Reduced Residue System under Multiplication forms Abelian Group

Theorem

Let $\Z_m$ be the set of set of residue classes modulo $m$.

Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$.

Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$.

Corollary

Let $p$ be a prime number.

Let $\Z_p$ be the set of integers modulo $p$.

Let $\struct {\Z'_p, \times}$ denote the multiplicative group of reduced residues modulo $p$.

Then $\struct {\Z'_p, \times}$ is an abelian group.

Proof 1

From Ring of Integers Modulo m is Ring, $\struct {\Z_m, +, \times}$‎ forms a (commutative) ring with unity.

Then we have that the units of a ring with unity form a group.

By Multiplicative Inverse in Ring of Integers Modulo m we have that the elements of $\struct {\Z'_m, \times}$ are precisely those that have inverses, and are therefore the units of $\struct {\Z_m, +, \times}$‎.

The fact that $\struct {\Z'_m, \times}$ is abelian follows from Restriction of Commutative Operation is Commutative.

$\blacksquare$

Proof 2

Taking the group axioms in turn:

G0: Closure

$\left({\Z'_m, \times}\right)$ is closed.

$\Box$

G1: Associativity

We have that Modulo Multiplication is Associative.

$\Box$

G2: Identity

From Modulo Multiplication has Identity, $\left[\!\left[{1}\right]\!\right]_m$ is the identity element of $\left({\Z'_m, \times}\right)$.

$\Box$

G3: Inverses

From Multiplicative Inverse in Monoid of Integers Modulo m, $\left[\!\left[{k}\right]\!\right]_m \in \Z_m$ has an inverse in $\left({\Z_m, \times_m}\right)$ if and only if $k$ is coprime to $m$.

Thus every element of $\left({\Z'_m, \times}\right)$ has an inverse.

$\Box$

All the group axioms are thus seen to be fulfilled, and so $\left({\Z'_m, \times}\right)$ is a group.

$\blacksquare$

Proof 3

Taking the finite group axioms in turn:

FG0: Closure

$\struct {\Z'_m, \times}$ is closed.

$\Box$

FG1: Associativity

We have that Modulo Multiplication is Associative.

$\Box$

FG2: Finiteness

The order of $\struct {\Z'_m, \times}$ is $\map \phi n$ by definition, where $\map \phi n$ denotes the Euler $\phi$ function.

As $\map \phi n < n$ it follows that $\struct {\Z'_m, \times}$ is of finite order.

$\Box$

FG3: Cancellability

$\Box$

Thus all the finite group axioms are fulfilled, and $\struct {\Z'_m, \times}$ is a group.

It remains to note that Modulo Multiplication is Commutative to confirm that $\struct {\Z'_m, \times}$ is abelian.

$\blacksquare$