True or false: Let $G$ be a finite group, and we'll denote $d(G)$ as the minimal size of the generating set of $G$. Let $Q$ be a quotient map of $G$, so it than it must be that $d(Q) \le d(G)$.
I believe that it is false – For abelian groups $G$ and subgroup $H\le G$, it is known that always $d(H) \le d(G)$. But for non abelian, it isn't always true -for example, $S_n$, and a subgroup of $S_n$ that is isomorphic to $H = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 $, have $d(S_n) = 2$ but $d(H) = 3$.
So my idea was to use that example ( $S_n$ and $H$), and the quotient group of that subgroup. But I'm not sure how to prove it formally?(Assuming it is correct).
Best Answer
Assume that $\{g_1,\ldots,g_n\}$ is a generating set for $G$ of the minimal size. If $Q$ is a quotient group $G/H$ for $H\triangleleft G$, note that $\{g_1H,\ldots,g_nH\}$ is a generating set for $Q$. Indeed, for every $x\in G$ you can write $x=s_1\ldots s_m$ where each $s_j\in \{g_i, g_i^{-1}\mid 1\leq i\leq n\}$, so $xH=(s_1H)\ldots(s_mH)$ and each $s_jH\in\{g_iH,g_i^{-1}H\mid 1\leq i\leq n\}$. So the minimal size of a generating set for $Q$, $d(Q)$, is less or equal to $n=d(G)$.