# 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 $n$.

From List of Elements in Finite Cyclic Group:

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

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.

$\blacksquare$

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): Chapter $\text{II}$: Groups: A Little Number Theory - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 63$