The Frattini subgroup is the set of all nongenerators

An element x of a group G is called a nongenerator if for all proper subgroups H \leq G, \langle x, H \rangle \leq G is also proper. Prove that \Phi(G) \leq G is the set \mathsf{ng}(G) of all nongenerators in G.

(\subseteq) [Not finished.] Let x \in \Phi(G). Now let H < G be a proper subgroup. If H is contained in some maximal subgroup M, then we have \Phi(G)H \leq M < G, so that \langle x, H \rangle < G is proper as desired. Suppose now that H is not contained in any maximal subgroup of G. Suppose further that \Phi(G)H = G. (@@@)

[Old proof, assumes G is finite.]
Let x \in \Phi(G). By a lemma we proved for a previous theorem, if H \leq G is proper then \Phi(G)H \leq G is proper. Now \langle x \rangle \leq \Phi(G)H and H \leq \Phi(G)H, so that \langle x,H \rangle \leq \Phi(G)H < G is proper. Thus x \in \mathsf{ng}(G) is a nongenerator.

(\supseteq) Let x \in \mathsf{ng}(G) be a nongenerator of G. If G has no maximal subgroups, then \Phi(G) = G, and we have \mathsf{ng}(G) \subseteq \Phi(G). Suppose now that M \leq G is maximal; in particular, M \leq G is proper, so that \langle x, M \rangle \leq G is proper. Thus \langle x, M \rangle \leq M, so that x \in M. Since x is contained in every maximal subgroup, x \in \Phi(G).

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  • Melanie  On September 11, 2011 at 10:16 pm

    The Lemma you refer to assumes that G is finite, yet this result should hold for infinite G as well.

    • nbloomf  On September 12, 2011 at 8:50 am

      Hmm… It appears that the lemma assumes G to be finite so that we can assume H is contained in some maximal subgroup. So we could replace the hypothesis “G is finite” by “every subgroup of G is contained in a maximal subgroup.” This statement is not necessarily true for all groups, as there are groups which have no maximal subgroups at all.

      I wonder if the following statement is true: “If G has a maximal subgroup, then every subgroup is contained in a maximal subgroup.” If so, then perhaps we could consider separately the cases where G has a maximal subgroup (which is covered by the amended lemma and this proof) and where G does not have a maximal subgroup (in which case the Frattini subgroup, being the intersection over an empty class of subgroups of G, is all of G.) I don’t see how to prove this, or even if it should be true.

      I’ll have to think about this some more.

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