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'.

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