Generator of Additive Group Modulo m iff Unit of Ring

Theorem
Let $$m \in \Z: m > 1$$

Let $$\left({\Z_m, +_m}\right)$$ be the Additive Group of Integers Modulo $m$.

Let $$\left({\Z_m, +_m, \times_m}\right)$$‎ be the ring of integers modulo $m$.

Let $$a \in \Z_m$$.

Then:
 * $$a$$ is a generator of $$\left({\Z_m, +_m}\right)$$

iff
 * $$a$$ is a unit of $$\left({\Z_m, \times_m}\right)$$

Proof
From Integers Infinite Cyclic Group, the identity element $$1_\Z$$ of the ring $$\left({\Z, +, \times}\right)$$ is a generator of the group $$\left({\Z, +}\right)$$.

Thus from Quotient Group of Cyclic Group, the identity element $$1_{\Z_m}$$ of the ring $$\left({\Z_m, +_m, \times_m}\right)$$ is a generator of the group $$\left({\Z_m, +_m}\right)$$.

Let $$a \in \Z_m$$.

Suppose $$1_{\Z_m} \in \left\langle{a}\right\rangle$$, where $$\left\langle{a}\right\rangle$$ signifies the group generated by $$a$$.

Then the smallest subgroup of $$\left({\Z_m, +_m}\right)$$ containing $$1_{\Z_m}$$, i.e. $$\left({\Z_m, +_m}\right)$$ itself, is contained in $$\left\langle{a}\right\rangle$$.

Thus $$\left\langle{a}\right\rangle = \left({\Z_m, +_m}\right)$$ iff $$1_{\Z_m} \in \left\langle{a}\right\rangle$$.

However, from Subgroup of Additive Group Modulo m is Ideal of Ring, $$\left\langle{a}\right\rangle$$ is an ideal of $$\left({\Z_m, +_m, \times_m}\right)$$, and hence is the principal ideal $$\left({a}\right)$$ generated by $$a$$.

But from Principal Ideal from Element in Center of Ring, $$1_{\Z_m} \in \left\langle{a}\right\rangle$$ iff $$a$$ is a unit of the ring $$\left({\Z_m, +_m, \times_m}\right)$$.

Hence the result.