Equivalence of definitions of Frattini subgroup

Definition in terms of maximal subgroups
For a group $$G$$, the Frattini subgroup $$\Phi(G)$$ is defined as the intersection of all maximal subgroups of $$G$$:

$$\Phi(G) = \bigcap_{M \le_{max} G} M$$

When there are no maximal subgroups, we define $$\Phi(G)$$ as the whole group $$G$$

Definition as the set of nongenerators
For a group $$G$$, the Frattini subgroup $$\Phi(G)$$ is defined as the set of all nongenerators. An element $$x \in G$$ is termed a nongenerator if, whenever $$S \subset G$$ is such that $$S \cup \{ x \}$$ generates $$G$$, $$S$$ itself generates $$G$$.

Facts used

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

The set of nongenerators is contained in every maximal subgroup
Given: A group $$G$$, a nongenerator $$x$$ for $$G$$, a maximal subgroup $$M$$ of $$G$$

To prove: $$x \in M$$

Proof: Suppose (contradiction point) $$x \notin M$$. Then, $$\langle M,x \rangle = G$$. Thus, the set $$M \cup \{ x \}$$ is a generating set for $$G$$. By the definition of nongenerator, we see that $$M$$, as a set, is a generating set for $$G$$. But this is a contradiction since the subgroup generated by $$M$$ is $$M$$ itself.

Thus, $$x \in M$$ is forced.

Any element in the intersection of all maximal subgroups is a nongenerator
Given: A group $$G$$, an element $$x$$ in the intersection of maximal subgroups of $$G$$, a subset $$S$$ such that $$S \cup \{ x \}$$ generates $$G$$

To prove: $$S$$ generates $$G$$

Proof: Let $$H$$ be the subgroup generated by $$S$$. Suppose (contradiction point) $$H$$ is proper. Then, $$x \notin H$$, since $$\langle H,x \rangle = G$$. Thus, $$H$$ is a 1-completed subgroup: it, along with one outside element, generates the whole group. By the above fact, there exists a maximal subgroup $$M$$ containing $$H$$. But by assumption, $$x$$ is in every maximal subgroup, so $$x \in M$$, so $$\langle H,x \rangle \le M$$, so $$G \le M$$, a contradiction.

Thus, the assumption that $$H$$ is proper is false, so $$H = G$$, so $$S$$ generates $$G$$.

Textbook references

 * , Exercise 25, Page 199 (Section 6.2)