Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements

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Statement

Suppose G is a primitive group with core-free maximal subgroup M. Then, if G has an Abelian minimal normal subgroup N, the following are true:

  1. NM = G
  2. N and M intersect trivially
  3. M \cong G/N
  4. The induced action of N on G/M by left multiplication, is equivalent to the regular action of a group on itself.

Definitions used

Core-free subgroup

Further information: Core-free subgroup

A subgroup of a group is termed core-free if its normal core in the whole group is trivial, or equivalently, if it does not contain any nontrivial normal subgroup of the whole group.

Maximal subgroup

Further information: Maximal subgroup

A maximal subgroup of a group is a proper subgroup that is not contained in any bigger proper subgroup.

Primitive group

Further information: Primitive group A primitive group is a group that possesses a core-free maximal subgroup.

Facts used

  1. Abelian normal subgroup and core-free subgroup generate whole group implies they intersect trivially

Related facts

Plinth theorem

Further information: plinth theorem

The plinth theorem is a similar result to the case of two distinct minimal nomral subgroups.

More results under similar hypotheses

Proceeding further along the lines of this proof, one can show that:

Breakdown of similar results

Proof

Given: A primitive group G with core-free maximal subgroup M, and an Abelian minimal normal subgroup N

To prove: N and M are permutable complements: NM = G and N \cap M is trivial

Proof: Since N is normal, NM is a subgroup of G containing M. Since M is maximal, either NM = M or NM = G. If NM = M, then N \le M, but this cannot happen because M does not contain any nontrivial normal subgroup of G. Hence NM = G.

Thus N is an Abelian normal subgroup and M is a core-free subgroup that together generate G. Applying the fact (1) stated above, N \cap M is trivial, so N and M are permutable complements.