Two groups \((G, \cdot)\) and \((H, \circ)\) are isomorphic if there exists a one-to-one and onto map \(\phi : G \rightarrow H\) such that the group operation is preserved; that is,
\begin{equation*}
\phi( a \cdot b) = \phi( a) \circ \phi( b)
\end{equation*}
for all \(a\) and \(b\) in \(G\text{.}\) If \(G\) is isomorphic to \(H\text{,}\) we write \(G \cong H\text{.}\) The map \(\phi\) is called an isomorphism.
Example6.1
To show that \({\mathbb Z}_4 \cong \langle i \rangle\text{,}\) define a map \(\phi: {\mathbb Z}_4 \rightarrow \langle i \rangle\) by \(\phi(n) = i^n\text{.}\) We must show that \(\phi\) is bijective and preserves the group operation. The map \(\phi\) is one-to-one and onto because
\begin{align*}
\phi(0) & = 1\\
\phi(1) & = i\\
\phi(2) & = -1\\
\phi(3) & = -i.
\end{align*}
Since
\begin{equation*}
\phi(m + n) = i^{m+n} = i^m i^n = \phi(m) \phi( n),
\end{equation*}
the group operation is preserved.
Example6.2
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{equation*}
\phi( x + y) = e^{x + y} = e^x e^y = \phi( x ) \phi( y).
\end{equation*}
Of course, we must still show that \(\phi\) is one-to-one and onto, but this can be determined using calculus.
Example6.3
The integers are isomorphic to the subgroup of \({\mathbb Q}^\ast\) consisting of elements of the form \(2^n\text{.}\) Define a map \(\phi: {\mathbb Z} \rightarrow {\mathbb Q}^\ast\) by \(\phi( n ) = 2^n\text{.}\) Then
\begin{equation*}
\phi( m + n ) = 2^{m + n} = 2^m 2^n = \phi( m ) \phi( n ).
\end{equation*}
By definition the map \(\phi\) is onto the subset \(\{2^n :n \in {\mathbb Z} \}\) of \({\mathbb Q}^\ast\text{.}\) To show that the map is injective, assume that \(m \neq n\text{.}\) If we can show that \(\phi(m) \neq \phi(n)\text{,}\) then we are done. Suppose that \(m \gt n\) and assume that \(\phi(m) = \phi(n)\text{.}\) Then \(2^m = 2^n\) or \(2^{m - n} = 1\text{,}\) which is impossible since \(m - n \gt 0\text{.}\)
Example6.4
The groups \({\mathbb Z}_8\) and \({\mathbb Z}_{12}\) cannot be isomorphic since they have different orders; however, it is true that \(U(8) \cong U(12)\text{.}\) We know that
\begin{align*}
U(8) & = \{1, 3, 5, 7 \}\\
U(12) & = \{1, 5, 7, 11 \}.
\end{align*}
An isomorphism \(\phi : U(8) \rightarrow U(12)\) is then given by
\begin{align*}
1 & \mapsto 1\\
3 & \mapsto 5\\
5 & \mapsto 7\\
7 & \mapsto 11.
\end{align*}
The map \(\phi\) is not the only possible isomorphism between these two groups. We could define another isomorphism \(\psi\) by \(\psi(1) = 1\text{,}\) \(\psi(3) = 11\text{,}\) \(\psi(5) = 5\text{,}\) \(\psi(7) = 7\text{.}\) In fact, both of these groups are isomorphic to \({\mathbb Z}_2 \times {\mathbb Z}_2\) (see Example 3.13 in Chapter 2).
Example6.5
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 \(ab \neq ba\text{.}\) Since \(\phi\) is an isomorphism, there exist elements \(m\) and \(n\) in \({\mathbb Z}_6\) such that
\begin{equation*}
\phi( m ) = a \quad \text{and} \quad \phi( n ) = b.
\end{equation*}
However,
\begin{equation*}
ab = \phi(m ) \phi(n) = \phi(m + n) = \phi(n + m) = \phi(n ) \phi(m) = ba,
\end{equation*}
which contradicts the fact that \(a\) and \(b\) do not commute.
Theorem6.6
Let \(\phi : G \rightarrow H\) be an isomorphism of two groups. Then the following statements are true.
\(\phi^{-1} : H \rightarrow G\) is an isomorphism.
\(|G| = |H|\text{.}\)
If \(G\) is abelian, then \(H\) is abelian.
If \(G\) is cyclic, then \(H\) is cyclic.
If \(G\) has a subgroup of order \(n\text{,}\) then \(H\) has a subgroup of order \(n\text{.}\)
Assertions (1) and (2) follow from the fact that \(\phi\) is a bijection. We will prove (3) here and leave the remainder of the theorem to be proved in the exercises.
(3) Suppose that \(h_1\) and \(h_2\) are elements of \(H\text{.}\) Since \(\phi\) is onto, there exist elements \(g_1, g_2 \in G\) such that \(\phi(g_1) = h_1\) and \(\phi(g_2) = h_2\text{.}\) Therefore,
\begin{equation*}
h_1 h_2 = \phi(g_1) \phi(g_2) = \phi(g_1 g_2) = \phi(g_2 g_1) = \phi(g_2) \phi(g_1) = h_2 h_1.
\end{equation*}
We are now in a position to characterize all cyclic groups.
