Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.16
Let $G$ be a group and $N \leq G$ a normal subgroup. Show that if $G = \langle S \rangle$, then $G/N = \langle \overline{S} \rangle$.
Solution: Let $x \in G/N$. Since $G = \langle S \rangle$, we have $x = s_1^{a_1}s_2^{a_2} \ldots s_k^{a_k}$ for some $s_i \in S$ and $a_i \in \mathbb{Z}$. Then $$xN = (s_1^{a_1}s_2^{a_2} \ldots s_k^{a_k})N = (s_1N)^{a_1}(s_2N)^{a_2} \ldots (s_kN)^{a_k} \in \langle \overline{S} \rangle.$$
