Vtyjkyuk

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 $Hle G$, 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 = mathbbZ_2 times mathbbZ_2 times mathbbZ_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).

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 $Htriangleleft G$, note that $g_1H,ldots,g_nH$ is a generating set for $Q$. Indeed, for every $xin G$ you can write $x=s_1ldots s_m$ where each $s_jin g_i, g_i^-1mid 1leq ileq n$, so $xH=(s_1H)ldots(s_mH)$ and each $s_jHing_iH,g_i^-1Hmid 1leq ileq n$. So the minimal size of a generating set for $Q$, $d(Q)$, is less or equal to $n=d(G)$.

Strong hint: Let $G=langle g_1,ldots,g_drangle$ where $d=d(G)$. If $Q=G/N$ then what is $langle g_1N,ldots,g_dNrangle$ (as a subgroup of $Q$)?

Your mistake is that $H$ is a subgroup not a quotient.