Show $U_t(st)$ isomorphic to $U(s)$.
I've shown that the mapping $x \rightarrow x \pmod{s}$ is an injection of $U_t(st)$ into $U(s)$ where $U_t(st)=\{x\in U(st) \mid x \pmod{t}=1\}$ and $s$ and $t$ are coprime. I now want to show that this mapping is also a surjection by showing that $|U_t(st)|=|U(s)|$. I know that $|U(s)|=\phi(s)$, and I know $|U(st)|=\phi(s)\phi(t)$, but I’m not sure what else to do.
Any help would be appreciated.
We will show by definition that the map $x \mapsto x \bmod{s}$ is surjective.
Let $y \in U(s)$. We want to find an $x \in U_t(st)$ such that $x$ should satisfy the following 3 properties:
For $y \in U(s)$, now we construct such an $x$.
Since $\gcd(s,t)=1$, so there $\exists \, a,b \in \Bbb{Z}$ such that $$as+bt=1 \tag{1}$$ Let us define, $$\color{magenta}{x':=y+sa(1-y)} \quad \text{ and } \quad \color{magenta}{x'=m(st)+x}, \quad \text{where }0 \leq x <st. $$ Then, $$\color{blue}{x' \equiv y \pmod{s}}. \tag{2}$$ Moreover we have \begin{align*} x'&=y+sa(1-y)\\ &=y+(1-bt)(1-y) & (\text{ from equ (1)})\\ &=1+t(by-b)\\ \color{blue}{x'} & \color{blue}{\equiv 1 \pmod{t}} \tag{3}. \end{align*} This means $\color{blue}{\gcd(x',t)=1}$. Suppose a prime $p \mid \gcd(x',s)$, then $y'$ being a linear combination of $x'$ and $s$ implies that $p \mid \gcd(y,s)$. But we know $y \in U(s)$, so $\gcd(y,s)=1$. Thus we have $\color{blue}{\gcd(x',s)=1}$ as well. Since $s$ and $t$ are relatively prime, so we can also claim that $$\color{blue}{\gcd(x', st)=1}.$$ But $x$ being a linear combination of $x'$ and $st$ implies that $\color{blue}{\gcd(x,st)=1}$ as well. So we got our condition 1. From equation (3) and the definition of $x$, we also get $\color{blue}{x \equiv 1 \pmod{t}}$. So we have our condition 2 as well. Lastly from the definition of $x$ and equation (2), we also get that $x \equiv y \pmod{s}$. This satisfies condition 3. So we have found our $x$. Thus the map is surjective.