7.1 Group Theory
Definition 7.1 (Group) A non-empty set \(G\), together with a binary operation, \(*\), is called a group if it satisfies the following conditions:
\(a*b \in G, \: \forall a,b \in G\) (Closure)
\((a*b)*c = a * (b*c), \forall a,b,c \in G\) (Associative)
There exists an element \(e \in G\) such that for all \(a\in G\), \(e*a=a*e=a\) (Identity)
For each \(a\in G\), there exists an element \(b\in G\) such that \(a*b=b*a=e\) (Inverse)
This concept of a group is a generalization of most of the properties that we noticed in Chapter 4 as we constructed the various number systems. We demonstrated the associative property for each of the number systems as we defined them. So we will be able to assume associative property when we are proving that a set and operation form a group.
Example 7.1 Some basic examples of groups that appear in the K-12 curriculum include:
\((\mathbb{Z},+)\), the set of integers under addition
\((2\mathbb{Z},+)\), the set of even integers under addition
\((\mathbb{Q},+)\), the set of rational numbers under addition
\((\mathbb{R},+)\), the set of real numbers under addition
\((\mathbb{Q}\setminus\{0\} , \cdot)\), the set of non-zero rational numbers under multiplication
\(( \mathbb{R}^+, \cdot)\), the set of positive real numbers under multiplication
\(\left( \mathbb{R}[x], +\right)\), the set of polynomials with real coefficients under addition
The set of one-to-one and onto functions, \(f:A\rightarrow A\) for some set \(A\), under function composition.
\(\left( M_{(m,n)}(\mathbb{R}), + \right)\), the set of \(m \times n\) matrices under addition.
\(GL_n(\mathbb{R})\), the set of \(n \times n\) invertible matrices under multiplication.
\(SL(n,\mathbb{R})\), the set of \(n \times n\) invertible matrices, with determinant of \(1\), under multiplication.
If we extend our examples to the calculus curriculum, we can include the differentiable (or integrable) functions with the binary operation of addition. For the group of differentiable functions under addition, the identity is the function \(f_0:\mathbb{R}\rightarrow \mathbb{R}\) defined by \(f_0(x)=0\) for all \(x\). If \(g:\mathbb{R}\rightarrow \mathbb{R}\) is differentiable, then its inverse in this group would be defined by \(h(x)=-g(x)\).
Example 7.2 Let \[C= \left\{ f:\mathbb{R}\rightarrow \mathbb{R} \vert f(x)=a\cdot \cos(x)+b; \: a,b\in \mathbb{R}\right\}.\] We will now show that \((C,+)\) is a group.
Closure: Let \(f_1(x)=a \cos(x)+b\) and \(f_2(x)= c \cos(x)+d\), with \(a,b,c,d\in \mathbb{R}\), be elements of \(C\). Then \[(f_1+f_2)(x)=f_1(x) + f_2(x) = (a+c) \cos(x) + (b+d).\] Since \((\mathbb{R},+)\) is closed, we see that \(a+c, b+d \in \mathbb{R}\) and so \(f_1+f_2\in C\) and so \(C\) is closed under addition.
Identity: The function \(f_0(x)=0 \cdot \cos(x) + 0\) acts as the identity since \(f_0+f=f\) for all \(f\in C\).
Inverse: Let \(f(x)=a \cos(x)+b\) be a generic function in \(C\). Then we can define \((-f)(x)=(-a)\cdot \cos(x) + (-b)\). \(-f\in C\) and so \[\begin{align*} (f+ (-f))(x) &= \left(a\cdot \cos(x)+b\right) + \left((-a) \cdot \cos(x) + (-b)\right) \\ &= (a+(-a)) \cdot \cos(x) + (b+(-b)) \\ &= f_0(x) \end{align*}\] So each element of \(C\) has an inverse in \(C\).
The cardinality of each of the sets for the groups in Example 7.1 is at least countable. We can also have groups with a finite number of elements as shown in the following example.
Example 7.3 Let \(\mathbb{Z}_2=\{0,1\}\) with addition defined by the following table (often referred to as a Cayley table) \[\begin{array}{c|cc} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array}\]
Then the identity is \(0\) and each element is its own inverse.
