normal.xml

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<chapter xmlns:xi="http://www.w3.org/2001/XInclude" xml:id="normal">
  <title>Normal Subgroups and Factor Groups</title>
  <introduction>
    <p>
      If <m>H</m> is a subgroup of a group <m>G</m>,
      then right cosets are not always the same as left cosets;
      that is, it is not always the case that <m>gH = Hg</m> for all <m>g \in G</m>.
      The subgroups for which this property holds play a critical role in group theory<mdash/>they allow for the construction of a new class of groups,
      called factor or quotient groups.
      Factor groups may be studied directly or by using homomorphisms,
      a generalization of isomorphisms.
      We will study homomorphisms in <xref ref="homomorph"/>.
    </p>
  </introduction>
    <!-- % TWJ, 2010/03/31 -->
    <!-- % The chapter HOMOMORPHISMS AND FACTOR GROUPS is now -->
    <!-- % two chapters: (10) NORMAL SUBGROUPS AND FACTOR GROUPS -->
    <!-- % (11) HOMOMORPHISMS -->
  <section xml:id="normal-section-factor-groups">
    <title>Factor Groups and Normal Subgroups</title>
    <subsection xml:id="normal-subsection-normal-subgroups">
      <title>Normal Subgroups</title>
      <p>
        A subgroup <m>H</m> of a group <m>G</m> is <term>normal</term><idx><h>Subgroup</h><h>normal</h></idx><idx><h>Normal subgroup</h></idx>
        in G if <m>gH = Hg</m> for all <m>g \in G</m>.
        That is, a normal subgroup of a group <m>G</m> is one in which the right and left cosets are precisely the same.
      </p>

      <example xml:id="normal-example-abelian">
        <p>
          Let <m>G</m> be an abelian group.
          Every subgroup <m>H</m> of <m>G</m> is a normal subgroup.
          Since <m>gh = hg</m> for all <m>g \in G</m> and <m>h \in H</m>,
          it will always be the case that <m>gH = Hg</m>.
        </p>
      </example>

      <example xml:id="normal-example-s3">
        <p>
          Let <m>H</m> be the subgroup of <m>S_3</m> consisting of elements <m>(1)</m> and <m>(12)</m>.
          Since
          <me>
            (123) H = \{ (1 \, 2 \, 3), (1 \, 3) \} \quad \text{and} \quad H (1 \, 2 \, 3) = \{ (1 \, 2 \, 3), (2 \, 3) \}
          </me>,
          <m>H</m> cannot be a normal subgroup of <m>S_3</m>.
          However, the subgroup <m>N</m>,
          consisting of the permutations <m>(1)</m>, <m>(1 \, 2 \, 3)</m>,
          and <m>(1 \, 3 \, 2)</m>, is normal since the cosets of <m>N</m> are
          <md>
            <mrow>N  =   \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}</mrow>
            <mrow>(1 \, 2) N =  N (1 \, 2)  =  \{ (1 \, 2), (1 \, 3), (2 \, 3) \}</mrow>
          </md>.
        </p>
      </example>

      <p>
        The following theorem is fundamental to our understanding of normal subgroups.
      </p>

      <theorem xml:id="normal-theorem-normal-equivalents">
        <statement>
          <p>
            Let <m>G</m> be a group and <m>N</m> be a subgroup of <m>G</m>.
            Then the following statements are equivalent.

            <ol>
              <li>
                <p>
                  The subgroup <m>N</m> is normal in <m>G</m>.
                </p>
              </li>

              <li>
                <p>
                  For all <m>g \in G</m>, <m>gNg^{-1} \subset N</m>.
                </p>
              </li>

              <li>
                <p>
                  For all <m>g \in G</m>, <m>gNg^{-1} = N</m>.
                </p>
              </li>
            </ol>
          </p>
        </statement>

        <proof>
          <p>
            (1) <m>\Rightarrow</m> (2).
            Since <m>N</m> is normal in <m>G</m>,
            <m>gN = Ng</m> for all <m>g \in G</m>.
            Hence, for a given <m>g \in G</m> and <m>n \in N</m>,
            there exists an <m>n'</m> in <m>N</m> such that <m>g n = n' g</m>.
            Therefore, <m>gng^{-1} = n' \in N</m> or <m>gNg^{-1} \subset N</m>.
          </p>

