Sometimes we wish to investigate smaller groups sitting inside a larger group. The set of even integers \(2{\mathbb Z} = \{\ldots, -2, 0, 2, 4, \ldots \}\) is a group under the operation of addition. This smaller group sits naturally inside of the group of integers under addition. We define a subgroup \(H\) of a group \(G\) to be a subset \(H\) of \(G\) such that when the group operation of \(G\) is restricted to \(H\), \(H\) is a group in its own right. Observe that every group \(G\) with at least two elements will always have at least two subgroups, the subgroup consisting of the identity element alone and the entire group itself. The subgroup \(H = \{ e \}\) of a group \(G\) is called the trivial subgroup. A subgroup that is a proper subset of \(G\) is called a proper subgroup. In many of the examples that we have investigated up to this point, there exist other subgroups besides the trivial and improper subgroups.

Example3.24

Consider the set of nonzero real numbers, \({\mathbb R}^*\), 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 \(1/a\). We will show that
\begin{equation*}{\mathbb Q}^* = \{ p/q : p \, {\rm and }\, q\, {\rm are\, nonzero\, integers} \}\end{equation*}
is a subgroup of \({\mathbb R}^*\). The identity of \({\mathbb R}^*\) is 1; however, \(1 = 1/1\) is the quotient of two nonzero integers. Hence, the identity of \({\mathbb R}^*\) is in \({\mathbb Q}^*\). Given two elements in \({\mathbb Q}^*\), say \(p/q\) and \(r/s\), their product \(pr/qs\) is also in \({\mathbb Q}^*\). The inverse of any element \(p/q \in {\mathbb Q}^*\) is again in \({\mathbb Q}^*\) since \((p/q)^{-1} = q/p\). Since multiplication in \({\mathbb R}^*\) is associative, multiplication in \({\mathbb Q}^*\) is associative.

Example3.25

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

Example3.26

Let \(SL_2( {\mathbb R})\) be the subset of \(GL_2( {\mathbb R })\)consisting of matrices of determinant one; that is, a matrix
\begin{equation*}A =
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}\end{equation*}
is in \(SL_2( {\mathbb R})\) exactly when \(ad - bc = 1\). 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 \(SL_2( {\mathbb R})\), as is the inverse of the matrix \(A\):
\begin{equation*}A^{-1} =
\begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}.\end{equation*}
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.

Example3.27

It is important to realize that a subset \(H\) of a group \(G\) can be a group without being a subgroup of \(G\). For \(H\) to be a subgroup of \(G\) it must inherit \(G\)'s binary operation. The set of all \(2 \times 2\) matrices, \({\mathbb M}_2(\mathbb R)\), 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 \({\mathbb M}_2(\mathbb R)\). If we add two invertible matrices, we do not necessarily obtain another invertible matrix. Observe that\begin{equation*}
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
+
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix},
\end{equation*}
but the zero matrix is not in \(GL_2( {\mathbb R })\).

Example3.28

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. From the group \({\mathbb Z}_2\), we can form another group of four elements as follows. As a set this group is \({\mathbb Z}_2 \times {\mathbb Z}_2\). We perform the group operation coordinatewise; that is, \((a,b) + (c,d) = (a+c, b+d)\). Table 3.29 is an addition table for \({\mathbb Z}_2 \times {\mathbb Z}_2\). Since there are three nontrivial proper subgroups of \({\mathbb Z}_2 \times {\mathbb Z}_2\), \(H_1 = \{ (0,0), (0,1) \}\), \(H_2 = \{ (0,0), (1,0) \}\), and \(H_3 = \{ (0,0), (1,1) \}\), \({\mathbb Z}_4\) and \({\mathbb Z}_2 \times {\mathbb Z}_2\) must be different groups.

First suppose that \(H\) is a subgroup of \(G\). We must show that the three conditions hold. Since \(H\) is a group, it must have an identity \(e_H\). We must show that \(e_H = e\), where \(e\) is the identity of \(G\). We know that \(e_H e_H = e_H\) and that \(ee_H = e_H e = e_H\); hence, \(ee_H = e_H e_H\). By right-hand cancellation, \(e =e_H\). The second condition holds since a subgroup \(H\) is a group. To prove the third condition, let \(h \in H\). Since \(H\) is a group, there is an element \(h' \in H\) such that \(hh' = h'h = e\). By the uniqueness of the inverse in \(G\), \(h' = h^{-1}\).

Conversely, if the three conditions hold, we must show that \(H\) is a group under the same operation as \(G\); however, these conditions plus the associativity of the binary operation are exactly the axioms stated in the definition of a group.

Proposition3.31

Let \(H\) be a subset of a group \(G\). Then \(H\) is a subgroup of \(G\) if and only if \(H \neq \emptyset\), and whenever \(g, h \in H\) then \(gh^{-1}\) is in \(H\).

First assume that \(H\) is a subgroup of \(G\). We wish to show that \(gh^{-1} \in H\) whenever \(g\) and \(h\) are in \(H\). Since \(h\) is in \(H\), its inverse \(h^{-1}\) must also be in \(H\). Because of the closure of the group operation, \(gh^{-1} \in H\).

Conversely, suppose that \(H \subset G\) such that \(H \neq \emptyset\) and \(g h^{-1} \in H\) whenever \(g, h \in H\). If \(g \in H\), then \(gg^{-1} = e\) is in \(H\). If \(g \in H\), then \(eg^{-1} = g^{-1}\) is also in \(H\). Now let \(h_1, h_2 \in H\). We must show that their product is also in \(H\). However, \(h_1(h_2^{-1})^{-1} = h_1 h_2 \in H\). Hence, \(H\) is a subgroup of \(G\).