Abstract Algebra: Theory and Applications, Remixed!

Example 2.24.

Consider the set of nonzero real numbers, \({\mathbb R}^*\text{,}\) with the group operation of multiplication. The identity of this group is \(1\) and the inverse of any element \(a \in {\mathbb R}^*\) is just \(1/a\text{.}\) We will show that

is a subgroup of \({\mathbb R}^*\text{.}\) The identity of \({\mathbb R}^*\) is \(1\text{;}\) however, \(1 = 1/1\) is the quotient of two nonzero integers. Hence, the identity of \({\mathbb R}^*\) is in \({\mathbb Q}^*\text{.}\) Given two elements in \({\mathbb Q}^*\text{,}\) say \(p/q\) and \(r/s\text{,}\) their product \(pr/qs\) is also in \({\mathbb Q}^*\text{.}\) The inverse of any element \(p/q \in {\mathbb Q}^*\) is again in \({\mathbb Q}^*\) since \((p/q)^{-1} = q/p\text{.}\) Since multiplication in \({\mathbb R}^*\) is associative, multiplication in \({\mathbb Q}^*\) is associative.

Example 2.25.

Recall that \({\mathbb C}^{\ast}\) is the multiplicative group of nonzero complex numbers. Let \(H = \{ 1, -1, i, -i \}\text{.}\) Then \(H\) is a subgroup of \({\mathbb C}^{\ast}\text{.}\) It is quite easy to verify that \(H\) is a group under multiplication and that \(H \subset {\mathbb C}^{\ast}\text{.}\)

Example 2.26.

Let \(SL_2( {\mathbb R})\) be the subset of \(GL_2( {\mathbb R })\) consisting of matrices of determinant one; that is, a matrix

is in \(SL_2( {\mathbb R})\) exactly when \(ad - bc = 1\text{.}\) To show that \(SL_2( {\mathbb R})\) is a subgroup of the general linear group, we must show that it is a group under matrix multiplication. The \(2 \times 2\) identity matrix is in \(SL_2( {\mathbb R})\text{,}\) as is the inverse of the matrix \(A\text{:}\)

It remains to show that multiplication is closed; that is, that the product of two matrices of determinant one also has determinant one. We will leave this task as an exercise. The group \(SL_2({\mathbb R})\) is called the special linear group.

Example 2.27.

It is important to realize that a subset \(H\) of a group \(G\) can be a group without being a subgroup of \(G\text{.}\) For \(H\) to be a subgroup of \(G\text{,}\) it must inherit the binary operation of \(G\text{.}\) The set of all \(2 \times 2\) matrices, \({\mathbb M}_2(\mathbb R)\text{,}\) forms a group under the operation of addition. The \(2 \times 2\) general linear group is a subset of \({\mathbb M}_2(\mathbb R)\) and is a group under matrix multiplication, but it is not a subgroup of \({\mathbb M}_2(\mathbb R)\text{.}\) If we add two invertible matrices, we do not necessarily obtain another invertible matrix. Observe that

but the zero matrix is not in \(GL_2( {\mathbb R })\text{.}\)

Example 2.28.

One way of telling whether or not two groups are the same is by examining their subgroups. Other than the trivial subgroup and the group itself, the group \({\mathbb Z}_4\) has a single subgroup consisting of the elements \(0\) and \(2\text{.}\) From the group \({\mathbb Z}_2\text{,}\) we can form another group of four elements as follows. As a set this group is \({\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\) We perform the group operation coordinatewise; that is, \((a,b) + (c,d) = (a+c, b+d)\text{.}\) Figure 2.29 is an addition table for \({\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\) Since there are three nontrivial proper subgroups of \({\mathbb Z}_2 \times {\mathbb Z}_2\text{,}\) \(H_1 = \{ (0,0), (0,1) \}\text{,}\) \(H_2 = \{ (0,0), (1,0) \}\text{,}\) and \(H_3 = \{ (0,0), (1,1) \}\text{,}\) \({\mathbb Z}_4\) and \({\mathbb Z}_2 \times {\mathbb Z}_2\) must be different groups.