          <p>
            (2) <m>\Rightarrow</m> (3).
            Let <m>g \in G</m>.
            Since <m>gNg^{-1} \subset N</m>,
            we need only show <m>N \subset gNg^{-1}</m>.
            For <m>n \in N</m>, <m>g^{-1}ng=g^{-1}n(g^{-1})^{-1} \in N</m>.
            Hence, <m>g^{-1}ng = n'</m> for some <m>n' \in N</m>.
            Therefore, <m>n = g n' g^{-1}</m> is in <m>g N g^{-1}</m>.
          </p>

          <p>
            (3) <m>\Rightarrow</m> (1).
            Suppose that <m>gNg^{-1} = N</m> for all <m>g \in G</m>.
            Then for any <m>n \in N</m> there exists an
            <m>n' \in N</m> such that <m>gng^{-1} = n'</m>.
            Consequently, <m>gn = n' g</m> or <m>gN \subset Ng</m>.
            Similarly, <m>Ng \subset gN</m>.
          </p>
        </proof>
      </theorem>
    </subsection>

    <subsection xml:id="normal-subsection-factor-groups">
      <title>Factor Groups</title>
      <p>
        If <m>N</m> is a normal subgroup of a group <m>G</m>,
        then the cosets of <m>N</m> in <m>G</m> form a group <m>G/N</m> under the operation <m>(aN) (bN) = abN</m>.
        This group is called the <term>factor</term><idx><h>Group</h><h>factor</h></idx>
        or <term>quotient group</term><idx><h>Group</h><h>quotient</h></idx>
        of <m>G</m> and <m>N</m>.
        <notation>
          <usage><m>G/N</m></usage>
          <description>factor group of <m>G</m> mod <m>N</m></description>
        </notation>
        Our first task is to prove that <m>G/N</m> is indeed a group.
      </p>

      <theorem>
        <statement>
          <p>
            Let <m>N</m> be a normal subgroup of a group <m>G</m>.
            The cosets of <m>N</m> in <m>G</m> form a group <m>G/N</m> of order <m>[G:N]</m>.
          </p>
        </statement>

        <proof>
          <p>
            The group operation on <m>G/N</m> is <m>(a N ) (b N)= a b N</m>.
            This operation must be shown to be well-defined;
            that is, group multiplication must be independent of the choice of  coset representative.
            Let <m>aN = bN</m> and <m>cN = dN</m>.
            We must show that
            <me>
              (aN) (cN) = acN = bd N = (b N)(d N)
            </me>.
            Then <m>a = b n_1</m> and <m>c = d n_2</m> for some <m>n_1</m> and <m>n_2</m> in <m>N</m>.
            Hence,
            <md>
              <mrow>acN &amp; = b n_1 d n_2 N</mrow>
              <mrow>&amp; = b n_1 d N</mrow>
              <mrow>&amp; = b n_1 N d</mrow>
              <mrow>&amp; = b N d</mrow>
              <mrow>&amp; = b d N</mrow>
            </md>.
            The remainder of the theorem is easy:
            <m>eN = N</m> is the identity and
            <m>g^{-1} N</m> is the inverse of <m>gN</m>.
            The order of <m>G/N</m> is,
            of course, the number of cosets of <m>N</m> in <m>G</m>.
          </p>
        </proof>
      </theorem>

      <p>
        It is very important to remember that the elements in a factor group are
        <em>sets of elements</em> in the original group.
      </p>

      <example xml:id="normal-example-factor-s3">
        <p>
          Consider the normal subgroup of <m>S_3</m>,
          <m>N = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}</m>.
          The cosets of <m>N</m> in <m>S_3</m> are <m>N</m> and <m>(12) N</m>.
          The factor group <m>S_3 / N</m> has the following multiplication table.
        </p>

        <p>
          <me>
            \begin{array}{c|cc}
            &amp; N &amp; (1 \, 2) N \\ \hline
            N &amp; N &amp; (1 \, 2) N \\
            (1 \, 2) N &amp; (1 \, 2) N &amp; N
            \end{array}
          </me>
        </p>

        <p>
          This group is isomorphic to <m>{\mathbb Z}_2</m>.
          At first, multiplying cosets seems both complicated and strange;
          however, notice that <m>S_3 / N</m> is a smaller group.
          The factor group displays a certain amount of information about <m>S_3</m>.
          Actually, <m>N = A_3</m>, the group of even permutations,
          and <m>(1 \, 2) N = \{ (1 \, 2), (1 \, 3), (2 \, 3) \}</m> is the set of odd permutations.
          The information captured in <m>G/N</m> is parity;
          that is, multiplying two even or two odd permutations results in an even permutation,
          whereas multiplying an odd permutation by an even permutation yields an odd permutation.
        </p>
      </example>

