Modulo Multiplication on Reduced Residue System is Closed

Theorem
Let $m \in \Z_{> 0}$ be a (strictly) positive integer.

Let $\Z'_m$ be the reduced residue system modulo $m$:


 * $\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$

Let $S = \struct {\Z'_m, \times_m}$ be the algebraic structure consisting of $\Z'_m$ under the modulo multiplication.

Then $S$ is closed, in the sense that:
 * $\forall a, b \in \Z'_m: a \times_m b \in \Z'_m$

Proof
Let $\eqclass r m, \eqclass s m \in \Z'_m$.

Then by definition of reduced residue system:
 * $r, s \perp m$

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

Then:

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

So the product of two elements of $\struct {\Z'_m, \times_m}$ is again in $\struct {\Z'_m, \times_m}$.

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

Proof

 * : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.31)$