The Frattini subgroup is inclusion-monotone on normal subgroups

When G is a finite group prove that if N \leq G is normal, then \Phi(N) \leq \Phi(G). Give an explicit example when this containment does not hold if N is not normal in G.

[With help from the GroupProps wiki.]

We begin with some lemmas.

Lemma 1: Let G be a group, and let A,B,C \leq G with A \leq C be subgroups. Then A(B \cap C) = AB \cap C. Proof: (\subseteq) Let x \in A(B \cap C). Then x = ad where d \in B and d \in C, so that x \in AB \cap AC = AB \cap C. (\supseteq) Let x \in AB \cap C. Then we have x = ab = c for some a \in A, b \in B, and c \in C. Then b = a^{-1}c \in C, so in fact b \in B \cap C. Thus x \in A(B \cap C). \square

Lemma 2: Let G be a finite group. If H < G is a proper subgroup, then \Phi(G)H < G is proper. Proof: \Phi(G) \leq G is characteristic, hence normal, in G, so that \Phi(G)H \leq G is a subgroup. Now H is contained in some maximal subgroup M \leq G since G is finite, and \Phi(G) \leq M by definition. Thus \Phi(G)H \leq M < G is proper. \square

Now to the main result.

Let N \leq G be normal and suppose by way of contradiction that \Phi(N) \not\leq M for some maximal subgroup M \leq G. Since \Phi(N) \leq N is characteristic and N \leq G is normal, \Phi(N) \leq G is normal. Thus \Phi(N)M \leq G is a subgroup, and in fact \Phi(N)M = G.

Now N = G \cap N = \Phi(N)M \cap N = \phi(N)(M \cap N). However, since \Phi(N) \not\leq M, then N \not\leq M, so that M \cap N \leq N is proper. But this implies (by Lemma 2) that N < N is proper, a contradiction. Thus no such subgroup M exists, and in fact \Phi(N) \leq M for all maximal subgroups M \leq G. Hence \Phi(N) \leq \Phi(G).

Now before we produce a counterexample, we prove another lemma.

Lemma 3: \Phi(S_5) = 1. Proof: \Phi(S_5) is a proper normal subgroup, so that \Phi(S_5) \in \{1, A_5\}. Now S_5 contains a subgroup H isomorphic to S_4; by Lagrange, we have that [S_5,H] = 5 is prime, so that H is maximal. Recall also that A_5 is maximal; thus |\Phi(S_5)| divides both |H| = 24 and |A_5| = 60; hence |\Phi(S_5)| divides 12, and we have \Phi(S_5) = 1. \square

Consider K = \langle (1\ 2\ 3\ 4) \rangle \leq S_5; this subgroup is cyclic of order 4, and thus has a unique maximal subgroup which is isomorphic to Z_2. Thus K \leq S_5 but \Phi(K) \not\leq \Phi(S_5).

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