Integer has Multiplicative Order Modulo n iff Coprime to n

Theorem
Let $a$ and $n$ be integers.

Let the multiplicative order of $a$ modulo $n$ exist.

Then $a \perp n$, that is, $a$ and $n$ are coprime.

Necessary Condition
Suppose $c \in \Z_{>0}$ is the multiplicative order of $a$ modulo $n$.

Then by definition:


 * $a^c \equiv 1 \pmod n$

Hence, by definition, $a^c = k n + 1$ for some $k \in \Z$.

Thus:


 * $a r + n s = 1$

where $r = a^{c-1}$ and $s = -k$.

It follows from Integer Combination of Coprime Integers that $a$ and $n$ are coprime.

Sufficient Condition
Suppose $a \perp n$.

Then by Euler's Theorem:
 * $a^{\phi \left({n}\right)} \equiv 1 \pmod n$

where $\phi \left({n}\right)$ is the Euler Phi Function of $n$.

Hence the multiplicative order of $a$ modulo $n$ exists, by taking $c = \phi \left({n}\right)$.