Multiplicative Inverse in Monoid of Integers Modulo m

Theorem
Let $\left({\Z_m, \times_m}\right)$ be the multiplicative monoid of integers modulo $m$.

Then:
 * $\left[\!\left[{k}\right]\!\right]_m \in \Z_m$ has an inverse in $\left({\Z_m, \times_m}\right)$


 * $k \perp m$
 * $k \perp m$

Proof
First, suppose $k \perp m$.

That is:
 * $\gcd \left\{{k, m}\right\} = 1$

By Bézout's Identity:
 * $\exists u, v \in \Z: u k + v m = 1$

Thus:
 * $\left[\!\left[{u k + v m}\right]\!\right]_m = \left[\!\left[{u k}\right]\!\right]_m = \left[\!\left[{u}\right]\!\right]_m \left[\!\left[{k}\right]\!\right]_m = \left[\!\left[{1}\right]\!\right]_m$

Thus:
 * $\left[\!\left[{u}\right]\!\right]_m$ is an inverse of $\left[\!\left[{k}\right]\!\right]_m$

Suppose that:
 * $\exists u \in \Z: \left[\!\left[{u}\right]\!\right]_m \left[\!\left[{k}\right]\!\right]_m = \left[\!\left[{u k}\right]\!\right]_m = 1$

Then:
 * $u k \equiv 1 \pmod m$

and:
 * $\exists v \in \Z: u k + v m = 1$

Thus from Bézout's Identity, $k \perp m$.