# Examples of Group Isomorphism

Example 1: Show that the multiplicative group $G$ consisting of three cube roots of unity $1,\omega ,{\omega ^2}$ is isomorphic to the group $G’$ of residue classes $\left( {\bmod 3} \right)$ under addition of residue classes $\left( {\bmod 3} \right)$.

Solution:
Let us consider the composition tables of two structures $G,\,G’$ as given below:

 $\times$ $1$ $\omega$ ${\omega ^2}$ $1$ $1$ $\omega$ ${\omega ^2}$ $\omega$ $\omega$ ${\omega ^2}$ $1$ ${\omega ^2}$ ${\omega ^2}$ $1$ $\omega$

 $+ \left( {\bmod 3} \right)$ $\left\{ 0 \right\}$ $\left\{ 1 \right\}$ $\left\{ 2 \right\}$ $\left\{ 0 \right\}$ $\left\{ 0 \right\}$ $\left\{ 1 \right\}$ $\left\{ 2 \right\}$ $\left\{ 1 \right\}$ $\left\{ 1 \right\}$ $\left\{ 2 \right\}$ $\left\{ 0 \right\}$ $\left\{ 2 \right\}$ $\left\{ 2 \right\}$ $\left\{ 0 \right\}$ $\left\{ 1 \right\}$

From this table it is evident that if $1,\omega ,{\omega ^2}$ are replaced by $\left\{ 0 \right\},\left\{ 1 \right\},\left\{ 2 \right\}$ respectively in the composition table for $G$, we get the composition table $G’$. This leads to the fact that mapping $f$ of $G$ onto $G’$ defined by $f\left( 1 \right) = \left\{ 0 \right\}$, $\,f\left( \omega \right) = \left\{ 1 \right\}$ , $f\left( {{\omega ^2}} \right) = \left\{ 2 \right\}$ is an isomorphism. Also:

$\begin{gathered} f\left( {\omega \cdot {\omega ^2}} \right) = f\left( 1 \right) = \left\{ 0 \right\} = \left\{ 1 \right\} + \left\{ 2 \right\} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( \omega \right) + f\left( {{\omega ^2}} \right) \\ \end{gathered}$

Example 2: Show that the additive group $G = \left\{ { \ldots , – 2, – 1,0,1,2, \ldots } \right\}$ is an isomorphic to the additive group $G’ = \left\{ { \ldots , – 2m, – m,0,m,2m, \ldots } \right\}$ for any given integer $m$.

Solution:
We define a mapping $f$ by $f:G \to G’:f\left( a \right) = ma$, where $a \in G,\,\,ma \in G’$ and show that $f$ is an isomorphism of $G$ onto $G’$.

We see that $f$ is one-one since two different elements of $G$ have two different $f –$ image in $G’$ is the $f –$ image of an element of $G$.

Again:
$\begin{gathered} f\left( {a + b} \right) = m\left( {a + b} \right) = ma + mb \\ \Rightarrow f\left( {a + b} \right) = f\left( a \right) + f\left( b \right) \\ \end{gathered}$

Thus $f$ is composition preserving as well. Hence $f$ is an isomorphic mapping of $G$ onto $G’$.