Maximal subgroup

A proper subgroup with no strictly intermediate subgroup between it and the ambient group. In a transitive action a point stabilizer is maximal exactly when the action is primitive.

Up to conjugation, the proper maximal subgroups of $S_n$ containing a three-cycle are $A_n$; the unequal-part subset stabilizers $S_k\times S_{n-k}$ for $1\le k<n/2$ with $n-k\ge 3$; and the full block stabilizers $S_a\wr S_b$ for $n=ab$, $a\ge 3$, $b\ge 2$. The primitive three-cycle criterion treats the primitive case. A nontrivial block system gives the full permutation wreath product in the imprimitive case. These block stabilizers are maximal because any extra permutation produces a transposition connecting two different blocks; conjugating by the block stabilizer then gives every cross-block transposition, hence the symmetric group. Unequal-part subset stabilizers are maximal by the same connected-transposition argument. Equal-part subset stabilizers are not maximal: they lie in the larger stabilizer that can interchange the two parts.

Two maximal subgroups of a finite soluble group either multiply to the whole group or are conjugate. Reduce modulo a nontrivial normal subgroup contained in either maximum. If no such subgroup exists, both maxima complement a minimal normal elementary abelian subgroup $K$. Their actions on $K$ are faithful and irreducible. A minimal normal subgroup of $G/K$ has characteristic different from that of $K$, since a normal p-subgroup has nonzero fixed points on a characteristic-p module. The intersections of the two maxima with its preimage are conjugate Sylow subgroups. Each maximum is the normalizer of its nontrivial intersection, proving conjugacy.

For a maximal subgroup $M$ of a finite nonabelian simple group, take a minimal nontrivial normal subgroup $N$ of $M$. It is a characteristically simple group. Its normalizer contains $M$ but cannot be the whole simple group, since $1<N\le M<G$. Maximality gives equality.