      <example xml:id="normal-example-factor-z3">
        <p>
          Consider the normal subgroup
          <m>3 {\mathbb Z}</m> of <m>{\mathbb Z}</m>.
          The cosets of <m>3 {\mathbb Z}</m> in <m>{\mathbb Z}</m> are
          <md>
            <mrow>0 + 3 {\mathbb Z} &amp; = \{ \ldots, -3, 0, 3, 6, \ldots \}</mrow>
            <mrow>1 + 3 {\mathbb Z} &amp; = \{ \ldots, -2, 1, 4, 7, \ldots \}</mrow>
            <mrow>2 + 3 {\mathbb Z} &amp; = \{ \ldots, -1, 2, 5, 8, \ldots \}</mrow>
          </md>.
          The group <m>{\mathbb Z}/ 3 {\mathbb Z}</m> is given by the Cayley table below.
        </p>

        <p>
          <me>
            \begin{array}{c|ccc}
            + &amp; 0 + 3{\mathbb Z} &amp; 1 + 3{\mathbb Z} &amp; 2 + 3{\mathbb Z} \\\hline
            0 + 3{\mathbb Z} &amp; 0 + 3{\mathbb Z} &amp; 1 + 3{\mathbb Z} &amp; 2 + 3{\mathbb Z} \\
            1 + 3{\mathbb Z} &amp; 1 + 3{\mathbb Z} &amp; 2 + 3{\mathbb Z} &amp; 0 + 3{\mathbb Z} \\
            2 + 3{\mathbb Z} &amp; 2 + 3{\mathbb Z} &amp; 0 + 3{\mathbb Z} &amp; 1 + 3{\mathbb Z}
            \end{array}
          </me>
        </p>

        <p>
          In general, the subgroup <m>n {\mathbb Z}</m> of <m>{\mathbb Z}</m> is normal.
          The cosets of <m>{\mathbb Z } / n {\mathbb Z}</m> are
          <md>
            <mrow>n {\mathbb Z}</mrow>
            <mrow>1 + n {\mathbb Z}</mrow>
            <mrow>2 + n {\mathbb Z}</mrow>
            <mrow>\vdots</mrow>
            <mrow>(n-1) + n {\mathbb Z}</mrow>
          </md>.
          The sum of the cosets <m>k + n{\mathbb Z}</m> and
          <m>l + n{\mathbb Z}</m> is <m>k+l + n{\mathbb Z}</m>.
          Notice that we have written our cosets additively,
          because the group operation is integer addition.
        </p>
      </example>

      <example xml:id="normal-example-factor_dn">
        <p>
          Consider the dihedral group <m>D_n</m>,
          generated by the two elements <m>r</m> and <m>s</m>, satisfying the relations
          <md>
            <mrow>r^n &amp; = \identity</mrow>
            <mrow>s^2 &amp; = \identity</mrow>
            <mrow>srs &amp; = r^{-1}</mrow>
          </md>.
          The element <m>r</m> actually generates the cyclic subgroup of rotations,
          <m>R_n</m>, of <m>D_n</m>.
          Since <m>srs^{-1} = srs = r^{-1} \in R_n</m>,
          the group of rotations is a normal subgroup of <m>D_n</m>;
          therefore, <m>D_n / R_n</m> is a group.
          Since there are exactly two elements in this group,
          it must be isomorphic to <m>{\mathbb Z}_2</m>.
        </p>
      </example>
    </subsection>

  </section>

  <section xml:id="normal-section-simplicity-of-an">
    <title>The Simplicity of the Alternating Group</title>
    <introduction>
      <p>
        Of special interest are groups with no nontrivial normal subgroups.
        Such groups are called <idx><h>Group</h><h>simple</h></idx><idx><h>Simple group</h></idx><term>simple groups</term>.
        Of course, we already have a whole class of examples of simple groups,
        <m>{\mathbb Z}_p</m>, where <m>p</m> is prime.
        These groups are trivially simple since they have no proper subgroups other than the subgroup consisting solely of the identity.
        Other examples of simple groups are not so easily found.
        We can, however, show that the alternating group,
        <m>A_n</m>, is simple for <m>n \geq 5</m>.
        The proof of this result requires several lemmas.
      </p>

