
## Section3.3Subgroups

### SubsectionDefinitions and Examples

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 $H\text{,}$ $H$ 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, ${\mathbb R}^*\text{,}$ 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

\begin{equation*} {\mathbb Q}^* = \{ p/q : p \, {\rm and }\, q\, {\rm are\, nonzero\, integers} \} \end{equation*}

is a subgroup of ${\mathbb R}^*\text{.}$ 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 ${\mathbb Q}^*\text{.}$ 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.

###### Example3.25

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 $H \subset {\mathbb C}^{\ast}\text{.}$

###### 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 $ad - bc = 1\text{.}$ 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 $A\text{:}$

\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 $G\text{.}$ For $H$ to be a subgroup of $G$ it must inherit $G$'s binary operation. The set of all $2 \times 2$ matrices, ${\mathbb M}_2(\mathbb R)\text{,}$ 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

\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 $GL_2( {\mathbb R })\text{.}$

###### 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 ${\mathbb Z}_2\text{,}$ 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, $(a,b) + (c,d) = (a+c, b+d)\text{.}$ Table 3.29 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.

### SubsectionSome Subgroup Theorems

Let us examine some criteria for determining exactly when a subset of a group is a subgroup.

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

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

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

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