K implies Frattini-free

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 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 $$H$$ &cap; $$L$$ 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

 * Every 1-completed subgroup is contained in a maximal subgroup

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 $$C$$ &cap; $$L$$ 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 $$x$$.