For Frattini subgroup $\Phi(G)$, if $N \vartriangleleft G$, then $\Phi(N) \leq \Phi(G)$. : Counter-example

Question

I'm trying chapter 6.1 of Dummit & Foote and the exercise 22 is the following:

Prove that if $N \vartriangleleft G$ then $\Phi(N) \leq \Phi(G)$.

Here $\Phi(G)$ is the Frattini subgroup. But let $G=S_3$, $N=A_3$, then $\Phi(A_3)=A_3$ since there are no maximal subgroups in $A_3$ and $\Phi(S_3)=1$.

What am I missing?

Answer 1

Every nontrivial finite group has maximal subgroups, because the proper subgroups form a finite nonempty lattice. A subgroup $H$ of $G$ is maximal if $H\ne G$ and there is no subgroup $K$ of $G$ such that $H\subsetneq K\subsetneq G$. Nowhere it is claimed that $\{1\}$ cannot be a maximal subgroup; in this context, proper just means “not equal to the whole group”.

Since $|A_3|=3$, it only has $\{1\}$ as a maximal subgroup. You can also see this directly, because $\{(123)\}$ and $\{132)\}$ are generating sets of $A_3$ and the single element cannot be removed, so no nonidentity element is a nongenerator.

The trivial group, consisting only of the identity, has no maximal subgroup; otherwise, only infinite groups can be equal to their Frattini subgroup; for instance the Prüfer $p$-group.

Answer 2

The trivial subgroup is maximal in $A_3$, and $\Phi(A_3)=1$.

Answer 3

You may prove the statement as follows:

Fact: Let $H \leq G $ then we have $\Phi(N) H \leq G$ because $\Phi(N) \ \text{char} \ N \trianglelefteq G $ and $\Phi(N) \trianglelefteq G.$

We want to prove that $\Phi(N)$ is contained in every maximal subgroup of $G.$ If there exists a maximal subgroup $M$ of $G$ such that $\Phi(N)$ is not contained in $M$ then the subgroup $\Phi(N) M$ (it is indeed a subgroup by the above fact) properly contains $M.$ By maximality of $M$ it follows that $\Phi(N) M=G.$ Now we have \begin{align} N &=N \cap G \\ &= N \cap \Phi(N) M \\ &= \Phi(N) (N \cap M). \end{align} where the last equality follows from Dedekind's Lemma. Now since $\Phi(N)$ is the set of non-generators we have $N=N \cap M,$ therefore $N$ must be contained in $M.$ This contradicts with the fact that $\Phi(N)$ is not contained in $M.$ So we must have $\Phi(N) \leq \Phi(G).$

If you want you may try to find a counterexample showing that $\Phi(H)$ need not be a subgroup of $\Phi(G)$ if $H$ is not normal in $G.$