Note that the complete definition of a group involves both the set \(G\) and the binary operation \(*\) and so the group is officially labeled as \((G,*)\). However, the binary operation is often inferred from the set itself. So we often just refer to the group as \(G\) with the operation implied.
When showing that a set and operation form a group, the most challenging property to show directly is often the associative property. Since almost all of the groups that we will encounter are derived from the elements and operations in the number systems constructed in Chapter 4, we will gain the associative property from those number systems.
7.1.1 Abelian Groups
In most of the above mentioned groups, the order of the operation does not matter. However, for groups such at \(GL_n(\mathbb{R})\), the order of the elements with the operation does matter.
Definition 7.2 (Abelian Groups) If a group \((G,*)\) also satisfies the property \(a*b=b*a, \: \forall a,b\in G\) (Commutative Property), then \((G,*)\) is called abelian.
Example 7.4 The non-abelian group whose set has the smallest cardinality is the dihedral group of order 6 with operation defined by the table: \[\begin{array}{c|cccccc} * & e & a & b & c & d & f \\ \hline e & e & a & b & c & d & f \\ a & a & e & d & f & b & c \\ b & b & f & e & d & c & a \\ c & c & d & f & e & a & b \\ d & d & c & a & b & f & e \\ f & f & b & c & a & e & d \\ \end{array}\]
These tables that define operations are called Cayley tables, first presented in a work by the British mathematician, Arthur Cayley (1854).
We can verify that this group is not abelian since \(a*b = d\) and \(b*a=f\).
It will be important to notice whether or not a group is abelian, as abelian groups have a simpler structure than those for which the operation does not commute.
7.1.2 Uniqueness of Identities and Inverses
The existence of identity and inverse elements does not guarantee uniqueness, a priori. However, the following two theorems verify the uniqueness of the identity and inverse elements within a group. This means that we can talk about the identity in a group instead of an identity.
Theorem 7.1 (Uniqueness of the Identity) For a group, \((G,*)\), the identity is unique.
Proof. Assume that there are two distinct identity elements that we will call \(e\) and \(\tilde{e}\). Then the following is true:\[\tilde{e}= e*\tilde{e} = e\] since both \(e\) and \(\tilde{e}\) are identities. Thus the identities are not distinct and so the identity is unique.
Theorem 7.2 (Uniqueness of the Inverse) For a group, \((G,*)\), for each \(a\in G\), the inverse of \(a\) is unique and so we call the inverse \(a^{-1}\).
Proof. Let \(a\in G\) and assume that there are distinct elements \(b,c \in G\) that are inverses of \(a\). Then \[b= b * (e)= b* (a*c)= (b*a)*c = (e)*c = c.\] Therefore, the inverse of \(a\) is unique.
7.1.3 Homomorphisms
Now that we have a better understanding of the definitions and structures of groups we turn our attention to looking at when two groups are really the `same’ group but viewed from different perspectives. To do that we must create a way to determine equivalence classes of groups that are based on a relationship between the two sets and how they interact with their binary operations.
Definition 7.3 A function, \(\phi\), from a group \((G,*)\) to a group \((G',\diamond)\) is called a homomorphism if \[\phi(a*b)=\phi(a) \diamond \phi(b)\] for every \(a,b\in G\).
In other words, a homomorphism is a function between groups that maintains the structure of the binary operation. Notice that this restricts the types of functions quite a bit since a simple function like \(f:\mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x)=x+1\) is not a homomorphism.
It will be helpful to understand this idea of maintaining the structure of the binary operation with a few examples of functions, some of which are homomorphisms and some are not.
Example 7.5 Let \(\phi: (\mathbb{Z},+) \rightarrow (\mathbb{Z},+)\) be defined by \(\phi(n)=2n\). In order to verify that \(\phi\) is a homomorphism, we see that for any integers \(m,n\) that \[\phi(m+n)= 2(m+n) = 2m + 2n = \phi(m)+\phi(n).\] So \(\phi\) is a homomorphism since \(\phi(m+n)=\phi(m)+\phi(n)\), maintaining the operation structure.
