The unit group $(\mathbb{Z}\mathrm{/}n{)}^{\times }$ and Euler's totient $\phi (n)=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid$ for $n\ge 1$

Let $n\ge 1$ be an integer. Multiplication makes $\mathbb{Z}\mathrm{/}n$ a commutative monoid with identity $[1{]}_n$ by For every natural $n$, $(\mathbb{Z}\mathrm{/}n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold. A class $u\in \mathbb{Z}\mathrm{/}n$ is a unit when it is invertible in that monoid (Left inverse, right inverse, and invertible element of a monoid). The set of all units is

$(\mathbb{Z}\mathrm{/}n{)}^{\times }:=\{u\in \mathbb{Z}\mathrm{/}n:\text{some }v\in \mathbb{Z}\mathrm{/}n\text{ satisfies }uv=[1{]}_n\}\mathrm{.}$

By The invertible elements of a monoid form a group under the restricted operation, multiplication restricts to a group operation on $(\mathbb{Z}\mathrm{/}n{)}^{\times }$, called the unit group modulo $n$.

The quotient $\mathbb{Z}\mathrm{/}n$ is finite with cardinality $n$ by For $n\ge 1$, every class in $\mathbb{Z}\mathrm{/}n$ has one representative $r$ with $0\le r<n$, so $\mid \mathbb{Z}\mathrm{/}n\mid =n$; while $\mathbb{Z}\mathrm{/}0$ is in bijection with $\mathbb{Z}$, and its unit set is a finite subset by A subset of a finite set is finite, with $\mid B\mid \le \mid A\mid$, and equality holds if and only if $B=A$. Euler's totient function is therefore defined for every positive integer $n$ by

$\phi (n):=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid \in \mathbb{N}$

(The cardinality $\mid A\mid$ of a finite set). For $n=1$, the quotient has one element, which is its multiplicative identity and hence a unit, so $\phi (1)=1$ follows from the definition.

• The domain of $\phi$ here is the positive integers. No value $\phi (0)$ is defined.
• The one-element multiplicative monoid is a group, even though its identity is also its additive zero. It is not a field because a field requires distinct elements $0$ and $1$ (Field).