2: Introduction to Groups

Definition 2.1: Groups

A group is an ordered pair \((G,*)\) where \(G\) is a set and \(*\) is a binary operation on \(G\) satisfying the following properties

1. \(x*(y*z) = (x*y)*z\) for all \(x\), \(y\), \(z\) in \(G\).
2. There is an element \(e \in G\) satisfying \(e*x=x\) and \(x*e=x\) for all \(x\) in \(G\).
3. For each element \(x\) in \(G\) there is an element \(y\) in \(G\) satisfying \(x*y = e\) and \(y*x=e\).

Thus, to describe a group one must specify two things:

1. a set, and
2. a binary operation on the set.

Then, one must verify that the binary operation is associative, that there is an identity in the set, and that every element in the set has an inverse.

Convention If it is clear what the binary operation is, then the group \((G,*)\) may be referred to by its underlying set \(G\) alone.

Examples of Groups:

1. \((\mathbb{Z},+)\) is a group with identity 0. The inverse of \(x \in \mathbb{Z}\) is \(-x\).
2. \((\mathbb{Q},+)\) is a group with identity 0. The inverse of \(x \in \mathbb{Q}\) is \(-x\).
3. \((\mathbb{R},+)\) is a group with identity 0. The inverse of \(x \in \mathbb{R}\) is \(-x\).
4. \((\mathbb{Q}-\{0\},\cdot)\) is a group with identity 1. The inverse of \(x \in \mathbb{Q}-\{0\}\) is \(x^{-1}\).
5. \((\mathbb{R}-\{0\},\cdot)\) is a group with identity 1. The inverse of \(x \in \mathbb{R}-\{0\}\) is \(x^{-1}\).
6. \((\mathbb{Z}_n,+)\) is a group with identity 0. The inverse of \(x \in \mathbb{Z}_n\) is \(n-x\) if \(x \ne 0\), the inverse of 0 is 0. See Corollary C.5 in Appendix C for a proof that this binary operation is associative.
7. \((\mathbb{R}^n,+)\) where \(+\) is vector addition. The identity is the zero vector \((0,0,\dots,0)\) and the inverse of the vector \(\mathbf{x}=(x_1,x_2,\dots,x_n)\) is the vector \(\mathbf{-x}=(-x_1,-x_2,\dots,-x_n)\).
8. \((\mathbb{Z}_2^n, +)\) where \(+\) is vector addition modulo 2. The identity is the zero vector \((0,0,\dots,0)\) and the inverse of the vector \(\mathbf{x}\) is the vector itself.
9. \((M_2(K),+)\) where \(K\) is any one of \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{Z}_n\) is a group whose identity is the zero matrix \[\nonumber \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right ]\] and the inverse of the matrix \[A=\left [ \nonumber \begin{array} {c c} a & b \\ c & d \end{array} \right]\] is the matrix \[\nonumber -A = \left [ \begin{array} {c c} -a & -b \\ -c & -d \end{array} \right].\]

Note that the binary operations in the above examples are all commutative. For historical reasons, there is a special name for such groups:

Definition 2.2: Abelian

A group \((G,*)\) is said to be abelian if \(x*y=y*x\) for all \(x\) and \(y\) in \(G\). A group is said to be non-abelian if it is not abelian.

Examples of Non-Abelian Groups:

1. For each \(n \in \mathbb{N}\), the set \(S_n\) of all permutations on \([n]= \{1,2,\dots, n\}\) is a group under compositions of functions. This is called the symmetric group of degree \(n\). We discuss this group in detail in the next chapter. The group \(S_n\) is non-abelian if \(n \ge 3\).
2. Let \(K\) be any one of \(\mathbb{Q}, \mathbb{R}\) or \(\mathbb{Z}_p\), where \(p\) is a prime number. Define \(GL(2,K)\) to be the set of all matrices in \(M_2(K)\) with non-zero determinant. Then \((GL(2,K), \cdot)\) is a group. Here \(\cdot\) represents matrix multiplication. The identity of \(GL(2,K)\) is the identity matrix \[\left[ \begin{array}{cc} 1&0\\0&1 \end{array} \right ] \nonumber\] and the inverse of \[\nonumber \left[ \begin{array}{cc} a&b\\c&d \end{array} \right ]\] is \[\nonumber \left[ \begin{array}{cc} \frac d{ad-bc}&\frac {-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc} \end{array} \right ].\]

Math Joke
Question: What’s purple and commutes? (For the answer see end of page.)

Theorem \(\PageIndex{1}\)

If \((G,*)\) is a group then:

(a) The identity of \(G\) is unique.

(b) The inverse of each element in \(G\) is unique. \(\blacksquare\)

Problem 2.1 Prove Theorem 2.1. Hints: To establish (a) assume that \(e\) and \(e'\) are identities of \(G\) and prove that \(e = e'\). [This was done in the previous chapter, but do it again anyhow.] To establish (b) assume that \(x\) and \(y\) are both inverses of some element \(a \in G\). Use the group axioms to prove that \(x = y\). Show carefully how each axiom is used. Don’t skip any steps.

Now we can speak of the identity of a group and the inverse of an element of a group. Since the inverse of \(a \in G\) is unique, the following definition makes sense:

Definition 2.3:

Let \((G,*)\) be a group. Let \(a\) be any element of \(G\). We define \(a^{-1}\) to be the inverse of \(a\) in the group \(G\).

The above definition is used when we think of the group’s operation as being a type of multiplication or product. If instead the operation is denoted by \(+\), we have instead the following definition.

Definition 2.4:

Let \((G,+)\) be a group. Let \(a\) be any element of \(G\). We define \(-a\) to be the inverse of \(a\) in the group \(G\).

