PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

<section xml:id="section-subgroups">
    <title>Subgroups</title>
    <subsection>
      <title>Definitions and Examples</title>
      <p>
        Sometimes we wish to investigate smaller groups sitting inside a larger group.
        The set of even integers <m>2{\mathbb Z} = \{\ldots, -2, 0, 2, 4, \ldots \}</m> is a group under the operation of addition.
        This smaller group sits naturally inside of the group of integers under addition.
        We define a <term>subgroup</term><idx><h>Subgroup</h><h>definition of</h></idx> <m>H</m> of a group <m>G</m> to be a subset <m>H</m> of <m>G</m> such that when the group operation of <m>G</m> is restricted to <m>H</m>,
        <m>H</m> is a group in its own right.
        Observe that every group <m>G</m> 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 <m>H = \{ e \}</m> of a group <m>G</m> is called the
        <term>trivial subgroup</term>.
            <idx><h>Subgroup</h><h>trivial</h></idx>
        A subgroup that is a proper subset of <m>G</m> is called a
        <term>proper subgroup</term>.
            <idx><h>Subgroup</h><h>proper</h></idx>
        In many of the examples that we have investigated up to this point,
        there exist other subgroups besides the trivial and improper subgroups.
      </p>
      <example xml:id="example-groups-multiplicative-subgroup">
        <title>A Subgroup of the Reals</title>
        <p>
          Consider the set of nonzero real numbers,
          <m>{\mathbb R}^*</m>, with the group operation of multiplication.
          <notation>
            <usage><m>\mathbb R^*</m></usage>
            <description>the multiplicative group of real numbers</description>
          </notation>
          The identity of this group is 1 and the inverse of any element <m>a \in {\mathbb R}^*</m> is just <m>1/a</m>.
          We will show that
          <me>
            {\mathbb Q}^* = \{ p/q : p\text{ and }q\text{ are nonzero integers}\}
          </me>
          is a subgroup of <m>{\mathbb R}^*</m>.
          <notation>
            <usage><m>\mathbb Q^*</m></usage>
            <description>the multiplicative group of rational numbers</description>
          </notation>
          The identity of <m>{\mathbb R}^*</m> is 1;
          however,  <m>1 = 1/1</m> is the quotient of two nonzero integers.
          Hence, the identity of <m>{\mathbb R}^*</m> is in <m>{\mathbb Q}^*</m>.
          Given two elements in <m>{\mathbb Q}^*</m>,
          say <m>p/q</m> and <m>r/s</m>,
          their product <m>pr/qs</m> is also in <m>{\mathbb Q}^*</m>.
          The inverse of any element <m>p/q \in {\mathbb Q}^*</m> is again in
          <m>{\mathbb Q}^*</m> since <m>(p/q)^{-1} = q/p</m>.
          Since multiplication in <m>{\mathbb R}^*</m> is associative,
          multiplication in <m>{\mathbb Q}^*</m> is associative.
        </p>
      </example>
      <example xml:id="example-groups-subgroup-h">
        <title>A Subgroup of the Nonzero Complex Numbers</title>
        <p>
          Recall that <m>{\mathbb C}^{\ast}</m> is the multiplicative group of nonzero complex numbers.
          Let <m>H = \{ 1, -1, i, -i \}</m>.
          Then <m>H</m> is a subgroup of <m>{\mathbb C}^{\ast}</m>.
          It is quite easy to verify that <m>H</m> is a group under multiplication and that <m>H \subset {\mathbb C}^{\ast}</m>.
        </p>
      </example>
      <example xml:id="example-groups-sl2">
        <title>A Subgroup of Matrices With Determinant One</title>
        <p>
          Let <m>SL_2( {\mathbb R})</m> be the subset of
          <m>GL_2( {\mathbb R })</m>consisting of matrices of determinant one;
          that is, a matrix
          <notation>
            <usage><m>SL_n(\mathbb R)</m></usage>
            <description>the special linear group</description>
          </notation>
          <me>
            A = \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix}
          </me>
          is in <m>SL_2( {\mathbb R})</m> exactly when <m>ad - bc = 1</m>.
