p-stubborn in G(qi). If N(Pi)=Pi is
p-reduced, then the wreath product (N(Pi)=Pi) o mi is p-reduced if and only if
either N(Pi)=Pi 6= 1, or mi is p-reduced. Finally, m fails to be p-reduced on*
*ly
when (p; m) is one of the pairs (2; 2), (2; 4), or (3; 3); and this finishes th*
*e proof of
point (ii).
For the sake of completeness, we also note the following conditions for when
for two p-stubborn subgroups of G(n), of the form described in Theorem 3, one is
contained (up to conjugacy) in the other.
Proposition 4. Let p, and G = G(n) be as in Theorem 3, and let P 0 P be a
pair of p-stubborn subgroups of G.
(i) If P is reducible _ if P = P1 x . .x.Ps where Pi G(ni) is 2-stubborn and
irreducible _ then P 0= P10x . .x.Ps0for some subgroups Pi0 Pi p-stubborn in
G(ni).
(ii) If P is irreducible and P 0is reducible, then P is a wreath product
P = o Eq1 o . .o.Eqr
8
for some G(q0) as in Theorem 3 and some r 1. And P 0= P10x. .x.Pt0, where
each Pi0is an irreducible subgroup of o Eq1 o . .o.Eqs P (one of the standard
subgroups of this form) for some 0 s < r.
(iii) Now assume that P and P 0are both irreducible, and that
P = o Eq1 o . .o.Eqr and P 0= 0o Eq01o . .o.Eq0s
(up to conjugacy), where G(q0) and 0 G(q00). Then either P 0= P , or s < r
and q0; : :;:qris a refinement of q00; : :;:q0s(i.e., q00= q0 . .q.i0, etc.).
Proof. Part (i) is clear. To see the other parts, define
A = is*
* a
p-stubborn subgroup of U (n); if and only if P = P \ SU (n) for some p-stubborn
subgroup P of U (n).
Proof. By Lemma 1(ii), for any connected G and any p-stubborn subgroup P G,
P CG (P ) Z(G). And for any P SU (n) such that P Z(SU (n)), if we set P =
, then P is p-toral if and only if P is, NSU (n)(P )=P ~=NU (n)(P *
*)=P ; and
so P is p-stubborn if and only if P is.
Note the following corollary to Theorem 5. For convenience, for any prime po*
*wer
pk, we write SUpk= SU (pk)\Upk. Note that there is a central extension 1!- Cpk*
*!-
SUpk-!(Cp)2k!- 1.
Corollary 6. If P SU (n) is a finite p-stubborn subgroup, then n = pk for so*
*me k,
and P is conjugate to SUpk. Any irreducible p-stubborn subgroup of SU (pk) cont*
*ains
a subgroup conjugate to SUpk.
Proof. Assume that P U(n) is p-stubborn: then P = P 0\SU (n) for some p-
stubborn P 0 U (n) by Theorem 5. In particular, if P is finite, then dim (P 0)*
* = 1
and P 0is irreducible. So n = pk for some k 0, P 0is conjugate to Upkby Theorem
3(i), and P is conjugate to SUpk= Upk\SU (pk).
The last statement follows easily from Theorems 3(i) and 5.
The relation between 2-stubborn subgroups of SO (n) and O (n) is more compli-
cated.
Proposition 7. For any 2-stubborn subgroup P SO (n), there is a unique 2-
stubborn subgroup P O (n) such that P = P \ SO(n) and NO (n)(P ) = NO (n)(P ).
If P1 P2 is a pair of 2-stubborn subgroups of SO (n), then P1 P2.
Proof. If P SO (n) is 2-stubborn, then NSO (n)(P )=P is 2-reduced, and so *
*the
intersection P =P of the 2-Sylow subgroups in NO (n)(P )=P has order at most
2. If P =P = 1, then P = P is 2-stubborn in O (n). Otherwise, P 6 SO (n)
and NO (n)(P )=P = NO (n)(P )=P is 2-reduced; and so P is 2-stubborn and P =
P \ SO (n). The uniqueness of P is clear.
Now assume that P1 P2 are 2-stubborn in SO (n), and that Pi Pi are as abov*
*e.
We must show that P1 P2 .
An inspection of the list in Theorem 3 shows that for any irreducible 2-stub*
*born
subgroup P O (m), either m = 2 (and P = O (2)), or P SO (m), or there
exist elements in P r SO (m) of nonzero trace. So in all cases, P = P \ SO (m)*
* is
irreducible. Upon extending this to arbitrary 2-stubborn subgroups, we see that
10
the Pi (for i = 1; 2) have the same decompositions into irreducible representat*
*ions
as the Pi. In particular, we may assume that P2 is irreducible.
Now define subgroups
Ai =