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

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?

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.

• Cool. Thank you. – LWW Apr 8 '18 at 21:47
• Since the problem was with the special role of the trivial subgroup, it might be worth to point out that the trivial group is an exception to your statements: it has no maximal subgroup and is its own Frattini. – verret Apr 8 '18 at 23:37
• @verret Right, I’ll do it. – egreg Apr 9 '18 at 8:02

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

• Oh, is the definition 'maximal' contains trivial case? Thank you. – LWW Apr 8 '18 at 21:25