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:
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$$ \times $$
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$$1$$
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$$\omega $$
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$${\omega ^2}$$
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$$1$$
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$$1$$
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$$\omega $$
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$${\omega ^2}$$
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|
$$\omega $$
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$$\omega $$
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$${\omega ^2}$$
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$$1$$
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$${\omega ^2}$$
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$${\omega ^2}$$
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$$1$$
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$$\omega $$
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| $$ + \left( {\bmod 3} \right)$$ |
$$\left\{ 0 \right\}$$
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$$\left\{ 1 \right\}$$
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$$\left\{ 2 \right\}$$
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$$\left\{ 0 \right\}$$
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$$\left\{ 0 \right\}$$
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$$\left\{ 1 \right\}$$
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$$\left\{ 2 \right\}$$
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|
$$\left\{ 1 \right\}$$
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$$\left\{ 1 \right\}$$
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$$\left\{ 2 \right\}$$
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$$\left\{ 0 \right\}$$
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$$\left\{ 2 \right\}$$
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$$\left\{ 2 \right\}$$
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$$\left\{ 0 \right\}$$
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$$\left\{ 1 \right\}$$
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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 $$G’$$.