Consider the group $D_4 = \{e, r, r^2, r^3, t, rt, (r^2)t, (r^3)t\}$ with $r^4 = t^2 = e$ and $t(r^i) = (r^{4 - i})t$. Verify that $N = \{e, r^2\}$ is a normal subgroup.
Here is my answer: Let $G = \langle r^2, (r^2)t \rangle$. The elements of $G$ are in the form $((r^2)t)a_1 \cdot ((r^2)t)b_1\cdots ((r^2)t)a_1=k((r^2)t)b_k$, where $a_1, ..., a_k, b_1, ... b_k \in\mathbb Z$.
One can check that $(r^2)t \cdot r^2 = t$, and $(r^2)t \cdot (r^2)t= e$. So the expression above simplifies to an expression of the form $$(r^2x)t$$ for some $x \in \mathbb Z$.
Suppose $n$ is even. Then $n = 2m$ for some $m \in\mathbb Z$. Thus $r^n = (r^{2m}) = e$, so the powers of $r^2 $ are all the even powers of $r$ up to $r^{2m−1}$. Thus $$G = \{e,r^2,\cdots, r^{2m−1}, (r^2)t,\cdots ,r^{2m−1})t\}.$$ Now suppose $n$ is odd. Then $n = 2m + 1$ for some $m\in \mathbb Z$, and $r^{2m+1} = e$. Since $r^{2m+2}$ is a power of $r^2$ and $r^{2m+2} = r$, we have that $r$ is in $G$. And since $(r^2)t \cdot r^2 = s$, $s \in G$. But $r$ and $t$ generate all of $D_n$, so $G = D_n$.
Appreciate if someone can see if I'm doing it right, or the wrong way, thanks in advance.