Minimum number of generators of $G$ and $G / \Phi(G)$ are equal

Question

Let $G$ be a finite group, and denote $\mathrm{d}(G)$ as the minimum number of generators of $G$. Prove that $\mathrm{d}(G) = \mathrm{d}\left(G/\Phi(G)\right)$ (where $\Phi(G$) denotes the Frattini subgroup of G)

Any hints for how to approach this problem are appreciated.

Answer 1

Clearly $d(G)\ge d(G/\Phi(G))$, since if $d(G)$ elements generate $G$, the images of those elements generate $G/\Phi(G)$. The goal then is to show that if $G/\Phi(G)$ can be generated by $n$ elements, then so can $G$.

Hint: Let $g_1,\ldots,g_n\in G$ such that their images generate $G/\Phi(G)$. Then $\{g_1,\ldots,g_n\}\cup \Phi(G)$ generates $G$. Can you prove that if $x\in \Phi(G)$, and $A$ is a set of generators for $G$, then $A\setminus \{x\}$ is still a set of generators for $G$? If so, then we know that we can remove $\Phi(G)$ from the list of generators and still have a generating set. Hence $g_1,\ldots, g_n$ generate $G$, so $d(G)\le d(G/\Phi(G))$ as well.

Solution:

Let $B=A\setminus\{x\}$, and let $H=\langle B\rangle$ be the subgroup generated by $B$. If $H\ne G$, then $H$ is contained in some maximal subgroup $M$ of $G$, but then $x\in M$ by definition of the Frattini subgroup, so $A\subset M$. Hence $A$ cannot generate $G$, contradiction. Therefore we must have $H=G$ as desired.