Let $X,Y \subset G$ and suppose that $K = \langle X \rangle$ is normal in $G$. Let $q: G → G/K$ be the quotient map. Show that $G = \langle X \cup Y \rangle$ iff $G/K = \langle q(Y)\rangle$.
I'm trying to look at how each element in $G$ is expressed and then mapped to $G/K$ and then see how that that could be generated by $q(Y)$. What I did was develop three cases for each element in $G$. 1. it could be the product of some finite number of elements in $X$, 2. it could be the product of some finite number of elements in $Y$ or 3. it could be the product of some finite number of elements in $X$ and some finite number of elements in Y. I can’t see how an element that is in the case 1 or 2 could be generated by $q(Y)$? Are these three cases correct? Like for G/K to be generated by $q(Y)$ wouldn’t $Y$ have to contain the identity element? But I don't know that it does.
$\Longrightarrow$: Suppose that $G=\langle X\cup Y\rangle$. Let $g\in G/K$; you want to prove that $g=q(h)$ for some $h\in\langle Y\rangle$. Since $g\in G/K$, $g=g'K$, for some $g'\in G$. Since $G=\langle X\cup Y\rangle$, $g'$ can be written as $x_1y_1x_2y_2\ldots x_my_m$, where each $x_i$ is either an element of $X$ or the inverse of an element of $X$ and each $y_i$ is either an element of $Y$ or the inverse of an element of $Y$ (actually, $x_1$, $y_m$ or both can be missing). But then$$\tag{1}g=x_1y_1x_2y_2\ldots x_my_mK.$$Note that$$x_my_mK=y_m{y_m}^{-1}x_my_mK\subset y_mK.$$So, one can deduce by induction from $(1)$ that $gK\in\bigl\langle q(Y)\big\rangle$.
$\Longleftarrow$: Let $g\in G$. Then $q(g)\in G/K=\bigl\langle q(Y)\bigr\rangle$ and therefore $q(g$) can be expressed as $y_1y_2\ldots y_mK$, where each $y_i$ is an element of $Y$ or the inverse of an element of $Y$. But then $g=y_1y_2\ldots y_mx_1x_2\ldots x_m$, where each $x_i$ is an element of $X$ or the inverse of an element of $X$. Therefore, $g\in\langle X\cup Y\rangle$.