Let $G$ be a group, and $A$, $B$ be subsets of $G$. Let $⟪A⟫$ mean the smallest subgroup containing $A$.
Prove that $G/⟪A\cup B⟫$ is isomorphic to $\bigl(G/⟪A⟫\bigr)/⟪p(B)⟫$ where $p:G\to G/⟪A⟫$, $p(g)=g⟪A⟫$.
I do not know how to write solution properly for this. It seems to be quite obvious that this is true, but I have problem with writing proper argument using basic definition of quotient groups.
I'll work under the assumption that $\newcommand{\minnorm}[1]{⟪#1⟫}\minnorm{A}$ denotes the smallest normal subgroup containing $A$. Clearly such a subgroup exists and also $\minnorm{A}\subseteq\minnorm{A\cup B}$.
Consider the canonical projection $q\colon G\to G/\minnorm{A\cup B}$; since its kernel contains $\minnorm{A}$, there is a unique homomorphism $f\colon G/\minnorm{A}\to G/\minnorm{A\cup B}$ such that $q=f\circ p$.
We want to show that the kernel of $f$ is $\minnorm{p(B)}$ so the homomorphism theorem yields the result you want to prove. Note that the kernel of $f$ is $\minnorm{A\cup B}/\minnorm{A}$ by construction.
A normal subgroup of $G/\minnorm{A}$ has the form $K/\minnorm{A}$, where $K$ is a uniquely determined normal subgroup of $G$ containing $\minnorm{A}$.
Note that $B\subseteq\ker q$, so $p(B)\subseteq\ker f$ and therefore $\minnorm{p(B)}\subseteq\ker f$.
Suppose $K/\minnorm{A}$ is a normal subgroup of $G/\minnorm{A}$ such that $p(B)\subseteq K/\minnorm{A}$. We want to show that $\ker f\subseteq K/\minnorm{A}$, which is the same as saying that $\minnorm{A\cup B}\subseteq K$ or, equivalently, that $A\cup B\subseteq K$.
Now it's just checking the definitions.