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 thatf 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'.