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 is a step on a bigger exercise to show that $G_{q^\alpha}$ is cyclic.
Someone gave me a hint to show that $\Psi$ is surjective, but how do I show that?
If for some $x \in \Bbb Z$, we have $\gcd(x,q)=1$, then $q$ does not divide $x$, so $x$ and $q^\alpha$ don't have any prime factor in common, thus $\gcd(x,q^\alpha)=1$, this implies that $\Psi$ is surjective.
Since you also tagged this with Galois theory, one can note that using the canonical isomorphisms $\mathrm{Gal}(\Bbb Q(\zeta_{q^\alpha})/\Bbb Q) \cong G_{q^{\alpha}}$ and $\mathrm{Gal}(\Bbb Q(\zeta_q)/\Bbb Q) \cong G_q$, one can identify $\Psi$ with the restriction map $\mathrm{Gal}(\Bbb Q(\zeta_{q^\alpha})/\Bbb Q) \to \mathrm{Gal}(\Bbb Q(\zeta_q)/\Bbb Q)$ which is surjective by general theory.
You don't really need the surjectivity of $\Psi$ to show that $|\ker \Psi|=q^{\alpha-1}$, though. You can just show that $1+qa$, where $a=0,1, \dots, q^{\alpha-1}-1$ is a system of representatives for $\ker \Psi$.