Example 7.6 Let \(f: (\mathbb{R}, +) \rightarrow (\mathbb{R}^+, \cdot)\) be defined by \(f(x)=2^x\). Then for real numbers \(x\) and \(y\) we have that \[f(x+y)=2^{(x+y)} = 2^x \cdot 2^y = f(x) \cdot f(y)\] making \(f\) a homomorphism.
Example 7.7 Let \(g: (\mathbb{R}, +) \rightarrow (\mathbb{R},+)\) be defined by \(g(x)=2x+3\). Then for real numbers \(x\) and \(y\), \[g(x+y) = 2\cdot (x+y)+3=2x+2y+3.\] However, \[g(x)+g(y)=(2x+3)+(2y+3)= 2x+2y+6.\] So \(g\) is not a homomorphism, since \(g(x+y)\neq g(x)+g(y)\).
For each of the examples above that generated a group homomorphism, notice that each of these homomorphisms mapped the identity element to the identity element and the mapped inverses to inverses. We see in the following two theorems that this is true for all homomorphisms.
Theorem 7.3 Let \(f:(G,*)\rightarrow (G',\cdot)\) be a group homomorphism. If \(e\) and \(e'\) are the identity elements of \((G,*)\) and \((G',\cdot)\), respectively, then \(f(e)=e'\).
Proof. Let \(a\) be any element of \(G\) and let \(e\) be the identity in \(G\). Then \[f(a) = f(a*e) = f(a) \cdot f(e)\] and so \[e' = f(a)^{-1}\cdot f(a) = f(a)^{-1}\cdot ( f(a) \cdot f(e) ) = (f(a)^{-1} \cdot f(a))\cdot f(e) = e' \cdot f(e) = f(e).\]
Theorem 7.4 Let \(f:(G,*)\rightarrow (G',\cdot)\) be a group homomorphism. Then for each \(a\in G\), \(f(a)^{-1} = f\left(a^{-1}\right)\).
Proof. Since \(f(a) \cdot f\left(a^{-1}\right) = f\left(a*a^{-1}\right) = f(e) = e'\) and since the inverse of an element is unique, we have that \(f(a)^{-1} = f\left(a^{-1}\right)\).
Just like with the functions in Chapter 5, we often find it useful to compose two homomorphisms. We see in the following theorem that such a composition is also a homomorphism.
Theorem 7.5 Let \(f: (A,*) \rightarrow (B,\times)\) and \(g:(B,\times) \rightarrow (C,\cdot)\) be two group homomorphisms. Then \[(g\circ f): (A,*) \rightarrow (C,\cdot)\] is a group homomorphism.
Proof. The proof of the theorem follows directly from the definition of \(f\) and \(g\) being group homomorphisms, with the difficulty lying in remembering the appropriate operation at each step in the process.
Let \(a_1\) and \(a_2\) be generic elements of \(A\). Then \[\begin{align*} (g\circ f)(a_1 * a_2) &= g \left( f(a_1*a_2)\right) = g\left( f(a_1)\times f(a_2)\right) \\ &= g(f(a_1)) \cdot g(f(a_2)) = (g\circ f)(a_1) \cdot (g\circ f)(a_2) \end{align*}\] and so \(g\circ f\) is a homomorphism.
7.1.4 Isomorphisms
When we combine the concept of a homomorphism with the idea of a bijection, we obtain a new type of function on groups.
Definition 7.4 A function \(\phi:(G,*)\rightarrow (G',\diamond)\) is called an isomorphism if it is a homomorphism and is also one-to-one and onto.
Since isomorphisms are bijections, they satisfy all of the same properties of bijections, including having an inverse. We also can deduce that an isomorphism can only exist between groups whose sets have the same cardinality. So there cannot exist an isomorphism from \((\mathbb{Z}_2,+)\) in Example 7.3 to the dihedral group of order 6 in Example 7.4.
Just as bijections create an equivalence relation in terms of cardinality, we will see that isomorphisms create an equivalence relation in terms of the group structure. In order to develop this equivalency it is helpful to understand the properties of the inverse function related to the group structures. As such, in the next theorem we prove that the inverse of an isomorphism is also an isomorphism.