Theorem \(\PageIndex{2}\)

Let \((G,*)\) be a group with identity \(e\). Then the following hold for all elements \(a,b,c,d\) in \(G\):

1. If \(a*c=a*b\), then \(c=b\).
2. If \(c*a=b*a\), then \(c=b\).
3. Given \(a\) and \(b\) in \(G\) there is a unique element \(x\) in \(G\) such that \(a*x=b\).
4. Given \(a\) and \(b\) in \(G\) there is a unique element \(x\) in \(G\) such that \(x*a=b\).
5. If \(a*b=e\) then \(a=b^{-1}\) and \(b=a^{-1}\). [Characterization of the inverse of an element.]
6. If \(a*b=a\) for just one \(a\), then \(b = e\).
7. If \(b*a=a\) for just one \(a\), then \(b = e\).
8. If \(a*a=a\), then \(a=e\). [The only idempotent in a group is the identity.]
9. \((a^{-1})^{-1} = a\).
10. \((a*b)^{-1}=b^{-1}*a^{-1}\).

Problem 2.2 Prove Theorem 2.2.

Problem 2.3 Restate Theorem 2.2 for a group \((G,+)\) with identity 0. (See Definition 2.4.)

Problem 2.4 Give a specific example of a group and two specific elements \(a\) and \(b\) in the group such that \((a*b)^{-1} \neq a^{-1}*b^{-1}\).

Problem 2.5 Let \(*\) be an associative binary operation on the set \(S\) and let \(a,b,c,d \in S\). Prove the following statements. [Be careful what you assume.]

1. \((a*b)*(c*d) =((a*b)*c)*d\).
2. \((a*b)*(c*d) = a*(b*(c*d))\).
3. In 1. and 2. we see three different ways to properly place parentheses in the product: \(a*b*c*d\)? Find all possible ways to properly place parentheses in the product \(a*b*c*d\) and show that all lead to the same element in \(S\).

Theorem \(\PageIndex{3}\)

Let \(*\) be an associative binary operation on a set \(S\). If \(a_1, a_2, \dots, a_n\) is a sequence of \(n \ge 3\) elements of \(S\), then the product \[a_1*a_2* \cdots*a_n\] is unambiguous; that is, the same element will be obtained regardless of how parentheses are inserted in the product (in a legal manner).

Proof: The case \(n=3\) is just the associative law itself. The case \(n=4\) is established in Problem 2.3. The general case can be proved by induction on \(n\). The details are quite technical, so to save time, we will omit them. One of the problems is stating precisely what is meant by “inserting the parentheses in a legal manner”. The interested reader can find a proof in most introductory abstract algebra books. See for example Chapter 1.4 of the book Basic Algebra I by Nathan Jacobson.

Remark

From now on, unless stated to the contrary, we will assume the Generalized Associative Law. That is, we will place parentheses in a product at will without a detailed justification. Note, however, the order may still be important, so unless the binary operation is commutative we must still pay close attention to the order of the elements in a product or sum.

Problem 2.6 Show that if \(a_1, a_2, a_3\) are elements of a group then \[\nonumber (a_1*a_2*a_3)^{-1}=a_3^{-1}*a_2^{-1}*a_1^{-1}.\] Show that in general if \(n \in \mathbb{N}\) and \(a_1, a_2, \dots, a_n\) are elements of a group then \[\nonumber (a_1*a_2*\cdots*a_n)^{-1}=a_n^{-1}*\cdots*a_2^{-1}*a_1^{-1}.\]

Now that we have the Generalized Associative Law, we can define \(a^n\) for \(n \in \mathbb{Z}\).

Definition 2.5:

Let \((G,*)\) be a group with identity \(e\). Let \(a\) be any element of \(G\). We define integral powers \(a^n\), \(n\in\mathbb{Z}\), as follows: \[\begin{aligned} a^0 &=& e \\ a^1 &=& a \\ a^{-1} &=& \mbox{the inverse of $a$} \end{aligned}\] and for \(n \ge 2\):
\(a^n=a^{n-1}*a\)
\(a^{-n}=(a^{-1})^n\)

Using this definition, it is easy to establish the following important theorem.

Theorem \(\PageIndex{4}\) (Laws of Exponents for Groups)

Let \((G,*)\) be a group with identity \(e\). Then for all \(n,m \in \mathbb{Z}\) we have \[a^n*a^m = a^{n+m} \quad \mbox{ for all $a \in G,$}\] \[(a^n)^m = a^{nm} \quad \mbox{ for all $a \in G,$}\] and whenever \(a,b \in G\) and \(a*b=b*a\) we have \[(a*b)^n = a^n*b^n. \rule{6pt}{6pt} \blacksquare\]

This theorem is easy to check for \(n, m \in \mathbb{N}\). A complete proof for \(n, m \in \mathbb{Z}\) involves a number of cases and is a little tedious, but the following problem gives some indication of how this could be done.

Problem 2.7 Let \((G,*)\) be a group with identity \(e\). Prove using Definition 2.3 the following special cases of Theorem 2.3. For \(a, b \in G\):

1. \(a^2*a^3 =a^5.\)
2. \(a^2*a^{-6} = a^{-4}.\)
3. \(a^{-2}*a^6=a^4.\)
4. \(a^{-2}*a^{-3} = a^{-5}\)
5. \(a^{-2}*a^{2} = a^0\).
6. Assuming \(a*b=b*a\), \(a^3*b^3=(a*b)^3\).
7. Assuming \(a*b=b*a\), \(a^{-3}*b^{-3}=(a*b)^{-3}\).

Problem 2.8 Restate Definition 2.3 for additive notation. (In this case \(a^n\) is replaced by \(na\).)

Problem 2.9 Restate Theorem 2.3 for a group whose operation is \(+\).

Answer to the joke: An abelian grape.