Frattini subgroup of a quotient

Question

Let $G$ be a finite group. The Frattini subgroup $\Phi(G)$ is the intersection all proper maximal subgroups.

If $K \lhd G$ is a normal subgroup, then it is easy to see that $\Phi(G) K/K \leq \Phi(G/K)$ as subgroups of $\Phi(G/K)$.

Question: What's an example of a finite $G$ such that this inclusion is strict; that is, such that $\Phi(G) K/K \neq \Phi(G/K)$?

Answer 1

As written in the comments, the $F_5$ group given by the presentation

$F_5 := \langle a,b \ | \ a^5=b^4=1, bab^{-1}=a^3 \rangle$

is a counterexample. It has a trivial Frattini subgroup, but it has the cyclic group $C_{4}$ as a quotient, which has a non-trivial Frattini subgroup given by $2C_{4} \leq C_{4}$.

Answer 2

Another counter-example: let $n\ge2$ and $q$ be a prime power. Then $\mathrm{PGL}_n(q)$ has trivial Frattini subgroup. Indeed, the subgroup of $\mathrm{GL}_n(q)$ of matrices of the form $\begin{pmatrix}g&x\\0&a\end{pmatrix}$ where $g\in\mathrm{GL}_{n-1}(q)$, $x\in\mathbb F_q^n$, and $a\in\mathbb F_q^\times$ (i.e., the stabilizer of a line in $\mathbb F_q^n$) form a maximal subgroup. But the intersection of its conjugates consists of matrices $A\in\mathrm{GL}_n(q)$ which preserves every line in $\mathbb F_q^n$, which must be scalar multiplication. But scalars are killed under the surjection $\mathrm{GL}_n(q)\to\mathrm{PGL}_n(q)$.

On the other hand, $\mathrm{PGL}_n(q)$ has a quotient $\mathbb F_q^\times/(\mathbb F_q^\times)^n$, descended from $\det\colon\mathrm{GL}_n(q)\to\mathbb F_q^\times$. With a clever choice of $q$ and $n$ (e.g., $q=5$ and $n=4$), $\mathbb F_q^\times/(\mathbb F_q^\times)^n$ will have non-trivial Frattini subgroup.