Frattini subgroup is normal-monotone

Verbal statement
The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.

(Note that this group property is always satisfied when the normal subgroup is a finite group, so for finite groups, the Frattini subgroup of a normal subgroup is always contained in the Frattini subgroup of the whole group).

Statement with symbols
Let $$N$$ be a normal subgroup of a group $$G$$, where $$N$$ satisfies the property that every proper subgroup is contained in a maximal subgroup. Then, $$\Phi(N)$$, the Frattini subgroup of $$N$$, is contained in $$\Phi(G)$$, the Frattini subgroup of $$G$$.

Property-theoretic statement
The subgroup-defining function that sends a group to its Frattini subgroup is a normal-monotone subgroup-defining function (with some assumptions on the nature of the groups).

Frattini subgroup
The Frattini subgroup of a group is the intersection of all its maximal subgroups. In other words, an element is in the Frattini subgroup if it is in every maximal subgroup.

Related facts

 * Frattini-embedded normal in subgroup and normal implies Frattini-embedded normal: This is the correct generalization, and no longer assumes anything about the group. It is also useful in proving a number of other results.

Proof outline
The proof uses four facts (for convenience, we denote the group by $$G$$ and normal subgroup by $$N$$):


 * 1) Characteristic subgroups of normal subgroups are normal: This fact helps us show that the Frattini subgroup of the subgroup $$N$$, is in fact normal in the whole group $$G$$
 * 2) For a group where every proper subgroup is contained in a maximal subgroup, the Frattini subgroup is a Frattini-embedded normal subgroup: a normal subgroup whose product with any proper subgroup is proper. Thus $$\Phi(N)$$ is a Frattini-embedded normal subgroup inside $$N$$.
 * 3) Frattini-embedded normal in subgroup and normal implies Frattini-embedded normal: This allows us to show that the Frattini subgroup of the subgroup, is Frattini-embedded normal inside the whole group. In other words, $$\Phi(N)$$ is a Frattini-embedded normal subgroup inside $$G$$
 * 4) For any group, any Frattini-embedded normal subgroup is contained inside the Frattini subgroup: This allows us to conclude that the Frattini subgroup of

Hands-on proof
Given: A group $$G$$,a normal subgroup $$N$$ of $$G$$ with the property that any proper subgroup of $$N$$ is contained in a maximal subgroup

To prove: $$\Phi(N) \le \Phi(G)$$

Proof: The Frattini subgroup $$\Phi(N)$$ is a characteristic subgroup of $$N$$. Since every characteristic subgroup of a normal subgroup is normal, $$\Phi(N)$$ is a normal subgroup of $$G$$.

We need to show that $$\Phi(N)$$ is contained in $$\Phi(G)$$. For this, it suffices to show that $$\Phi(N)$$ is contained in every maximal subgroup of $$G$$. We prove this by contradiction.

Suppose $$M$$ is a maximal subgroup of $$G$$ not containing $$\Phi(N)$$. Then, the subgroup generated by $$M$$ and $$\Phi(N)$$ is the whole of $$G$$. Since $$\Phi(N)$$ is normal in $$G$$, $$M\Phi(N) = G$$. From that, it follows that $$\Phi(N)(M \cap N) = N$$ (this is a particular example of the modular property of groups).

Since $$M$$ does not contain $$\Phi(N)$$, $$M$$ does not contain $$N$$ either and hence $$M \cap N$$ is a proper subgroup of $$N$$. Since every proper subgroup is contained in a maximal subgroup of $$N$$, there is a maximal subgroup of $$N$$ containing both $$M \cap N$$ and $$\Phi(N)$$. This contradicts the fact that their product is $$N$$.

Normality of $$N$$ is thus crucial because it guarantees normality of $$\Phi(N)$$. This in turn is crucial in converting a subgroup-generated statement to a product of subgroups statement.

Textbook references

 * , Exercise 22, Page 199 (Section 6.2)