It is easy to show that in general $\phi(m n) \neq \phi(m) \phi(n)$ for general $m, n$, but what is remarkable is that when $\operatorname{gcd}(m, n)=1$, $\phi(m n)=\phi(m) \phi(n)$. The function $\phi$ is an example of a multiplicative function in number theory. Perhaps more surprising is that this is a direct consequence of the Chinese Remainder Theorem. Give a proof that $\phi$ is multiplicative using the following idea. Suppose that $m, n \geq 2$ and $\operatorname{god}(m, n)=1$. Show that there is a bijection between the sets $U_{m n}$ and $U_m \times U_n$ (ordered pairs $(a, b)$ with $a \in U_m$, $\left.b \in U_n\right)$. Note that $U_{m n}$ has cardinality $\phi(m n)$ and $U_m \times U_n$ has cardinality $\phi(m) \cdot \phi(n)$. To establish the bijection, define a map $F: U_{m n} \rightarrow U_m \times U_n$ by $F\left([a]_{m n}\right)=\left([a]_m,[a]_n\right)$. You need to show this map is well-defined, one-to-one, and onto. Then deduce the result.
Some of these words may be new to you, so here are some definitions.
- We have encountered the term "well-defined" before. In this context it means that if $[a]_{m n}=[b]_{m n}$, then $F([a])=$ $F([b])$.
- The map $F$ is one-to-one (injective) if $F([a])=F([b])$ implies $[a]_{m n}=[b]_{m n}$.
- The map $F$ is onto (surjective) if given $\left([b]_m,[c]_n\right) \in U_m \times$ $U_n$, there exists $[a]_{m n} \in U_{m n}$ so that $F([a])=([b],[c])$.
- A map is bijective if it is one-to-one and onto.
- If $f: S \rightarrow T$ is a bijection, then $S$ and $T$ are said to have the same cardinality (size), and the result you are to prove is simply that when $\operatorname{gcd}(m, n)=1$, the size of $U_{m n}$ and $U_m \times U_n$ is the same.