Finite Cyclic Group is Isomorphic to Integers under Modulo Addition

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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 $n$ $\struct {\Z_n, +_n}$.


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.


\(\ds \map \phi {a^r \circ a^s}\) \(=\) \(\ds \map \phi {a^{r + s} }\) 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.


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 $\eqclass k n \in \Z_n$.


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

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

So by definition $G$ is cyclic.