Reduced Residue System under Multiplication forms Abelian Group/Proof 2

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, precisely equal to the group of units of $\Z_m$.

Proof
Taking the group axioms in turn:

G0: Closure
Let $r, s \in \Z'_m$.

By Bézout's Lemma:
 * $\exists u_1, v_1 \in \Z: u_1 r + v_1 m = 1$
 * $\exists u_2, v_2 \in \Z: u_2 s + v_2 m = 1$

Then:

So, again by Bézout's Lemma, $r s$ is coprime to $m$.

So the product of two elements of $\left({\Z'_m, \times}\right)$ is again in $\left({\Z'_m, \times}\right)$.

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

G1: Associativity
We have that Modulo Multiplication is Associative.

G2: Identity
From Modulo Multiplication has Identity, $\left[\!\left[{1}\right]\!\right]_m$ is the identity element of $\left({\Z'_m, \times}\right)$.

G3: Inverses
From Multiplicative Inverse in Monoid of Integers Modulo m, $\left[\!\left[{k}\right]\!\right]_m \in \Z_m$ has an inverse in $\left({\Z_m, \times_m}\right)$ iff $k$ is coprime to $m$.

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

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