Consider $G_n=\{ \overline{x}\in\mathbb{Z}/n\mathbb{Z}; \gcd(x,n)=1 \}$, and $q$ an odd prime number.
For all $x\in \mathbb{Z}$, $\overline{x}$ represents, in this question, the class of $x$ modulo $q^\alpha$, and let's use the notation $\dot{x}$ for the class of $x$ modulo $q$ and consider the morphism of groups \begin{align*} \Psi : G_{q^\alpha} &\rightarrow G_q \\ \overline{x}&\mapsto \dot{x} \end{align*} Prove that $|\operatorname{Ker}\Psi|=q^{\alpha-1}$ (with $|\operatorname{Ker}\Psi|$ the notation for order).
This question is a step by step to prove that $G_{q^\alpha}$ is cyclic, but I can't solve this part.
If you can show that the above group homomorphism is surjective, then you have $G_q \cong G_{q^\alpha} / \ker \Psi$.
Now, $|G_q| = q-1$ and $|G_{q^\alpha}| = q^{\alpha-1}(q-1)$. Hence $|\ker \psi| = q^{\alpha-1}$.