Show that $\pi$ is onto.

Question

Let $q_1\mid q$. Consider the group $U_{q_1},U_{q}$ where $U_q=\{x\in \Bbb Z_q:\gcd(x,q)=1\}$ and similarly $U_{q_1}$ is defined.

Define a map $\pi :U_q\to U_{q_1}$ by $\pi(x)$ to be that unique element in $U_{q_1}$ such that $\pi(a)=a\mod q_1$

Show that $\pi$ is onto.

My try:

Let us suppose that $\pi$ is not onto. Then there exists $y\in U_{q_1}$ such that $\pi (x)\equiv y \mod q_1$ does not hold for any $x\in U_q$.

Now $\pi (x)=x\mod q_1\implies x_1\equiv y$ does not hold for any $x\in U_q$.

Now since $q_1\mid q,x\in U_q\implies x\in U_{q_1}$ for $x<q_1$.

But I am unable to complete the proof.

Can I get some help please.

Answer 1

Indeed, do it constructively.

Take an $x'\in U_{q_1}$, that is, $(x',q_1)=1$. Let $\mathcal{P}$ be the set of primes, dividing $q$ and coprime to $q_1$. Take an $x$, such that, $x\equiv x'\pmod{q_1}$, and $x\equiv 1\pmod{p}$, for each $p\in\mathcal{P}$. Such an $x$ exists due to Chinese remainder theorem, and is clearly coprime to $q$, and can be made less than $q$, and thus it corresponds (in mod $q$) to a number, belonging to $U_q$.