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.
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 =
= ftQ = {a, 7} and
= a. So
= 7, and
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.
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.
Let ft^ be the set of /c-tuples of distinct elements of ft, so that X acts
transitively on
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.
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
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 =
= 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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