Minimal normal subgroup and core-free maximal subgroup need not be permutable complements

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Statement

Suppose G is a primitive group, M is a core-free maximal subgroup of G, and N is a minimal normal subgroup of G. Then, M and N need not be permutable complements. Specifically, either of these cases is possible:

  • G itself is a simple group and N = G.
  • G is not a simple group, and N is a proper subgroup of G, but M \cap N is trivial.

Proof

Example of situation where the whole group is simple

Further information: alternating group:A5, A5 is simple

Let G be the alternating group on \{ 1,2,3,4,5 \}. Let N = G. Let M be the subgroup of G comprising those permutations that fix \{ 5 \}.

Note that:

  • M is maximal, since it has index five.
  • Since G is simple and M is a proper subgroup, M is a core-free subgroup.
  • Since G is simple, N = G is a minimal normal subgroup.

Example of situation where the whole group is not simple

Further information: symmetric group:S5

Let G be the symmetric group on \{ 1,2,3,4,5 \}. Let M be the subgroup comprising those permutations that fix \{ 5 \}. Let N be the subgroup comprising the even permutations, i.e., the alternating group on \{ 1,2,3,4,5 \}.

  • M is maximal, since it has index five, which is prime.
  • M is a core-free subgroup, since its conjugates are precisely the subgroups that fix the elements \{ 1 \}, \{ 2 \}, \{ 3 \}, \{ 4 \}, \{ 5 \}, and the intersection of all these is the identity element.
  • N is a minimal normal subgroup, since N is a simple normal subgroup.