Theorem6.7
All cyclic groups of infinite order are isomorphic to \({\mathbb Z}\text{.}\)
Let \(G\) be a cyclic group with infinite order and suppose that \(a\) is a generator of \(G\text{.}\) Define a map \(\phi : {\mathbb Z} \rightarrow G\) by \(\phi : n \mapsto a^n\text{.}\) Then
\begin{equation*}
\phi( m+n ) = a^{m+n} = a^m a^n = \phi( m ) \phi( n ).
\end{equation*}
To show that \(\phi\) is injective, suppose that \(m\) and \(n\) are two elements in \({\mathbb Z}\text{,}\) where \(m \neq n\text{.}\) We can assume that \(m \gt n\text{.}\) We must show that \(a^m \neq a^n\text{.}\) Let us suppose the contrary; that is, \(a^m = a^n\text{.}\) In this case \(a^{m - n} = e\text{,}\) where \(m - n \gt 0\text{,}\) which contradicts the fact that \(a\) has infinite order. Our map is onto since any element in \(G\) can be written as \(a^n\) for some integer \(n\) and \(\phi(n) = a^n\text{.}\)
Theorem6.8
If \(G\) is a cyclic group of order \(n\text{,}\) then \(G\) is isomorphic to \({\mathbb Z}_n\text{.}\)
Let \(G\) be a cyclic group of order \(n\) generated by \(a\) and define a map \(\phi : {\mathbb Z}_n \rightarrow G\) by \(\phi : k \mapsto a^k\text{,}\) where \(0 \leq k \lt n\text{.}\) The proof that \(\phi\) is an isomorphism is one of the end-of-chapter exercises.
Corollary6.9
If \(G\) is a group of order \(p\text{,}\) where \(p\) is a prime number, then \(G\) is isomorphic to \({\mathbb Z}_p\text{.}\)
The proof is a direct result of Corollary 7.12.
The main goal in group theory is to classify all groups; however, it makes sense to consider two groups to be the same if they are isomorphic. We state this result in the following theorem, whose proof is left as an exercise.
Theorem6.10
The isomorphism of groups determines an equivalence relation on the class of all groups.
Hence, we can modify our goal of classifying all groups to classifying all groups up to isomorphism; that is, we will consider two groups to be the same if they are isomorphic.
Subsection6.3.1Cayley's Theorem
¶Cayley proved that if \(G\) is a group, it is isomorphic to a group of permutations on some set; hence, every group is a permutation group. Cayley's Theorem is what we call a representation theorem. The aim of representation theory is to find an isomorphism of some group \(G\) that we wish to study into a group that we know a great deal about, such as a group of permutations or matrices.
Example6.11
Consider the group \({\mathbb Z}_3\text{.}\) The Cayley table for \({\mathbb Z}_3\) is as follows.
\begin{equation*}
\begin{array}{c|ccc}
+ & 0 & 1 & 2 \\ \hline
0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1
\end{array}
\end{equation*}
The addition table of \({\mathbb Z}_3\) suggests that it is the same as the permutation group \(G = \{ (0), (0 1 2), (0 2 1) \}\text{.}\) The isomorphism here is
\begin{align*}
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{align*}
Theorem6.12Cayley
Every group is isomorphic to a group of permutations.
Let \(G\) be a group. We must find a group of permutations \(\overline{G}\) that is isomorphic to \(G\text{.}\) For any \(g \in G\text{,}\) define a function \(\lambda_g : G \rightarrow G\) by \(\lambda_g(a) = ga\text{.}\) We claim that \(\lambda_g\) is a permutation of \(G\text{.}\) To show that \(\lambda_g\) is one-to-one, suppose that \(\lambda_g(a) = \lambda_g(b)\text{.}\) Then
\begin{equation*}
ga =\lambda_g(a) = \lambda_g(b) = gb.
\end{equation*}
Hence, \(a = b\text{.}\) To show that \(\lambda_g\) is onto, we must prove that for each \(a \in G\text{,}\) there is a \(b\) such that \(\lambda_g (b) = a\text{.}\) Let \(b = g^{-1} a\text{.}\)
Now we are ready to define our group \(\overline{G}\text{.}\) Let
\begin{equation*}
\overline{G} = \{ \lambda_g : g \in G \}.
\end{equation*}
We must show that \(\overline{G}\) is a group under composition of functions and find an isomorphism between \(G\) and \(\overline{G}\text{.}\) We have closure under composition of functions since
\begin{equation*}
(\lambda_g \circ \lambda_h )(a) = \lambda_g(ha) = gha = \lambda_{gh} (a).
\end{equation*}
Also,
\begin{equation*}
\lambda_e (a) = ea = a
\end{equation*}
and
\begin{equation*}
(\lambda_{g^{-1}} \circ \lambda_g) (a) = \lambda_{g^{-1}} (ga) = g^{-1} g a = a = \lambda_e (a).
\end{equation*}
We can define an isomorphism from \(G\) to \(\overline{G}\) by \(\phi : g \mapsto \lambda_g\text{.}\) The group operation is preserved since
\begin{equation*}
\phi(gh) = \lambda_{gh} = \lambda_g \lambda_h = \phi(g) \phi(h).
\end{equation*}
It is also one-to-one, because if \(\phi(g)(a) = \phi(h)(a)\text{,}\) then
\begin{equation*}
ga = \lambda_g a = \lambda_h a= ha.
\end{equation*}
Hence, \(g = h\text{.}\) That \(\phi\) is onto follows from the fact that \(\phi( g ) = \lambda_g\) for any \(\lambda_g \in \overline{G}\text{.}\)
The isomorphism \(g \mapsto \lambda_g\) is known as the left regular representation of \(G\text{.}\)