      <lemma xml:id="normal-lemma-1-normal-3cycle">
        <statement>
          <p>
            The alternating group <m>A_n</m> is generated by <m>3</m>-cycles for <m>n \geq 3</m>.
          </p>
        </statement>

        <proof>
          <p>
            To show that the <m>3</m>-cycles generate <m>A_n</m>,
            we need only show that any pair of transpositions can be written as the product of <m>3</m>-cycles.
            Since <m>(a, b) = (b, a)</m>,
            every pair of transpositions must be one of the following:
            <md>
              <mrow>(a,b)(a,b) &amp; = \identity</mrow>
              <mrow>(a,b)(c,d) &amp; = (a,c,b)(a,c,d)</mrow>
              <mrow>(a,b)(a,c) &amp; = (a,c,b)</mrow>
            </md>.
          </p>
        </proof>
      </lemma>

      <lemma xml:id="normal-lemma-2-normal-3cycle">
        <statement>
          <p>
            Let <m>N</m> be a normal subgroup of <m>A_n</m>, where <m>n \geq 3</m>.
            If <m>N</m> contains a <m>3</m>-cycle, then <m>N = A_n</m>.
          </p>
        </statement>

        <proof>
          <p>
            We will first show that <m>A_n</m> is generated by <m>3</m>-cycles of the specific form <m>(i,j,k)</m>,
            where <m>i</m> and <m>j</m> are fixed in 
            <m>\{ 1, 2, \ldots,
            n \}</m> and we let <m>k</m> vary.
            Every <m>3</m>-cycle is the product of <m>3</m>-cycles of this form, since
            <md>
              <mrow>(i, a, j) &amp; = (i, j, a)^2</mrow>
              <mrow>(i, a, b) &amp; = (i, j, b) (i, j, a)^2</mrow>
              <mrow>(j, a, b) &amp; = (i, j, b)^2 (i, j, a)</mrow>
              <mrow>(a, b, c) &amp; = (i, j, a)^2 (i, j, c) (i, j, b)^2 (i, j, a)</mrow>
            </md>.
            Now suppose that <m>N</m> is a nontrivial normal subgroup of <m>A_n</m> for
            <m>n  \geq 3</m> such that <m>N</m> contains a <m>3</m>-cycle of the form <m>(i, j, a)</m>.
            Using the normality of <m>N</m>, we see that
            <me>
              [(i, j)(a, k)](i, j, a)^2 [(i, j)(a, k)]^{-1} = (i, j, k)
            </me>
            is in <m>N</m>.
            Hence, <m>N</m> must contain all of the <m>3</m>-cycles <m>(i, j, k)</m>  for <m>1 \leq k \leq n</m>.
            By <xref ref="normal-lemma-1-normal-3cycle"/>, these <m>3</m>-cycles generate <m>A_n</m>;
            hence, <m>N = A_n</m>.
          </p>
        </proof>
      </lemma>

      <lemma xml:id="normal-lemma-3-normal-3cycle">
        <statement>
          <p>
            For <m>n \geq 5</m>,
            every nontrivial normal subgroup <m>N</m> of <m>A_n</m> contains a <m>3</m>-cycle.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>\sigma</m> be an arbitrary element in a normal subgroup <m>N</m>.
            There are several possible cycle structures for <m>\sigma</m>.

            <ul>
              <li>
                <p>
                  <m>\sigma</m> is a <m>3</m>-cycle.
                </p>
              </li>

              <li>
                <p>
                  <m>\sigma</m> is the product of disjoint cycles,
                  <m>\sigma = \tau(a_1, a_2, \ldots, a_r) \in N</m>, where <m>r \gt 3</m>.
                </p>
              </li>

              <li>
                <p>
                  <m>\sigma</m> is the product of disjoint cycles,
                  <m>\sigma = \tau(a_1, a_2, a_3)(a_4, a_5, a_6)</m>.
                </p>
              </li>

              <li>
                <p>
                  <m>\sigma = \tau(a_1, a_2, a_3)</m>,
                  where <m>\tau</m> is the product of disjoint 2-cycles.
                </p>
              </li>

              <li>
                <p>
                  <m>\sigma = \tau (a_1, a_2) (a_3, a_4)</m>,
                  where <m>\tau</m> is the product of an even number of disjoint 2-cycles.
                </p>
              </li>
            </ul>