          To show that <m>SL_2( {\mathbb R})</m> is a subgroup of the general linear group,
          we must show that it is a group under matrix multiplication.
          The <m>2 \times 2</m> identity matrix is in <m>SL_2( {\mathbb R})</m>,
          as is the inverse of the matrix <m>A</m>:
          <me>
            A^{-1} = \begin{pmatrix} d  &amp; -b \\ -c &amp;  a \end{pmatrix}
          </me>.
          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 <m>SL_2({\mathbb R})</m> is called the
          <term>special linear group</term>.
              <idx><h>Group</h><h>special linear</h></idx>
        </p>
      </example>
      <example xml:id="example-groups-gl2-subgroup">
        <title>Groups, Subsets, Operations</title>
        <p>
          It is important to realize that a subset <m>H</m> of a group <m>G</m> can be a group without being a subgroup of <m>G</m>.
          For <m>H</m> to be a subgroup of <m>G</m> it must inherit <m>G</m>'s binary operation.
          The set of all <m>2 \times 2</m> matrices,
          <m>{\mathbb M}_2(\mathbb R)</m>,
          forms a group under the operation of addition.
          The <m>2 \times 2</m> general linear group is a subset of
          <m>{\mathbb M}_2(\mathbb R)</m> and is a group under matrix multiplication,
          but it is not a subgroup of <m>{\mathbb M}_2(\mathbb R)</m>.
          If we add two invertible matrices,
          we do not necessarily obtain another invertible matrix.
          Observe that
          <me>
            \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} + \begin{pmatrix} -1 &amp;  0 \\ 0  &amp; -1 \end{pmatrix} = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 0 \end{pmatrix}
          </me>,
          but the zero matrix is not in <m>GL_2( {\mathbb R })</m>.
        </p>
      </example>
      <example xml:id="example-groups-z2xz2">
        <title>Structurally Different Groups</title>
        <p>
          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 <m>{\mathbb Z}_4</m> has a single subgroup consisting of the elements 0 and 2.
          From the group <m>{\mathbb Z}_2</m>,
          we can form another group of four elements as follows.
          As a set this group is <m>{\mathbb Z}_2 \times {\mathbb Z}_2</m>.
          We perform the group operation coordinatewise;
          that is, <m>(a,b) + (c,d) = (a+c, b+d)</m>.
          <xref ref="figure-z2xz2" /> is an addition table for <m>{\mathbb Z}_2 \times {\mathbb Z}_2</m>.
          Since there are three nontrivial proper subgroups of <m>{\mathbb Z}_2 \times {\mathbb Z}_2</m>,
          <m>H_1 = \{ (0,0), (0,1) \}</m>,
          <m>H_2 = \{ (0,0), (1,0) \}</m>,
          and <m>H_3 = \{ (0,0), (1,1) \}</m>,
          <m>{\mathbb Z}_4</m> and <m>{\mathbb Z}_2 \times {\mathbb Z}_2</m> must be different groups.
        </p>
        <figure xml:id="figure-z2xz2">
          <caption>Addition table for <m>{\mathbb Z}_2 \times {\mathbb Z}_2</m></caption>
          <p>
            <me>
              \begin{array}{c|cccc} +     &amp; (0,0) &amp; (0,1) &amp; (1,0) &amp; (1,1) \\ \hline (0,0) &amp; (0,0) &amp; (0,1) &amp; (1,0) &amp; (1,1) \\ (0,1) &amp; (0,1) &amp; (0,0) &amp; (1,1) &amp; (1,0) \\ (1,0) &amp; (1,0) &amp; (1,1) &amp; (0,0) &amp; (0,1) \\ (1,1) &amp; (1,1) &amp; (1,0) &amp; (0,1) &amp; (0,0) \end{array}
            </me>
          </p>
        </figure>
      </example>
    </subsection>
    <subsection>
      <title>Some Subgroup Theorems</title>
      <p>
        Let us examine some criteria for determining exactly when a subset of a group is a subgroup.
