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**.

######
Example9.1

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.

######
Example9.2

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.

######
Example9.3

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 \(m-n \gt 0\text{.}\)

######
Example9.4

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 3.28 in Chapter 3).

######
Example9.5

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.

######
Theorem9.6

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 \(n\text{.}\)

###### Proof

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.

######
Theorem9.7

All cyclic groups of infinite order are isomorphic to \({\mathbb Z}\text{.}\)

###### Proof

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 \(\phi(n) = a^n\text{.}\)

######
Theorem9.8

If \(G\) is a cyclic group of order \(n\text{,}\) then \(G\) is isomorphic to \({\mathbb Z}_n\text{.}\)

###### Proof

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.

######
Corollary9.9

If \(G\) is a group of order \(p\text{,}\) where \(p\) is a prime number, then \(G\) is isomorphic to \({\mathbb Z}_p\text{.}\)

###### Proof

The proof is a direct result of Corollary 6.12.

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.

######
Theorem9.10

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.

###
SubsectionCayley'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.

######
Example9.11

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*}
######
Theorem9.12Cayley

Every group is isomorphic to a group of permutations.

###### Proof

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 \(G\text{.}\)