            If <m>\sigma</m> is a <m>3</m>-cycle, then we are done.
            If <m>N</m> contains a product of disjoint cycles, <m>\sigma</m>,
            and at least one of these cycles has length greater than 3, say <m>\sigma = \tau(a_1, a_2, \ldots, a_r)</m>, then
            <me>
              (a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}
            </me>
            is in <m>N</m> since <m>N</m> is normal; hence,
            <me>
              \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}
            </me>
            is also in <m>N</m>.
            Since
            <md>
              <mrow>\sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} &amp; = \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_3, a_2)</mrow>
              <mrow>&amp; = (a_1, a_2, \ldots, a_r)^{-1}\tau^{-1}(a_1, a_2, a_3) \tau(a_1, a_2, \ldots, a_r)(a_1, a_3, a_2)</mrow>
              <mrow>&amp; = (a_1, a_r, a_{r-1}, \ldots, a_2 )(a_1, a_2, a_3) (a_1, a_2, \ldots, a_r)(a_1, a_3, a_2)</mrow>
              <mrow>&amp; = (a_1, a_3, a_r)</mrow>
            </md>,
            <m>N</m> must contain a <m>3</m>-cycle;
            hence, <m>N = A_n</m>.
          </p>

          <p>
            Now suppose that <m>N</m> contains a disjoint product of the form
            <me>
              \sigma = \tau(a_1, a_2, a_3)(a_4, a_5, a_6)
            </me>.
            Then
            <me>
              \sigma^{-1}(a_1, a_2, a_4)\sigma(a_1, a_2, a_4)^{-1} \in N
            </me>
            since
            <me>
              (a_1, a_2, a_4)\sigma(a_1, a_2, a_4)^{-1} \in N
            </me>.
            So
            <md>
              <mrow>\sigma^{-1}(a_1, a_2, a_4) \sigma(a_1, a_2, a_4)^{-1} &amp; = [ \tau (a_1, a_2, a_3) (a_4, a_5, a_6) ]^{-1} (a_1, a_2, a_4) \tau (a_1, a_2, a_3) (a_4, a_5, a_6) (a_1, a_2, a_4)^{-1}</mrow>
              <mrow>&amp; = (a_4, a_6, a_5) (a_1, a_3, a_2) \tau^{-1}(a_1, a_2, a_4) \tau (a_1, a_2, a_3) (a_4, a_5, a_6) (a_1, a_4, a_2)</mrow>
              <mrow>&amp; = (a_4, a_6, a_5)(a_1, a_3, a_2) (a_1, a_2, a_4) (a_1, a_2, a_3) (a_4, a_5, a_6)(a_1, a_4, a_2)</mrow>
              <mrow>&amp; = (a_1, a_4, a_2, a_6, a_3)</mrow>
            </md>.
            So <m>N</m> contains a disjoint cycle of length greater than 3, and we can apply the previous case.
          </p>

          <p>
            Suppose <m>N</m> contains a disjoint product of the form <m>\sigma = \tau(a_1, a_2, a_3)</m>,
            where <m>\tau</m> is the product of disjoint 2-cycles.
            Since <m>\sigma \in N</m>, <m>\sigma^2 \in N</m>, and
            <md>
              <mrow>\sigma^2 &amp; = \tau(a_1, a_2, a_3)\tau(a_1, a_2, a_3)</mrow>
              <mrow>&amp; =(a_1, a_3, a_2)</mrow>
            </md>.
            So <m>N</m> contains a <m>3</m>-cycle.
          </p>