      </p>
      <proposition>
        <statement>
          <p>
            A subset <m>H</m> of <m>G</m> is a subgroup if and only if it satisfies the following conditions.
            <ol>
              <li>
                <p>
                  The identity <m>e</m> of <m>G</m> is in <m>H</m>.
                </p>
              </li>
              <li>
                <p>
                  If <m>h_1, h_2 \in H</m>, then <m>h_1h_2 \in H</m>.
                </p>
              </li>
              <li>
                <p>
                  If <m>h \in H</m>, then <m>h^{-1} \in H</m>.
                </p>
              </li>
            </ol>
          </p>
        </statement>
        <proof>
          <p>
            First suppose that <m>H</m> is a subgroup of <m>G</m>.
            We must show that the three conditions hold.
            Since <m>H</m> is a group, it must have an identity <m>e_H</m>.
            We must show that <m>e_H = e</m>,
            where <m>e</m> is the identity of <m>G</m>.
            We know that <m>e_H e_H = e_H</m> and that <m>ee_H = e_H e = e_H</m>;
            hence, <m>ee_H = e_H e_H</m>.
            By right-hand cancellation, <m>e =e_H</m>.
            The second condition holds since a subgroup <m>H</m> is a group.
            To prove the third condition, let <m>h \in H</m>.
            Since <m>H</m> is a group,
            there is an element <m>h' \in H</m> such that <m>hh' = h'h = e</m>.
            By the uniqueness of the inverse in <m>G</m>, <m>h' = h^{-1}</m>.
          </p>
          <p>
            Conversely, if the three conditions hold,
            we must show that <m>H</m> is a group under the same operation as <m>G</m>;
            however, these conditions plus the associativity of the binary operation are exactly the axioms stated in the definition of a group.
          </p>
        </proof>
      </proposition>
      <proposition xml:id="proposition-subgroup">
        <statement>
          <p>
            Let <m>H</m> be a subset of a group <m>G</m>.
            Then <m>H</m> is a subgroup of <m>G</m> if and only if <m>H \neq \emptyset</m>,
            and whenever <m>g, h \in H</m> then <m>gh^{-1}</m> is in <m>H</m>.
          </p>
        </statement>
        <proof>
          <p>
            First assume that <m>H</m> is a subgroup of <m>G</m>.
            We wish to show that <m>gh^{-1} \in H</m> whenever <m>g</m> and <m>h</m> are in <m>H</m>.
            Since <m>h</m> is in <m>H</m>,
            its inverse <m>h^{-1}</m> must also be in <m>H</m>.
            Because of the closure of the group operation, <m>gh^{-1} \in H</m>.
          </p>
          <p>
            Conversely, suppose that <m>H \subset G</m> such that <m>H \neq \emptyset</m> and
            <m>g h^{-1} \in H</m> whenever <m>g, h \in H</m>.
            If <m>g \in H</m>, then <m>gg^{-1} = e</m> is in <m>H</m>.
            If <m>g \in H</m>, then <m>eg^{-1} = g^{-1}</m> is also in <m>H</m>.
            Now let <m>h_1, h_2 \in H</m>.
            We must show that their product is also in <m>H</m>.
            However, <m>h_1(h_2^{-1})^{-1} = h_1 h_2 \in H</m>.
            Hence, <m>H</m> is a subgroup of <m>G</m>.
          </p>
        </proof>
      </proposition>



      <remark xml:base="sage/groups-info.xml">
  <title>Sage</title>
  <p>
    The first half of this text is about group theory.
    Sage includes Groups, Algorithms and Programming (GAP), a program designed primarly for just group theory,
    and in continuous development since 1986.
    Many of Sage's computations for groups ultimately are performed by GAP.
  </p>
</remark>
    </subsection>
  </section>