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

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Statement

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

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

Related facts

Definitions used

Frattini-embeddded normal subgroup

Further information: Frattini-embedded 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:

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.