          <p>
            The only remaining possible case is a disjoint product of the form
            <me>
              \sigma = \tau (a_1, a_2) (a_3, a_4)
            </me>,
            where <m>\tau</m> is the product of an even number of disjoint <m>2</m>-cycles.
            But
            <me>
              \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}
            </me>
            is in <m>N</m> since <m>(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}</m> is in <m>N</m>; and so
            <md>
              <mrow>\sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} &amp; = \tau^{-1} (a_1, a_2) (a_3, a_4) (a_1, a_2, a_3) \tau (a_1, a_2)(a_3, a_4)(a_1, a_2, a_3)^{-1}</mrow>
              <mrow>&amp; = (a_1, a_3)(a_2, a_4)</mrow>
            </md>.
            Since <m>n \geq 5</m>, we can find
            <m>b \in \{1, 2, \ldots,
            n \}</m> such that <m>b \neq a_1, a_2, a_3, a_4</m>.
            Let <m>\mu = (a_1, a_3, b)</m>.
            Then
            <me>
              \mu^{-1} (a_1, a_3)(a_2, a_4) \mu (a_1, a_3)(a_2, a_4) \in N
            </me>
            and
            <md>
              <mrow>\mu^{-1} (a_1, a_3)(a_2, a_4) \mu (a_1, a_3)(a_2, a_4) &amp; = (a_1, b a_3)(a_1, a_3)(a_2, a_4) (a_1, a_3, b)(a_1, a_3)(a_2, a_4)</mrow>
              <mrow>&amp; = (a_1 a_3 b )</mrow>
            </md>.
            Therefore, <m>N</m> contains a <m>3</m>-cycle.
            This completes the proof of the lemma.
          </p>
        </proof>
      </lemma>
            <!-- TWJ - 1/13/2014 -->
            <!-- nontrivial added to the lemma statement.  Suggested by M. Faucette. -->
      <theorem xml:id="normal-theorem-an-simple">
        <statement>
          <p>
            The alternating group, <m>A_n</m>,
            is simple for <m>n \geq 5</m>.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>N</m> be a normal subgroup of <m>A_n</m>.
            By <xref ref="normal-lemma-3-normal-3cycle"/>, <m>N</m> contains a <m>3</m>-cycle.
            By <xref ref="normal-lemma-2-normal-3cycle"/>, <m>N = A_n</m>;
            therefore, <m>A_n</m> contains no proper nontrivial normal subgroups for <m>n \geq 5</m>.
          </p>
        </proof>
      </theorem>

      <paragraphs component="sage-blurb">
        <title>Sage</title>
        <p>
          Sage can easily determine if a subgroup is normal or not.
          If so, it can create the quotient group.
          However, the construction creates a new permuation group,
          isomorphic to the quotient group,
          so its utility is limited.
        </p>
      </paragraphs>
    </introduction>

    <subsection xml:id="normal-subsection-historical-note">
      <title>Historical Note</title>
      <p>
        One of the foremost problems of group theory has been to classify all <idx><h>Group</h><h>simple</h></idx>simple finite groups.
        This problem is over a century old and has been solved only in the last few decades of the twentieth century.
        In a sense, finite simple groups are the building blocks of all finite groups.
        The first nonabelian simple groups to be discovered were the alternating groups.
        Galois was the first to prove that <m>A_5</m> was simple.
        Later, mathematicians such as C. Jordan<idx><h>Jordan, C.</h></idx>
        and L. E. Dickson<idx><h>Dickson, L. E.</h></idx>
        found several infinite families of matrix groups that were simple.
        Other families of simple groups were discovered in the 1950s.
        At the turn of the century, William Burnside<idx><h>Burnside, William</h></idx>
        conjectured that all nonabelian simple groups must have even order.
        In 1963, W. Feit<idx><h>Feit, W.</h></idx>
        and J. Thompson<idx><h>Thompson, J.</h></idx>
        proved Burnside's conjecture and published their results in the paper
        <q>Solvability of Groups of Odd Order,</q>
        which appeared in the <pubtitle>Pacific Journal of Mathematics</pubtitle>.
        Their proof, running over 250 pages,
        gave impetus to a program in the 1960s and 1970s to classify all finite simple groups.
        Daniel Gorenstein<idx><h>Gorenstein, Daniel</h></idx>
        was the organizer of this remarkable effort.
        One of the last simple groups was the
        <q>Monster,</q> discovered by <idx><h>Greiss, R.</h></idx>R. Greiss.
        The Monster, a <m>196{,}833 \times 196{,}833</m> matrix group,
        is one of the 26 sporadic, or special, simple groups.
        These sporadic simple groups are groups that fit into no infinite family of simple groups.
        Some of the sporadic groups play an important role in physics.
      </p>
    </subsection>

