Multiplicativity of the totient function

Statement

Let $m$ and $n$ be positive integers that are coprime. Then $\varphi(m n) = \varphi(m) \varphi(n)$, where $\varphi$ is Euler's totient function.

Remarks

The proof uses the Chinese remainder theorem. It gives a bijection between the residues modulo $m n$ and the pairs of residues modulo $m$ and modulo $n$. A residue is coprime to $m n$ if and only if both parts are coprime to their moduli. The hypothesis is necessary: $\varphi(4) = 2$, but $\varphi(2) \varphi(2) = 1$. So $\varphi$ is a multiplicative function. For example, $\varphi(36) = \varphi(4) \varphi(9) = 2 \cdot 6 = 12$.

Explanations

L1 human

When $m$ and $n$ have no common factor, a number is coprime to $m n$ exactly when it is coprime to $m$ and to $n$. The two conditions are independent, so the counts multiply. For example, $\varphi(15) = \varphi(3) \varphi(5) = 2 \cdot 4 = 8$.