Isomorphism of Cyclic Groups

Theorem 1:
Cyclic group of 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 groups, 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 groups, 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, give 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.

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