K implies Frattini-free

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Verbal statement

Every K-group is a Frattini-free group. In other words, the Frattini subgroup of any K-group is trivial.

Symbolic statement

Let G be a K-group. Then the Frattini subgroup \Phi(G) is trivial.

Property-theoretic statement

The group property of being a K-group is stronger than the group property of being a Frattini-free group.

Definitions used

Definitions for K-group

A K-group is a group wih the property that every subgroup has a lattice complement. That is, a group G is a K-group if for any subgroup H of G, there is a subgroup L such that HL is trivial and the subgroup generated by H and L is the whole of G.

Definitions for Frattini-free group

A Frattini-free group is a group with the property that the intersection of all its maximal subgroups (called its Frattini subgroup) is trivial. Equivalently, for any nontrivial element, there is a maximal subgroup not containing that element.

Facts used


Symbol-free proof

  • A K-group has the property that every cyclic subgroup has a lattice complement.
  • The set of subgroups that do not contain the generator of the cyclic subgroup, ordered by subgroup inclusion, satisfies the conditions needed for Zorn's lemma, and hence there is a maximal element in this
  • Any subgroup maximal with respect to not containing this element, must be a maximal subgroup of the group.
  • Thus, given any element, there is a maximal subgroup not containing it.

Proof with symbols

Let G be a K-group and x a nontrivial element of G. We need to show that there is a maximal subgroup of G not containing x.

Let C be the cyclic sugbroup generated by x. Since G is a K-group, there exists a proper subgroup L of G such that CL is trivial and the subgroup generated by C and L is the whole of G.

Consider the family of subgroups of G that contain L do not contain x. Under subgroup inclusion, this family forms a partially ordered set that satisfies the conditions for Zorn's lemma. Hence, by Zorn's lemma, there exists a subgroup M maximal with respect to the property of not containing x.

Since L along with x generates G, M along with x also generates G. Hence, any proper subgroup of G containing M must not contain x. Since M is maximal among such subgroups, M is a maximal subgroup.

Thus, starting with an arbitrary nontrivial x in G, we have produced a maximal subgroup M not containing x.