Definition and Examples

To show that \({\mathbb Z}_4 \cong \langle i \rangle\), define a map \(\phi: {\mathbb Z}_4 \rightarrow \langle i \rangle\) by \(\phi(n) = i^n\). We must show that \(\phi\) is bijective and preserves the group operation. The map \(\phi\) is one-to-one and onto because \[\begin{aligned}\phi(0) & = 1 \\ \phi(1) & = i \\ \phi(2) & = -1 \\ \phi(3) & = -i\end{aligned}\]. Since \[\begin{aligned}\end{aligned}\], the group operation is preserved.

We can define an isomorphism \(\phi\) from the additive group of real numbers \(( {\mathbb R}, + )\) to the multiplicative group of positive real numbers \(( {\mathbb R^+}, \cdot )\) with the exponential map; that is, \[\begin{aligned}\end{aligned}\]. Of course, we must still show that \(\phi\) is one-to-one and onto, but this can be determined using calculus.

The integers are isomorphic to the subgroup of \({\mathbb Q}^\ast\) consisting of elements of the form \(2^n\). Define a map \(\phi: {\mathbb Z} \rightarrow {\mathbb Q}^\ast\) by \(\phi( n ) = 2^n\). Then \[\begin{aligned}\end{aligned}\]. By definition the map \(\phi\) is onto the subset \(\{2^n :n \in {\mathbb Z} \}\) of \({\mathbb Q}^\ast\). To show that the map is injective, assume that \(m \neq n\). If we can show that \(\phi(m) \neq \phi(n)\), then we are done. Suppose that \(m \gt n\) and assume that \(\phi(m) = \phi(n)\). Then \(2^m = 2^n\) or \(2^{m - n} = 1\), which is impossible since \(m - n \gt 0\).

The groups \({\mathbb Z}_8\) and \({\mathbb Z}_{12}\) cannot be isomorphic since they have different orders; however, it is true that \(U(8) \cong U(12)\). We know that \[\begin{aligned}U(8) & = \{1, 3, 5, 7 \} \\ U(12) & = \{1, 5, 7, 11 \}\end{aligned}\]. An isomorphism \(\phi : U(8) \rightarrow U(12)\) is then given by \[\begin{aligned}1 & \mapsto 1 \\ 3 & \mapsto 5 \\ 5 & \mapsto 7 \\ 7 & \mapsto 11\end{aligned}\]. The map \(\phi\) is not the only possible isomorphism between these two groups. We could define another isomorphism \(\psi\) by \(\psi(1) = 1\), \(\psi(3) = 11\), \(\psi(5) = 5\), \(\psi(7) = 7\). In fact, both of these groups are isomorphic to \({\mathbb Z}_2 \times {\mathbb Z}_2\) (see in ).

Even though \(S_3\) and \({\mathbb Z}_6\) possess the same number of elements, we would suspect that they are not isomorphic, because \({\mathbb Z}_6\) is abelian and \(S_3\) is nonabelian. To demonstrate that this is indeed the case, suppose that \(\phi : {\mathbb Z}_6 \rightarrow S_3\) is an isomorphism. Let \(a , b \in S_3\) be two elements such that \(ab \neq ba\). Since \(\phi\) is an isomorphism, there exist elements \(m\) and \(n\) in \({\mathbb Z}_6\) such that \[\begin{aligned}\end{aligned}\]. However, \[\begin{aligned}\end{aligned}\], which contradicts the fact that \(a\) and \(b\) do not commute.

Consider the group \({\mathbb Z}_3\). The Cayley table for \({\mathbb Z}_3\) is as follows.

\[\begin{aligned}\end{aligned}\]

The addition table of \({\mathbb Z}_3\) suggests that it is the same as the permutation group \(G = \{ (0), (0 1 2), (0 2 1) \}\). The isomorphism here is \[\begin{aligned}0 & \mapsto \begin{pmatrix} 0 & 1 & 2 \\ 0 & 1 & 2 \end{pmatrix} = (0) \\ 1 & \mapsto \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 0 \end{pmatrix} = (0 \, 1 \, 2) \\ 2 & \mapsto \begin{pmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \end{pmatrix} = (0 \, 2 \, 1)\end{aligned}\].