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:

$r, s \perp m$
$\exists u_1, v_1 \in \Z: u_1 r + v_1 m = 1$
$\exists u_2, v_2 \in \Z: u_2 s + v_2 m = 1$

Then:

 $\displaystyle \paren {u_1 r + v_1 m} \paren {u_2 s + v_2 m}$ $=$ $\displaystyle u_1 u_2 r s + v_1 u_2 s m + u_1 v_2 r m + v_1 v_2 m^2$ $\displaystyle$ $=$ $\displaystyle \paren ({u_1 u_2} r s + \paren {v_1 u_2 s + u_1 v_2 r + v_1 v_2 m} m$ $\displaystyle$ $=$ $\displaystyle 1$

So, again by Bézout's Lemma, $r s$ 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.

$\blacksquare$