Cyclic Group 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 $\left({\Z_n, +_n}\right)$.

Proof

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \phi \left({a^r \circ a^s}\right)\)	\(=\)	\(\displaystyle \phi \left({a^{r + s} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Powers of Group Elements: Sum of Indices
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{r + s}\right]\!\right]_n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $\phi$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{r}\right]\!\right]_n +_n \left[\!\left[{s}\right]\!\right]_n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of modulo addition
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({a^r}\right) \circ \phi \left({a^s}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by 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 $a = \phi \left({\left[\!\left[{1}\right]\!\right]_n}\right)$.

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g\)	\(=\)	\(\displaystyle \phi \left({\left[\!\left[{k}\right]\!\right]_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({\left[\!\left[{1}\right]\!\right]_n +_n \cdots (k) \cdots +_n \left[\!\left[{1}\right]\!\right]_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

So by definition $G$ is cyclic.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 63$