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 $m$:


 * $\Z'_m = \left\{{\left[\!\left[{k}\right]\!\right]_m \in \Z_m: k \perp m}\right\}$

Let $S = \left({\Z'_m, \times_m}\right)$ 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 $\left[\!\left[{r}\right]\!\right]_m, \left[\!\left[{s}\right]\!\right]_m \in \Z'_m$.

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

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 $\left({\Z'_m, \times_m}\right)$.

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

Proof

 * : $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.31)$