For which permutation groups does there exist a subset of the permuted set whose stabilizer in the group is trivial?The permuted set has so many subsets that one might expect that subsets with trivial stabilizer usually exist. The symmetric and alternating groups are obvious exceptions to this expectation. Another, more interesting, infinite family of exceptions are the 2-Sylow subgroups of the symmetric groups on 2n symbols, in their natural representations on 2n points.One of our main results, Corollary 1, sheds some light on this last family of groups. We show that when the permutation group has odd order, there is indeed a subset of the permuted set whose stabilizer in the group is trivial. Corollary 1 follows easily from Theorem 1, which completely classifies the primitive solvable permutation groups in which every subset of the permuted set has non-trivial stabilizer.

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A base of a permutation group G is a sequence B of points from the permutation domain such that only the identity of G fixes B pointwise. We show that primitive permutation groups with no alternating composition factors of degree greater than d and no classical composition factors of rank greater than d have a base of size bounded above by a function of d. This confirms a conjecture of Babai.

Much information about a finite group is encoded in its character table. Indeed even a small portion of the character table may reveal significant information about the group. By a famous theorem of Jordan, knowing the degree of one faithful irreducible character of a finite group gives an upper bound for the index of its largest normal abelian subgroup.Here we consider b(G), the largest irreducible character degree of the group G. A simple application of Frobenius reciprocity shows that b(G) ≧ |G:A| for any abelian subgroup A of G. In light of this fact and Jordan's theorem, one might seek to bound the index of the largest abelian subgroup of G from above by a function of b(G). If is G is nilpotent, a result of Isaacs and Passman (see [7, Theorem 12.26]) shows that G has an abelian subgroup of index at most b(G)4.