Theorem 7.6 Let \(f:(G,*)\rightarrow (G',\cdot)\) be an isomorphism. Then \(f^{-1}:(G',\cdot)\rightarrow (G,*)\) is also an isomorphism.
Proof. Since \(f\) is a bijection, we know from Chapter 5 that \(f^{-1}\) is also a bijection. So in order to prove that \(f^{-1}\) is an isomorphism it suffices to prove that \(f^{-1}\) is a homomorphism.
If we let \(b_1\) and \(b_2\) be two generic elements of \(G'\). Then since \(f\) is a bijection, there exist unique \(a_1\) and \(a_2\) in \(G\) such that \(b_1=f(a_1)\) and \(b_2=f(a_2)\). Then \[\begin{align*} f^{-1}(b_1 \cdot b_2) &= f^{-1}\left( f(a_1)\cdot f(a_2)\right) \\ &= f^{-1} \left(f(a_1*a_2)\right) \: \mbox{ (since } f \mbox{ is a homomorphism)} \\ &= (f^{-1}\circ f) (a_1*a_2) = a_1*a_2 \\ &= f^{-1}(b_1)* f^{-1}(b_2) \end{align*}\] and so \(f^{-1}\) is a homomorphism, and since it is a bijection it is an isomorphism.
Theorem 7.7 We say that a group \((G,*)\) is isomorphic to a group \((G',\cdot)\) if there is an isomorphism from \((G,*)\) to \((G',\cdot)\). This relation of \((G,*)\) isomorphic to \((G',\cdot)\) is an equivalence relation.
Proof. Recall that in order to prove that a relation is an equivalence relation we have to prove that the relation is reflexive, symmetric, and transitive.
Reflexive. Let \((A,*)\) be a group and define \(\phi:(A,*)\rightarrow (A,*)\) by \(\phi(a)=a\) for all \(a\in A\). Since \(\phi\) is the identity function on \((A,*)\) we see that \(\phi\) is an isomorphism and so \((A,*)\) is isomorphic to itself.
Symmetric. Let \((A,*)\) and \((B,+)\) be groups such that \((A,*)\) is isomorphic to \((B,+)\). Let \(f:(A,*)\rightarrow (B,+)\) be such an isomorphism. From the previous theorem we know that \(f^{-1}: (B,+) \rightarrow (A,*)\) is also an isomorphism. Therefore, \((B,+)\) is isomorphic to \((A,*)\).
Transitive. The proof that isomorphisms are transitive follows from Theorem 5.3 that the composition of two bijections is a bijection and Theorem 7.5 that the composition of two homomorphisms is a homomorphism. Therefore, the composition of two isomorphisms is again an isomorphism. Therefore, group isomorphism creates a transitive relation.
The concept of isomorphism in algebra is similar to the concept of congruence in geometry where two triangles being congruent means that they share all of the related properties of triangles. Two groups are considered the “same” group if there is an isomorphism between them and are distinct only in the perspective in which one is viewing the groups. Such groups will be referred to as isomorphic groups. If \(G\) and \(G'\) are isomorphic groups, then we will denote this by \(G \cong G'\).
7.1.5 Finite Groups
To better understand this relationship between isomorphic groups, let’s look at groups with two elements. If \((G,*)\) is a group with two elements, the identity \(e\), and an element \(a\neq e\). We know that in order for \((G,*)\) to be a group, we know that each element must have a unique inverse. Since \(a*e=a\), we know that \(e\) is not the inverse of \(a\). So the inverse of \(a\) must be \(a\) and we have the Cayley table
\[\begin{array}{c|cc} * & e & a \\ \hline e & e & a \\ a & a & e \end{array}\]
We can then see that the function \(\phi:(\mathbb{Z}_2,+)\rightarrow (G,*)\) defined by \(\phi(0)=e\) and \(\phi(1)=a\), that \(\phi\) is an isomorphism and so every group with two elements is isomorphic to \((\mathbb{Z}_2,+)\).
