# Isomorphism of Cyclic Groups

Theorem 1:
Cyclic groups of the same order are isomorphic.

Proof: Let $G$ and$\,G’$ be two cyclic groups of order $n$, which are generated by $a$ and $b$ respectively. Then
$G = \left\{ {a,{a^2},{a^3}, \ldots ,{a^n} = e} \right\}$ and $G’ = \left\{ {b,{b^2},{b^3}, \ldots ,{b^n} = e’} \right\}$

The mapping $f:G \to G’$, defined by $f\left( {{a^r}} \right) = {b^r}$, is isomorphism.

For, $f\left( {{a^r} \cdot {a^s}} \right) = f\left( {{a^{r + s}}} \right) = {b^{r + s}} = {b^r} \cdot {b^s} = f\left( {{a^r}} \right) \cdot f\left( {{a^s}} \right)$

Therefore the groups are isomorphic.

Theorem 2:
An infinite cyclic group is isomorphic to the additive group of integers.

Proof: Let $G$ be an infinite cyclic group, generated by $a$, then
$G = \left\{ { \ldots ,{a^{ – 2}},{a^{ – 1}},{a^0} = e,{a^1},{a^2},{a^3}, \ldots } \right\} = \left\{ {{a^r}:r\,{\text{is}}\,{\text{an}}\,{\text{integer}}} \right\}$

The mapping $f:G \to \mathbb{Z}$, defined by $f\left( {{a^r}} \right) = r$ is an isomorphism, for it is one-one onto, and further,
$\begin{gathered} f\left( {{a^r} \cdot {a^s}} \right) = f\left( {{a^{r + s}}} \right) = r + s \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {{a^r}} \right) + f\left( {{a^s}} \right) \\ \end{gathered}$

It follows that $G$ is isomorphic to $\mathbb{Z}$.

Theorem 3:
A cyclic group of order $n$ is isomorphic to the additive group of residue classes modulo $n$.

Proof: Let $G$ be an infinite cyclic group, generated by $a$, then
$G = \left\{ {a,{a^2},{a^3}, \ldots ,{a^{n – 1}},{a^n} = e} \right\}$

Let $G’$ be the additive group or residue classes $\left( {\bmod n} \right)$, i.e.
$G’ = \left\{ {\left[ 1 \right],\left[ 2 \right],\left[ 3 \right], \ldots ,\left[ n \right] = \left[ 0 \right]} \right\}$

The mapping $f:G \to G’$, defined by $f\left( {{a^r}} \right) = \left[ r \right]$, is isomorphism, for it is one-one onto, and further,
$\begin{gathered} f\left( {{a^r} \cdot {a^s}} \right) = f\left( {{a^{r + s}}} \right) = \left[ {r + s} \right] = \left[ r \right] + \left[ s \right] \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {{a^r}} \right) + f\left( {{a^s}} \right) \\ \end{gathered}$

It follows that $G$ is isomorphic to $G’$.

Theorem 4:
A subgroup of the infinite cyclic group is isomorphic to the additive group of integral multiples of an integer.

Proof: Let $G = \left\{ { \ldots ,{a^{ – 2}},{a^{ – 1}},{a^0} = e,{a^1},{a^2},{a^3}, \ldots } \right\}$ and let $H$ be a subgroup of $G$, given by,
$H = \left\{ { \ldots ,{a^{ – 2m}},{a^{ – m}},{a^0} = e,{a^m},{a^{2m}}, \ldots } \right\} = \left\{ {{{\left( {{a^m}} \right)}^n}:n \in \mathbb{Z}} \right\}$

Then $H$ is isomorphic to the additive group $H’$, given by
$H’ = \left\{ {0, \pm m, \pm 2m, \pm 3m, \ldots } \right\} = \left\{ {nm:n \in \mathbb{Z}} \right\}$

The mapping $f:H \to H’$, defined by $f\left( {{a^{mn}}} \right) = nm$, is isomorphism, for it is one-one onto, and if $r,s \in \mathbb{Z}$, then
$\begin{gathered} f\left( {{a^{rm}} \cdot {a^{sm}}} \right) = f\left( {{a^{\left( {r + s} \right)m}}} \right) = {\left( {r + s} \right)^m} = rm + sm \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {{a^{rm}}} \right) + f\left( {{a^{sm}}} \right) \\ \end{gathered}$

It will be observed that $H$ is itself an infinite cyclic group, and as such it is isomorphic to $G$. Thus a subgroup of an infinite cyclic group is isomorphic to the group itself.