Definitions and Examples

The integers \({\mathbb Z } = \{ \ldots , -1, 0, 1, 2, \ldots \}\) form a group under the operation of addition. The binary operation on two integers \(m, n \in {\mathbb Z}\) is just their sum. Since the integers under addition already have a well-established notation, we will use the operator \(+\) instead of \(\circ\); that is, we shall write \(m + n\) instead of \(m \circ n\). The identity is \(0\), and the inverse of \(n \in {\mathbb Z}\) is written as \(-n\) instead of \(n^{-1}\). Notice that the set of integers under addition have the additional property that \(m + n = n + m\) and therefore form an abelian group.

The integers mod \(n\) form a group under addition modulo \(n\). Consider \({\mathbb Z}_5\), consisting of the equivalence classes of the integers \(0\), \(1\), \(2\), \(3\), and \(4\). We define the group operation on \({\mathbb Z}_5\) by modular addition. We write the binary operation on the group additively; that is, we write \(m + n\). The element \(0\) is the identity of the group and each element in \({\mathbb Z}_5\) has an inverse. For instance, \(2 + 3 = 3 + 2 = 0\). is a Cayley table for \({\mathbb Z}_5\). By , \({\mathbb Z}_n = \{0, 1, \ldots, n-1 \}\) is a group under the binary operation of addition mod \(n\).

Not every set with a binary operation is a group. For example, if we let modular multiplication be the binary operation on \({\mathbb Z}_n\), then \({\mathbb Z}_n\) fails to be a group. The element 1 acts as a group identity since \(1 \cdot k = k \cdot 1 = k\) for any \(k \in {\mathbb Z}_n\); however, a multiplicative inverse for \(0\) does not exist since \(0 \cdot k = k \cdot 0 = 0\) for every \(k\) in \({\mathbb Z}_n\). Even if we consider the set \({\mathbb Z}_n \setminus \{0 \}\), we still may not have a group. For instance, let \(2 \in {\mathbb Z}_6\). Then \(2\) has no multiplicative inverse since \[\begin{aligned}0 \cdot 2 & = 0 \qquad 1 \cdot 2 = 2 \\ 2 \cdot 2 & = 4 \qquad 3 \cdot 2 = 0 \\ 4 \cdot 2 & = 2 \qquad 5 \cdot 2 = 4\end{aligned}\]. By , every nonzero \(k\) does have an inverse in \({\mathbb Z}_n\) if \(k\) is relatively prime to \(n\). Denote the set of all such nonzero elements in \({\mathbb Z}_n\) by \(U(n)\). \(U(n)\) group of units in \(\mathbb Z_n\) Then \(U(n)\) is a group called the group of units of \({\mathbb Z}_n\). is a Cayley table for the group \(U(8)\).

The symmetries of an equilateral triangle described in form a nonabelian group. As we observed, it is not necessarily true that \(\alpha \beta = \beta \alpha\) for two symmetries \(\alpha\) and \(\beta\). Using , which is a Cayley table for this group, we can easily check that the symmetries of an equilateral triangle are indeed a group. We will denote this group by either \(S_3\) or \(D_3\), for reasons that will be explained later.

We use \({\mathbb M}_2 ( {\mathbb R})\) to denote the set of all \(2 \times 2\) matrices. Let \(GL_2({\mathbb R})\) be the subset of \({\mathbb M}_2 ( {\mathbb R})\) consisting of invertible matrices; that is, a matrix \(\mathbb M_n(\mathbb R)\) the \(n \times n\) matrices with entries in \(\mathbb R\) \[\begin{aligned}\end{aligned}\] is in \(GL_2( {\mathbb R})\) if there exists a matrix \(A^{-1}\) such that \(A A^{-1} = A^{-1} A = I\), where \(I\) is the \(2 \times 2\) identity matrix. For \(A\) to have an inverse is equivalent to requiring that the determinant of \(A\) be nonzero; that is, \(\det A = ad - bc \neq 0\). \(\det A\) the determinant of \(A\) The set of invertible matrices forms a group called the general linear group. \(GL_n(\mathbb R)\) the general linear group The identity of the group is the identity matrix \[\begin{aligned}\end{aligned}\]. The inverse of \(A \in GL_2( {\mathbb R})\) is \[\begin{aligned}\end{aligned}\]. The product of two invertible matrices is again invertible. Matrix multiplication is associative, satisfying the other group axiom. For matrices it is not true in general that \(AB = BA\); hence, \(GL_2({\mathbb R})\) is another example of a nonabelian group.

Let \[\begin{aligned}1 & = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \qquad I = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \\ J & = \begin{pmatrix} 0 & i\\ i & 0 \end{pmatrix} \qquad K = \begin{pmatrix} i & 0\\ 0 & -i \end{pmatrix}\end{aligned}\], where \(i^2 = -1\). Then the relations \(I^2 = J^2 = K^2 = -1\), \(IJ=K\), \(JK = I\), \(KI = J\), \(JI = -K\), \(KJ = -I\), and \(IK = -J\) hold. The set \(Q_8\) the group of quaternions \(Q_8 = \{\pm 1, \pm I, \pm J, \pm K \}\) is a group called the quaternion group. Notice that \(Q_8\) is noncommutative.

Let \({\mathbb C}^\ast\) be the set of nonzero complex numbers. \(\mathbb C^*\) the multiplicative group of complex numbers Under the operation of multiplication \({\mathbb C}^\ast\) forms a group. The identity is \(1\). If \(z = a+bi\) is a nonzero complex number, then \[\begin{aligned}\end{aligned}\] is the inverse of \(z\). It is easy to see that the remaining group axioms hold.