Finite Cyclic Group is Isomorphic to Integers under Modulo Addition

Theorem
Let $\left({G, \circ}\right)$ be a finite group whose identity element is $e$.

Then $\left({G, \circ}\right)$ is cyclic of order $n$ iff $\left({G, \circ}\right)$ is isomorphic with the additive group of integers modulo n $\left({\Z_n, +_n}\right)$.

Necessary Condition
Let $\left({G, \circ}\right)$ be a cyclic group of order $n$.

From List of Elements in Finite Cyclic Group:
 * $G = \left\{{a^0, a^1, a^2, \ldots, a^n}\right\}$

where $a^0= e, a^1 = a$.

From the definition of integers modulo n, $\Z_n$ can be expressed as:


 * $\displaystyle \Z_n = \left\{{\left[\!\left[{0}\right]\!\right]_n, \left[\!\left[{1}\right]\!\right]_n, \ldots, \left[\!\left[{n-1}\right]\!\right]_n}\right\}$

where $\left[\!\left[{x}\right]\!\right]_n$ is the residue class of $x$ modulo $n$.

Let $\phi: G \to \Z_n$ be the mapping defined as:
 * $\forall k \in \left\{{0, 1, \ldots, n-1}\right\}: \phi \left({a^k}\right) = \left[\!\left[{k}\right]\!\right]_n$

By its definition it is clear that $\phi$ is a bijection.

Also:

Thus the morphism property of $\phi$ is demonstrated, and $\phi$ is thus a group homomorphism.

By definition, a group isomorphism is a group homomorphism which is also a bijection.

Sufficient Condition
Now suppose $G$ is a group such that $\phi: \Z_n \to G$ is a group isomorphism.

Let $a = \phi \left({\left[\!\left[{1}\right]\!\right]_n}\right)$.

Let $g \in G$.

Then $g = \phi \left({\left[\!\left[{k}\right]\!\right]_n}\right)$ for some $\left[\!\left[{k}\right]\!\right]_n \in \Z_n$.

Therefore:

So every element of $G$ is a power of $a$.

So by definition $G$ is cyclic.