Is every finite 2-generated group a quotient of a Frattini-free 2-generated group?

Question

Let $G$ be a finite 2-generated group. Is $G$ the quotient of a finite 2-generated group $G'$ with $\Phi(G') = 1$?, where $\Phi(G')$ is its Frattini subgroup.

Answer 1

Yes. here is a sketch of a proof. I can fill in any details of necessary.

Let $p$ be a prime not dividing $|G|$, and let $M_1,\ldots,M_r$ be a set of non-isomorphic irreducible modules for $G$ over ${\mathbb F}_p$ such that no element of $G$ acts trivially on every $M_i$. (So, if $G$ has a faithful irreducible module, then we can take $r=1$.)

Let $M=M_1 \oplus \cdots \oplus M_r$ be the direct sum of the $M_i$ and let $G'$ be the semidirect product $M \rtimes G$.

It is is not hard to show that the semidirect products $M_i \rtimes G$ are all $2$-generated (since otherwise $M_i$ would have at least $|M_i|^2$ distinct complements in $M_i \rtimes G$, but all such conjugates are conjugate, so it has at most $|M_i|$), and then the fact that the $M_i$ are mututally non-isomorphic implies that $G'$ is $2$-generated.

Let $1 \ne g \in G'$. We need to show that $g \not\in \Phi(G')$. If $g \in M$, then $g \not\in M_i$ for some $i$, say $i=r$, and then $g$ is not in the maximal subgroup $(M_1 \oplus \cdots \oplus M_{r-1}) \rtimes G$ of $G'$.

Otherwise, some power $g^k$ of $g$ has order coprime to $p$. We know that $g^k$ acts non-trivially on some $M_i$, say $M_r$, and by Schur-Zassenhaus $g^k$ is contained in some conjugate $H$ of $G$, and hence in the maximal subgroup $(M_1 \oplus \cdots \oplus M_{r-1}) \rtimes H$ of $G'$. By conjugating this maximal subgroup by an element of $M_r$ that does not centralize $g^k$, we get a maximal subgroup of $G'$ not containing $g^k$ and hence not containing $g$.