Self-centralizing and minimal normal implies strictly characteristic

From Groupprops
Jump to: navigation, search
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., self-centralizing minimal normal subgroup) must also satisfy the second subgroup property (i.e., strictly characteristic subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about self-centralizing minimal normal subgroup|Get more facts about strictly characteristic subgroup

Statement

Verbal statement

Any Minimal normal subgroup (?) that is also self-centralizing (i.e., contains its centralizer in the whole group) is strictly characteristic: it is invariant under any surjective endomorphism of the group.

Definitions used

Minimal normal subgroup

Further information: minimal normal subgroup

A minimal normal subgroup of a group is a nontrivial normal subgroup that does not properly contain any other nontrivial normal subgroup.

Self-centralizing subgroup

Further information: self-centralizing subgroup

A subgroup of a group is termed self-centralizing if it contains its centralizer in the whole group.

Strictly characteristic subgroup

Further information: strictly characteristic subgroup

A subgroup H of a group G is termed strictly characteristic if for every surjective endomorphism \sigma of G, \sigma(H) \le H.

Related facts

Facts used

  1. Self-centralizing and minimal normal implies monolith: In other words, a self-centralizing minimal normal subgroup is contained in every nontrivial normal subgroup.
  2. Monolith is strictly characteristic: A subgroup contained in every nontrivial normal subgroup is strictly characteristic.

Proof

Given: A group G, a minimal normal subgroup N such that C_G(N) \le N. A surjective endomorphism \sigma of G.

To prove: \sigma(N) \le N.

Proof: Consider the subgroup \sigma^{-1}(N). This is normal by fact (2), so by fact (1) either \sigma^{-1}(N) is trivial or N \le \sigma^{-1}(N). Since \sigma is surjective and N is nontrivial, \sigma^{-1}(N) cannot be trivial. Thus, N \le \sigma^{-1}(N). This forces that \sigma(N) \le N, as desired.