Remark 3.1.
It is worth pointing out that we are developing the rudiments of a general theory of mathematical structures called Category Theory. This course will only scratch the surface of this beautiful framework.
Chapter 3 Morphisms of Groups
Section 3.1 Morphisms
In general, algebra is the study of mathematical structure. For this course, structure is imposed by equipping a set with one (or more) operations to form algebraic objects. An incredibly important way to understand these structures is to study the functions that preserve the structure. In the modern lexicon, one typically refers to these functions as a morphism of a certain type.
The most basic structure we have at our disposal is that of a set. As such, a morphism of sets is another term for a set function that helps to standardize the language in many contexts. The other morphisms we encounter (primarily for groups and rings) will be morphisms of sets with additional axioms, analogous to how groups are sets with additional axioms.
Definition 3.2. Morphism of Groups.
Let \((G,\ast)\) and \((H,\star)\) be groups. A morphism of groups (this is the modern form of the classical homomorphism of groups), is a set function \(f \colon G \to H\) that satisfies the condition for all \(x,y \in G\text{,}\)
\begin{equation*}
f(x \ast y) = f(x) \star f(y)\text{.}
\end{equation*}
We refer to the set of all such morphisms as \(\Grp(G,H)\text{.}\)
Except in situations where we require extra clarity, we will generally omit the group operations and simply write
\begin{equation*}
f(xy) = f(x)f(y)\text{.}
\end{equation*}
Warning 3.3.
Pay close attention to the group operations! The first uses the binary operation on \(G\)—\(x \ast y\)—while the second uses the binary operation on \(H\)—\(f(x) \star f(y)\text{.}\)
Example 3.4. Canonical Inclusion Morphisms.
The set inclusions \(\Z \subseteq \Q \subseteq \R \subseteq \C\) are all morphisms of additive groups because addition takes the same meaning in any one of those sets. For example, adding the integers \(2\) and \(3\) produces the integer \(5\text{;}\) adding the real numbers \(2\) and \(3\) produces the real number \(5\text{.}\) This says the function
\begin{align*}
\phi \colon \Z \amp\to \R\\
x \amp\mapsto x
\end{align*}
is a morphism of groups because
\begin{equation*}
\phi(x + y) = x + y = \phi(x) + \phi(y)\text{.}
\end{equation*}
Example 3.5. Matrices are Morphism.
Fix \(m,n \in \N\text{.}\) An \(m \times n\) matrix
\begin{equation*}
M = \begin{bmatrix}
a_{1,1} \amp a_{1,2} \amp \cdots \amp a_{1,n}\\
a_{2,1} \amp a_{2,2} \amp \cdots \amp a_{2,n}\\
\vdots \amp \vdots \amp \ddots \amp \vdots\\
a_{m,1} \amp a_{m,2} \amp \cdots \amp a_{m,n}
\end{bmatrix}
\end{equation*}
defines a function \(T \colon \R^n \to \R^m\) via the matrix-vector multiplication
\begin{align*}
T(x_1,x_2,\ldots,x_n) \amp=
\begin{bmatrix}
a_{1,1} \amp a_{1,2} \amp \cdots \amp a_{1,n}\\
a_{2,1} \amp a_{2,2} \amp \cdots \amp a_{2,n}\\
\vdots \amp \vdots \amp \ddots \amp \vdots\\
a_{m,1} \amp a_{m,2} \amp \cdots \amp a_{m,n}
\end{bmatrix}
\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}\\
\amp= \begin{bmatrix}
a_{1,1}x_1 + a_{1,2}x_2 + \cdots a_{1,n}x_n\\
a_{2,1}x_1 + a_{2,2}x_2 + \cdots a_{2,n}x_n\\
\vdots\\
a_{m,1}x_1 + a_{m,2}x_2 + \cdots a_{m,n}x_n)
\end{bmatrix}
\end{align*}
We know from linear algebra that matrix-vector multiplication is linear. In particular, if \(\mathbf{v},\mathbf{w} \in \R^n\text{,}\) then
\begin{align*}
T(\mathbf{v} + \mathbf{w}) \amp= M(\mathbf{v} + \mathbf{w})\\
\amp= M\mathbf{v} + M\mathbf{w}\\
\amp= T(\mathbf{v}) + T(\mathbf{w})
\end{align*}
Hence the transformation \(T \colon \R^n \to \R^m\) induced by a matrix \(M \in \operatorname{M}_{m \times n}(\R)\) is a morphism of (additive) groups.
Example 3.6. The Exponential.
Consider the set \((0,\infty) \subseteq \R\text{.}\) Since the product of two positive numbers is again a positive number, multiplication is a binary operation on this set. The number \(1 \in (0,\infty)\) serves as a multiplicative inverse and every positive number, \(x \in (0,\infty)\text{,}\) has a positive inverse, \(1/x \in (0, \infty)\text{.}\) Hence \((0,\infty)\) is a group under multiplication.
