## 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 $\Phi(K) \not\leq \Phi(S_5)$.