# Finite Cyclic Group is Isomorphic to Integers under Modulo Addition

## Theorem

Let $\struct {G, \circ}$ be a finite group whose identity element is $e$.

Then $\struct {G, \circ}$ is cyclic of order $n$ if and only if $\struct {G, \circ}$ is isomorphic with the additive group of integers modulo $n$ $\struct {\Z_n, +_n}$.

## Proof

### Necessary Condition

Let $\struct {G, \circ}$ be a cyclic group of order $n$.

$G = \set {a^0, a^1, a^2, \ldots, a^n}$

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

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

$\Z_n = \set {\eqclass 0 n, \eqclass 1 n, \ldots, \eqclass {n - 1} n}$

where $\eqclass x n$ is the residue class of $x$ modulo $n$.

Let $\phi: G \to \Z_n$ be the mapping defined as:

$\forall k \in \set {0, 1, \ldots, n - 1}: \map \phi {a^k} = \eqclass k n$

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

Also:

 $\ds \map \phi {a^r \circ a^s}$ $=$ $\ds \map \phi {a^{r + s} }$ Powers of Group Elements: Sum of Indices $\ds$ $=$ $\ds \eqclass {r + s} n$ Definition of $\phi$ $\ds$ $=$ $\ds \eqclass r n +_n \eqclass s n$ Definition of Modulo Addition $\ds$ $=$ $\ds \map \phi {a^r} +_n \map \phi {a^s}$ Definition of $\phi$

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.

$\Box$

### Sufficient Condition

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

Let $a = \map \phi {\eqclass 1 n}$.

Let $g \in G$.

Then $g = \map \phi {\eqclass k n}$ for some $\eqclass k n \in \Z_n$.

Therefore:

 $\ds g$ $=$ $\ds \map \phi {\eqclass k n}$ $\ds$ $=$ $\ds \map \phi {\eqclass 1 n +_n \cdots (k) \cdots +_n \eqclass 1 n}$ $\ds$ $=$ $\ds a^k$

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

So by definition $G$ is cyclic.

$\blacksquare$