4 CHAPTE R 1. GENERAL LEMMAS
LEMMA
2.2
(BENDER'S CRITERION).
Let X be transitive on ft with |ft| 2
and let p be an odd prime. Suppose that for each a, /3 G ft with a ^ j3 and for each
P G Sylp(Xap), ftp {a, (3}. Then X is 2-transitive on ft.
PROOF.
Fix a G ft and let /?, 7 G ft - {a}, P G Sy\p(Xap) and Q G Sylp(Xa7).
By Lemma 2.1, P,Q G Sylp(X) and so P,Q G Sylp(Xa). So there exists g e Xa
with P? = Q. Then {a,/?p =
ft9p
= ftQ = {a, 7} and
a9
= a. So
(39
= 7, and
X
a
is transitive on ft {ct}. Since |ft| 2 and X is transitive, this immediately
implies the desired 2-transitivity.
Next we prove a Prattini-like argument due to Witt.
LEMMA
2.3
(WITT' S LEMMA).
Suppose that X is k-transitive on ft, k 1.
Let A be a subset of ft of cardinality k, and D the pointwise stabilizer of A. Let S
be any subset of D. Then the following are equivalent:
(a) Nx{S) is k-transitive on fts; and
(b) IfgeX with S9 D, then there exists deD with S9 = Sd.
In particular, if S G Sylp(D) for some prime p or if D is solvable and S is a
Hall 7r-subgroup of D for some set TT of primes, then Nx(S) is k-transitive on O5.
PROOF.
Let ft^ be the set of /c-tuples of distinct elements of ft, so that X acts
transitively on
ft^k\
and assertion (a) is equivalent to the transitivity of Nx(S) on
(ft^)s- Thus replacing ft by ft^ we reduce the proof to the case that k = 1.
Now set T = {(/3, T)\p G ft, T is an X-conjugate of 5, and T Xp}. Then X
clearly acts on T and it is straightforward to check that conditions (a) and (b) are
each equivalent to the transitivity of X on T, so they are equivalent to each other.
Finally if S G Sylp(I}) or if S is a Hall subgroup of the solvable group D, then
(b) holds by Sylow's Theorem or by [IQ] 11.22]. Hence (a) holds as well and the
proof is complete.
The following extension of Witt's Lemma will be important in the study of
groups with strongly embedded subgroups.
LEMMA
2.4
(BENDER).
Let X be k-transitive on ft and D be the pointwise
stabilizer of the k-element subset A of ft. Letp be a prime and let S be a p-subgroup
of D maximal subject to the condition \fts\ k. Then Nx(S) is k-transitive on
ft5.
PROOF.
First suppose that k = 1, so that D = Xa for some a G ft. Choose
R G Sylp(Xa) with S R. If R = S then the desired conclusion follows from Witt's
Lemma, so assume that R S. By the maximahty of 5, ft# = {a}. In particular,
|ft| = |ftjj| = 1 mod p, whence R G Sylp(X). Thus each Sylow p-subgroup of X
has a unique fixed point on ft.
Now let p G fts with a ^ (3. There is Q G Sylp(X^) such that S Q. Set
Qx = NQ(S) and = NR(S). For some g G NX(S), 5* = (Qf ,i?i) is a p-group
and so fixes a point 6 G ft. Both Ri and Q\ contain S properly and hence by
maximahty we conclude first that R\ has a unique fixed point, which must be
6 = a. Thus 5* fixes a, so 5* D and the maximahty again implies that Q\ has
a unique fixed point, which must be 6 =
(39.
Thus
f39
= a, proving the lemma in
the case k = 1.
If k 1, choose a G A and set ^ = ft {a}. Thus Xa is (k l)-transitive on ^
and S is a p-subgroup of D = (Xa)^_^aj which is maximal subject to |\ls| k 1.
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