The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$

Let $n\ge1$ be an integer. Multiplication makes $\mathbb Z/n$ a commutative monoid with identity $[1]_n$ by For every natural $n$, $(\mathbb{Z}/n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold. A class $u\in\mathbb Z/n$ 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

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

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

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

$\varphi(n):=\big|(\mathbb Z/n)^\times\big|\in\mathbb N$

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

• The domain of $\varphi$ here is the positive integers. No value $\varphi(0)$ 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).