Frattini subgroup is ACIC
This fact is related to the problem of realization related to the following subgroup-defining function: Frattini subgroup
Realization problems are usually about which groups can be realized as subgroups/quotients related to a subgroup-defining function.
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- 1 Statement
- 2 Definitions used
- 3 Facts used
- 4 Generalizations
- 5 Proof
Let be a finite group (or more generally, a group where every subgroup is contained in a maximal subgroup), and be its Frattini subgroup. Then, is an ACIC-group: whenever is an automorph-conjugate subgroup of , is characteristic in .
Further information: Frattini subgroup
The Frattini subgroup of a group is defined as the intersection of all its maximal subgroups.
Group where every subgroup is contained in a maximal subgroup
For infinite groups, it could happen that there are no maximal subgroups, and it could also happen that not every proper subgroup is a maximal subgroup. The proof we give here does not work for such groups. We require our group to have the property that every proper subgroup is contained in a maximal subgroup.
Further information: automorph-conjugate subgroup
A subgroup in a group is automorph-conjugate if for any automorphism of , and are conjugate.
Any characteristic subgroup is automorph-conjugate.
Further information: ACIC-group
An ACIC-group is a group in which every automorph-conjugate subgroup is characteristic. Equivalently, every automorph-conjugate subgroup is normal (The two ar eequivalent because normal automorph-conjugate subgroups are characteristic).
The proof is a generalized version of Frattini's argument.
To Frattini-embedded normal subgroups
Further information: Frattini-embedded normal-realizable implies ACIC The result generalizes to the following result for arbitrary groups (with no finiteness assumptions): any Frattini-embedded normal subgroup of a group is ACIC. A Frattini-embedded normal subgroup is a normal subgroup whose product with any proper subgroup is proper.
Given: a group with the property that every proper subgroup is contained in a maximal subgroup, and . is an automorph-conjugate subgroup of .
To prove: is a normal subgroup of
Proof: We will in fact show that is normal in . Normality in will then follow.
Frattini's argument step
Further information: Frattini's argument
We first show that . This part is called the Frattini's argument. It only uses the hypothesis that is automorph-conjugate in and is normal in .
Suppose is any element. Then, since is normal in , conjugation by restricts to an automorphism of . Thus, and are related via an automorphism in .
Using the hypothesis that is automorph-conjugate inside , we see that there exists such that . Hence, the element lies in the normalizer . Rearranging, we see that any element of can be written as a product of an element of and an element of . So .
This part uses the assumption we made about (that every proper subgroup is contained in a maximal subgroup) and the fact that is the Frattini subgroup of .
We assume that and obtain a contradiction. If , i.e. it is a proper subgroup, then by hypothesis it is contained in some maximal subgroup . By definition of Frattini subgrou, . So , yielding , a contradiction.
Thus, , hence is normal in .