Equivalence of Definitions of Finite Cyclic Group

Theorem

The following definitions of the concept of Finite Cyclic Group are equivalent:

Definition $1$

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

Then $\struct {G, \circ}$ is a finite cyclic group if and only if it is a finite group.

Definition $2$

Let $\struct {G, \circ}$ be a cyclic group generated by $a \in G$.

Then $\struct {G, \circ}$ is a finite cyclic group if and only if:

$\exists n \in \N: a^n = e$

Proof

Definition $(1)$ implies Definition $(2)$

Let $\struct {G, \circ}$ be a finite cyclic group by definition $1$.

Then by definition, $\struct {G, \circ}$ is both cyclic and finite.

By definition of cyclic group:

$G = \gen a$

for some $a \in G$.

Hence by definition of group generator:

$\forall n \in \N: a^n \in G$

As $G$ is finite:

$\exists r, s \in \N: a^r = a^s$

such that $r \ne s$.

Without loss of generality, suppose $r > s$.

Then:

\(\ds a^r\)

\(=\)

\(\ds a^s\)

\(\ds \leadsto \ \ \)

\(\ds a^{r - s}\)

\(=\)

\(\ds a^{s - s}\)

\(\ds \leadsto \ \ \)

\(\ds a^{r - s}\)

\(=\)

\(\ds a^0\)

\(\ds \leadsto \ \ \)

\(\ds a^n\)

\(=\)

\(\ds e\)

where $n = r - s$ and $a^0 = e$ by definition

Thus $\struct {G, \circ}$ is a finite cyclic group by definition $2$.

$\Box$

Definition $(2)$ implies Definition $(1)$

Let $\struct {G, \circ}$ be a finite cyclic group by definition $2$.

Then, by definition:

$\exists n \in \N: a^n = e$

Again, by definition of cyclic group and group generator:

$\forall g \in G: \exists m \in \N: g = a^m$

By the Division Theorem:

$m = q n + r$

for some $q \in \N$ and $0 \le r < n$.

Thus:

\(\ds a^m\)

\(=\)

\(\ds a^{q n + r}\)

\(\ds \)

\(=\)

\(\ds \paren {a^n}^q a^r\)

\(\ds \)

\(=\)

\(\ds e^q a^r\)

\(\ds \)

\(=\)

\(\ds a^r\)

That is, every element of $G$ is in the form $a^r$ where $0 \le r < n$.

That is, there exist no more than $n$ elements in $G$.

Thus $\struct {G, \circ}$ is a finite cyclic group by definition $1$.

$\blacksquare$