The exponential function \(\exp \colon \R \to (0,\infty)\) is a morphism of groups because for all \(x,y \in \R\)
\begin{equation*}
\exp(x + y) = \exp(x)\exp(y)\text{.}
\end{equation*}
Example 3.7. The Logarithm.
The logarithm function \(\log \colon (0,\infty) \to \R\) is also a moprhism of groups because for all \(x,y \in (0,\infty)\text{,}\)
\begin{equation*}
\log(xy) = \log(x) + \log(y)
\end{equation*}
Example 3.8. The Determinant.
The determinant is a morphism of groups
\begin{equation*}
\det \colon \operatorname{GL}_n(\R) \to \R^\times
\end{equation*}
because for all \(M,N \in \operatorname{GL}_n(\R)\text{,}\)
\begin{equation*}
\det{MN} = \det{M}\det{N}\text{.}
\end{equation*}
Section 3.2 Isomorphisms and Automorphisms
Definition 3.9. Endomorphism of Groups.
We say a morphism of groups is an endomorphism of groups if its domain and codomain are the same group. We refer to the collection of all endomorphisms of groups as \(\End_{\Grp}(G)\text{.}\)
Warning 3.10.
It is important to note that not all endomorphisms are the same! For example, the function
\begin{align*}
f \colon \Z \amp\to \Z\\
x \amp\mapsto \begin{cases} 0 \amp\text{if}\ x\ \text{is even}\\1 \amp\text{if}\ x\ \text{is odd}\end{cases}
\end{align*}
is a well-defined endomorphism of sets, but is not an endomorphism of groups because
\begin{equation*}
f(1 + 1) = f(2) = 0 \neq 2 = 1 + 1 = f(1) + f(1)\text{.}
\end{equation*}
Definition 3.11. Isomorphism of Groups.
Let \(G\) and \(H\) be groups. We say a morphism of groups \(f \colon G \to H\) is an isomorphism of groups if there exists a morphism of groups \(f^{-1} \colon H \to G\) such that
\begin{equation*}
f^{-1} \circ f = \id_G
\quad\text{and}\quad
f \circ f^{-1} = \id_H
\end{equation*}
We refer to the morphism \(f^{-1}\) as the inverse of \(f\).
We say the groups \(G\) and \(H\) are isomorphic if there exists an isomorphism \(G \to H\) (or, equivalently, an isomorphism \(H \to G\)) and write \(G \cong H\text{.}\)
Definition 3.12. Automorphism of Groups.
We call an endomorphism that is also an isomorphism an automorphism of groups. We denote the set of all automorphism of groups \(\Aut_{\Grp}(G)\)
Example 3.13. Isomorphic Groups are “The Same”.
As a general rule, asking for two groups to be equal is too restrictive. It requires that they have the exact same elements and binary operation. However, the actual elements of the set are not intrinsic to the structure—that data is housed, at least partially, in the operation and can really only be properly accessed via morphisms.
First, consider the set \(\{1,-1\} \times \{1,-1\}\) equipped with the operation
\begin{equation*}
(a,b)(c,d) = (ac, bd)\text{,}
\end{equation*}
where \(ac\) and \(bd\) are computed using the usual integer multiplication. We can summarize the operation succinctly using the following group table.
While we could prove that this is a group with identity element \((1,1)\text{,}\) the necessary computations are tedious and unenlightening. With the tools we develop later, it will be trivial to prove this is a group, so we omit the details here.
Now, take the set \(\{1,a,b,c\}\) and define the operation by the relations
\begin{align*}
aa \amp= bb = cc = 1,\\
ab \amp= ba = c,\\
ac \amp= ca = b,\,\text{and}\\
bc \amp= cb = a\text{.}
\end{align*}
This yields a group, called the Klein 4-group, with operation table given below.
If one stares carefully at the two tables for long enough, a few mappings between them jump out. One particularly simple mapping identifies the entries along the header row of the two tables in order:
\begin{align*}
f \colon \{1,a,b,c\} \amp\to \{1,-1\} \times \{1,-1\}\\
1 \amp\mapsto (1,1)\\
a \amp\mapsto (1,-1)\\
b \amp\mapsto (-1,1)\\
c \amp\mapsto (-1,-1)
\end{align*}
This map is bijective, so it is an isomorphism as long as it is a morphism of groups. It is easy (albeit tedious) to check that this is the case. This tells us these two very different sets with seemingly different operations are, in fact, encoding the same basic algebraic structure.
Example 3.16.
The exponential and logarithm functions provide an isomorphism between the groups \((\R,+)\) and \(((0,\infty), \times)\) because for all \(x \in \R\) and for all \(y \in (0,\infty)\text{,}\)
\begin{equation*}
\ln \circ \exp(x) = x \quad\text{and}\quad \exp \circ \ln(y) = y\text{.}
\end{equation*}
Section 3.3 Basic Properties of Morphisms of Groups
Theorem 3.17.
A morphism \(f \in \Grp(G,H)\) is an isomorphism if and only if it is bijective.
