The question below is Exercise $1.12$ from here. I think I have an example that does not require $G$ to be generated by two elements. I do not know what the question is trying to ask if my solution is correct.
Let $G = \mathbb{Z}, H = 2\mathbb{Z}$. Then, $H \unlhd G$, and $H = \langle2\rangle$, so H is cyclic. Similarly, $G/H = \{2\mathbb{Z}, 1+2\mathbb{Z} \}$, which is also cyclic since it is an order $2$ group. But clearly, $G = \langle1\rangle$.
Is my solution wrong?
They mean "Prove that you can always find two elements $g,h\in G$ such that $G=\langle g,h\rangle$". There is no requirement that this is optimal. Cyclic groups can also be generated by two elements if you want. And clearly we have $G=\langle 2,3\rangle$ in your case.