K implies Frattini-free

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property must also satisfy the second group property
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Statement

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 $\mathrm{\Phi }(G)$ 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 $H$ ∩ $L$ 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

Proof

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 $C$ ∩ $L$ 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 $x$.