Proof.
Assume \(f\) is an isomorphism. By Definition 3.11, \(f\) has an inverse and hence \(f\) is a bijective morphism of groups.
Conversely, assume that \(f\) is a bijective morphism of groups. Since \(f\) is bijective, there exists an inverse \(f^{-1} \colon H \to G\) as morphisms of sets. We must prove that \(f^{-1}\) is a morphism of groups.
Towards that end, let \(y_1, y_2 \in H\) be given. Since \(f\) is bijective, there exist unique \(x_1, x_2 \in G\) such that
\begin{equation*}
f(x_1) = y_1 \iff x_1 = f^{-1}(y_1)
\quad\text{and}\quad
f(x_2) = y_2 \iff x_y = f^{-1}(y_2)
\end{equation*}
Hence
\begin{align*}
f^{-1}(y_1y_2) \amp= f^{-1}\left(f(x_1)f(x_2)\right)\\
\amp= f^{-1}\left(f(x_1x_2)\right)\\
\amp= f^{-1}\circ f (x_1x_2)\\
\amp= x_1x_2\\
\amp= f^{-1}(y_1) f^{-1}(y_2)
\end{align*}
implies \(f^{-1}\) is a morphism of groups. Therefore \(f\) is an isomorphism of groups if and only if \(f\) is a bijective morphism of groups.
Lemma 3.18. Morphisms are Closed Under Composition.
Assume \(G\text{,}\) \(H\text{,}\) and \(K\) are groups. If \(f \colon G \to H\) is a morphism of groups and \(g \colon H \to K\) is a morphism of groups, then so is \(g \circ f \colon G \to K\text{.}\)
Proof.
Let \(a,b \in G\) be given. Since \(f\) and \(g\) are assumed to be groups,
\begin{align*}
g \circ f(ab) \amp= g\left(f(ab)\right)\\
\amp= g\left(f(a)f(b)\right)\\
\amp= g\left(f(a)\right)g\left(f(b)\right)\\
\amp= \left(g \circ f(a)\right)\left(g \circ f(b)\right)
\end{align*}
Therefore \(g \circ f \colon G \to K\) is a morphism of groups.
Lemma 3.19. Isomorphisms are Closed Under Composition.
Assume \(G\text{,}\) \(H\text{,}\) and \(K\) are groups. If \(f \colon G \to H\) is a morphism of groups and \(g \colon H \to K\) is a morphism of groups, then so is \(g \circ f \colon G \to K\text{.}\)
Proof.
Since the composition of bijective functions is again a bijective function, Lemma 3.18 implies that \(g \circ f\) is a bijective homomorphism. Therefore \(g \circ f\) is an isomorphism by Theorem 3.17.
Theorem 3.20. Isomorphism is an Equivalence Relation.
Isomorphism is an equivalence relation on the class of all groups.
Proof.
We must show that equivalence satisfies the following properties:
- Reflexive
- Symmetric
- Transitive
- For all groups \(G\text{,}\) \(H\text{,}\) and \(K\text{,}\) if \(G \cong H\) and \(H \cong K\text{,}\) then \(G \cong K\text{.}\)
Let \(G\) be a group. To see that isomorphism is reflexive, observe that \(\id_G \colon G \to G\) is a morphism of groups because for all \(a,b \in G\)
\begin{equation*}
\id_G(ab) = ab = \id_G(a)\id_G(b)\text{,}
\end{equation*}
injective because for all \(a,b \in G\text{,}\) \(\id_G(a) = \id_G(b)\) implies
\begin{equation*}
a = \id_G(a) = \id_G(b) = b\text{,}
\end{equation*}
and surjective because for all \(a \in G\text{,}\) \(\id_G(a) = a\text{.}\) Hence \(\id_G \colon G \to G\) is an isomorphism and \(G \cong G\text{.}\) Therefore isomorphism is reflexive.
Let \(G\) and \(H\) be groups. Assume \(G \cong H\text{.}\) By Definition 3.11, there exist morphisms \(f \colon G \to H\) and \(f^{-1} \colon H \to G\) such that \(f^{-1} \circ f = \id_G\) and \(f\circ f^{-1} = \id_H\text{.}\) If we interchange the roles of these two functions, then we see \(f^{-1} \colon H \to G\) and \(f \colon G \to H\) satisfies \(f \circ f^{-1} = \id_H\) and \(f^{-1} \circ f = \id_G\text{.}\) Hence \(H \cong G\text{.}\) Therefore isomorphism is symmetric.
Let \(G\text{,}\) \(H\text{,}\) and \(K\) be groups. Assume \(G \cong H\) and \(H \cong K\text{.}\) By Definition 3.11, there exist isomorphisms \(f \colon G \to H\) and \(g \colon H \to K\text{.}\) By Lemma 3.19, \(g \circ f \colon G \to K\) is an isomorphism. Hence \(G \cong K\text{.}\) Therefore isomorphism is transitive.
