Frattini-embedded normal in subgroup and normal implies Frattini-embedded normal

Statement
Suppose $$H \le K \le G$$ are groups, such that:


 * $$H$$ is a normal subgroup of $$G$$
 * $$H$$ is a Frattini-embedded normal subgroup of $$K$$. In other words, $$H$$ is normal in $$K$$ and for any proper subgroup $$M$$ of $$K$$, $$HM$$ is a proper subgroup

Then $$H$$ is a Frattini-embedded normal subgroup of $$G$$.

Related facts

 * Frattini subgroup is normal-monotone: The Frattini subgroup of a normal subgroup is contained in the Frattini subgroup of the whole group, when the group satisfies the property that every proper subgroup is contained in a maximal subgroup. The proof of this monotonicity uses the statement of this article, along with the fact that a characteristic subgroup of a normal subgroup must be normal.

Frattini-embeddded normal subgroup
A subgroup $$H$$ of a group $$G$$ is termed Frattini-embedded normal in $$G$$ if $$H \triangleleft G$$ and, for any proper subgroup $$N$$ of $$G$$, $$HN$$ is proper.

Proof
Given: $$H \le K \le G$$ are groups, such that:


 * $$H$$ is a normal subgroup of $$G$$
 * $$H$$ is a Frattini-embedded normal subgroup of $$K$$. In other words, $$H$$ is normal in $$K$$ and for any proper subgroup $$M$$ of $$K$$, $$HM$$ is a proper subgroup

To prove: For any proper subgroup $$N$$ of $$G$$, $$HN$$ is a proper subgroup of $$G$$

Proof: Pick a proper subgroup $$N$$ of $$G$$. We need to show that $$HN$$ is proper in $$G$$. Suppose $$HN = G$$. We'll derive a contradiction.

We have $$HN \cap K = K$$, which, by the modular property of groups, gives:

$$H(N \cap K) = K$$

Now, clearly, $$H$$ is not contained in $$N$$ (otherwise $$HN = N \ne G$$), and since $$H \le K$$, $$K$$ is not contained in $$N$$. So $$N \cap K$$ is a proper subgroup of $$K$$, and thus we have found a proper subgroup of $$K$$ whose product with $$H$$ equals $$K$$. This contradicts the assumption that $$H$$ is a Frattini-embedded normal subgroup of $$K$$.