In order to study groups with three elements, similarly to how we defined \((\mathbb{Z}_2,+)\) in Example 7.3, we can define \((\mathbb{Z}_3,+)\) as the set \(\{0,1,2\}\) with the operation defined by the Cayley table, \[\begin{array}{c|ccc} + & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array}\]
If we let \((G,*)\) be another group with three distinct elements, \(\{e,a,b\}\), with identity, \(e\), we know that \(e*a=a\), \(a*e=a\), \(e*b=b\), and \(b*e=b\) giving the first row and first column of the Cayley table,
\[\begin{array}{c|ccc} * & e & a & b \\ \hline e & e & a & b \\ a & a & & \\ b & b & & \end{array}\]
Since the group is closed under the operation, \(a*b\in G\). If \(a*b=a\), since \(a\) has an inverse, we have \(b=a^{-1}*(a*b) = a^{-1}*a = e\). Similarly, if \(a*b=b\), we would conclude that \(a=e\). Since the identity is unique, neither of these could be true. So \(a*b=e\) and \(b*a=e\), and we have filled in another two positions in the Cayley table. \[\begin{array}{c|ccc} * & e & a & b \\ \hline e & e & a & b \\ a & a & & e \\ b & b & e & \end{array}\]
We now need to determine \(a*a\) and \(b*b\). Since inverses are unique, we know that \(a*a\neq e\) and \(b*b\neq e\). If \(a*a=a\), we have that \(a = a*e=a*a*b=a*b=e\), which is not possible. So \(a*a=b\), and similarly, \(b*b=a\). So we see that \((G,*)\) has the Cayley table \[\begin{array}{c|ccc} * & e & a & b \\ \hline e & e & a & b \\ a & a & b & e \\ b & b & e & a \end{array}\]
We can now define a function \(\phi:(\mathbb{Z}_3,+) \rightarrow (G,*)\) by the correspondence \(0\mapsto e\), \(1\mapsto a\), and \(2\mapsto b\), and one can verify that such a function is an isomorphism. So we can now conclude that all groups with three elements are isomorphic to \((\mathbb{Z}_3,+)\). One cannot prove the same type of result for groups with four elements, since \((\mathbb{Z}_4,+)\) is not isomorphic to \((\mathbb{Z}_2 \times \mathbb{Z}_2, +)\).
\[\begin{array}{ccc} (\mathbb{Z}_4,+) & & (\mathbb{Z}_2 \times \mathbb{Z}_2, +) \\ \begin{array}{c|cccc} + & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \end{array} & & \begin{array}{c|cccc} + & (0,0) & (0,1) & (1,0) & (1,1) \\ \hline (0,0) & (0,0) & (0,1) & (1,0) & (1,1) \\ (0,1) & (0,1) & (0,0) & (1,1) & (1,0) \\ (1,0) & (1,0) & (1,1) & (0,0) & (0,1) \\ (1,1) & (1,1) & (1,0) & (0,1) & (0,0) \end{array} \end{array}\]
The classification of the groups with a finite number of elements was a significant endeavor throughout the twentieth century, with a complete classification verified in the early twenty-first century using computers to check the validity of the proof.
7.1.6 Exercises
For each of the groups listed in Example 7.1:
What is the identity element?
For a generic element \(a\), what would \(a^{-1}\) look like?
Is the group abelian?
In what ways are the groups related to one another? (Think about subsets or homomorphisms)
Find a homomorphism from \(\left(\mathbb{Z}, +\right)\) to \(\left(5\mathbb{Z}, +\right)\).
Are \((\mathbb{Z},+)\) and \((\mathbb{R},+)\) isomorphic groups? Why or why not?
Let \((G,*)\) and \((H,*)\) be two groups such that \(H\subseteq G\) and the operation is the same for both groups.
What are some possible homomorphisms \(\phi:(H,*)\rightarrow (G,*)\)?
What are some possible homomorphisms \(\psi:(G,*) \rightarrow (H,*)\)?
Let \(\phi:(G,*)\rightarrow (H,\diamond)\) be a group homomorphism.
- Define \[\mbox{Ker}(\phi)= \left\{g\in G \: \vert \: \phi(g)=e_H\right\}\] and prove that \(\mbox{Ker}(\phi)\) is a group.
- Define \[\mbox{Ran}(\phi) = \left\{h\in H\: \vert \: h=\phi(g) \mbox{ for some } g\in G\right\}\] and prove that \(\mbox{Ran}(\phi)\) is a group.