Multiplicative Group of Integers Modulo m

Theorem

Let $\Z_m$ be the set of integers modulo $m$.

Let $\Z'_m$ be the set of integers coprime to $m$ in $\Z_m$.

Then the structure $\left({\Z'_m, \times}\right)$ is an abelian group.

Corollary

Let $p$ be a prime number.

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

Let $\Z'_p = \Z_p \setminus \left\{{\left[\!\left[{0}\right]\!\right]}\right\}$ be the set of non-zero residue classes modulo $p$.

Then the structure $\left({\Z'_p, \times}\right)$ is an abelian group.

Proof

We have that the structure $\left({\Z'_m, +, \times}\right)$‎ forms a (commutative) ring with unity.

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

We also have that all the elements of $\left({\Z'_m, \times}\right)$ have inverses, and are therefore units.

The fact that $\left({\Z'_m, \times}\right)$ is abelian follows from Restriction of Operation Commutativity.

$\blacksquare$

Proof of Corollary

Suppose $p \in \Z$ be a prime number.

From the definition of Set of Coprime Integers, as $p$ is prime, $\Z'_p$ becomes $\left\{{\left[\!\left[{1}\right]\!\right]_p, \left[\!\left[{2}\right]\!\right]_p, \ldots, \left[\!\left[{p-1}\right]\!\right]_p}\right\}$.

This is precisely $\Z_p \setminus \left\{{\left[\!\left[{0}\right]\!\right]_p}\right\}$ which is what we wanted to show.

$\blacksquare$

Sources

• Iain T. Adamson: Introduction to Field Theory (1964)... (previous)... (next): $\S 1.1$: Example $4$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 34$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 34 \ (3)$