  </section>

    <reading-questions>

    <exercise label="rq-normal-equilateral">
      <statement>
        <p>
          Let <m>G</m> be the group of symmetries of an equilateral triangle,
          expressed as permutations of the vertices numbered <m>1,2,3</m>.
          Let <m>H</m> be the subgroup <m>H=\langle (1\,2) \rangle</m>.
          Build the left and right cosets of <m>H</m> in <m>G</m>.
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-normal-is-normal">
      <statement>
        <p>
          Based on your answer to the previous question,
          is <m>H</m> normal in <m>G</m>?
          Explain why or why not.
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-normal-8Z">
      <statement>
        <p>
          The subgroup <m>8\mathbb Z</m> is normal in <m>\mathbb Z</m>.
          In the factor group <m>\mathbb Z/8\mathbb Z</m>
          perform the computation <m>(3+8\mathbb Z)+(7+8\mathbb Z)</m>.
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-normal-equivalent">
      <statement>
        <p>
          List two statements about a group <m>G</m> and a subgroup <m>H</m> that are equivalent to
          <q><m>H</m> is normal in <m>G</m>.</q>
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-normal-factor">
      <statement>
        <p>
          In your own words, what is a factor group?
        </p>
      </statement>
      <response/>
    </exercise>

  </reading-questions>
    <!-- Exercises with Solutions  -->
    <!-- File: normal.xml  -->
    <!-- Title: Normal Subgroups and Factor Groups -->
  <exercises xml:id="normal-exercises" filenamebase="normal">
    <title>Exercises</title>

    <exercise number="1" xml:id="normal-exercise-normal-subgroup">
      <statement>
        <p>
          For each of the following groups <m>G</m>,
          determine whether <m>H</m> is a normal subgroup of <m>G</m>.
          If <m>H</m> is a normal subgroup,
          write out a Cayley table for the factor group <m>G/H</m>.

          <ol>
            <li>
              <p>
                <m>G = S_4</m> and <m>H = A_4</m>
              </p>
            </li>

            <li>
              <p>
                <m>G = A_5</m> and <m>H = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}</m>
              </p>
            </li>

            <li>
              <p>
                <m>G = S_4</m> and <m>H = D_4</m>
              </p>
            </li>

            <li>
              <p>
                <m>G = Q_8</m> and <m>H = \{ 1, -1, I, -I \}</m>
              </p>
            </li>

            <li>
              <p>
                <m>G = {\mathbb Z}</m> and <m>H = 5 {\mathbb Z}</m>
              </p>
            </li>
          </ol>
        </p>
      </statement>
      <hint>
        <p>
          (a)
          <me>
            \begin{array}{c|cc}
            &amp; A_4 &amp; (1 \, 2)A_4 \\ \hline
            A_4 &amp; A_4 &amp; (1 \, 2) A_4 \\
            (1 \, 2) A_4 &amp; (1 \, 2) A_4 &amp; A_4
            \end{array}
          </me>
        </p>

        <p>
          (c) <m>D_4</m> is not normal in <m>S_4</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="2" xml:id="normal-exercise-D4">
      <statement>
        <p>
          Find all the subgroups of <m>D_4</m>.
          Which subgroups are normal?
          What are all the factor groups of <m>D_4</m> up to isomorphism?
        </p>
      </statement>
    </exercise>

    <exercise number="3" xml:id="normal-exercise-Q8">
      <statement>
        <p>
          Find all the subgroups of the quaternion group, <m>Q_8</m>.
          Which subgroups are normal?
          What are all the factor groups of <m>Q_8</m> up to isomorphism?
        </p>
      </statement>
    </exercise>

    <exercise number="4" xml:id="normal-exercise-triangular-matrices">
      <statement>
        <p>
          Let <m>T</m> be the group of nonsingular upper triangular
          <m>2 \times 2</m> matrices with entries in <m>{\mathbb R}</m>;
          that is, matrices of the form
          <me>
            \begin{pmatrix}
            a &amp; b \\
            0 &amp; c
            \end{pmatrix}
          </me>,
          where <m>a</m>,
          <m>b</m>, <m>c \in {\mathbb R}</m> and <m>ac \neq 0</m>.
          Let <m>U</m> consist of matrices of the form
          <me>
            \begin{pmatrix}
            1 &amp; x \\
            0 &amp; 1
            \end{pmatrix}
          </me>,
          where <m>x \in {\mathbb R}</m>.

          <ol>
            <li>
              <p>
                Show that <m>U</m> is a subgroup of <m>T</m>.
              </p>
            </li>

            <li>
              <p>
                Prove that <m>U</m> is abelian.
              </p>
            </li>

            <li>
              <p>
                Prove that <m>U</m> is normal in <m>T</m>.
              </p>
            </li>

            <li>
              <p>
                Show that <m>T/U</m> is abelian.
              </p>
            </li>

            <li>
              <p>
                Is <m>T</m> normal in <m>GL_2( {\mathbb R})</m>?
              </p>
            </li>
          </ol>
        </p>
      </statement>
    </exercise>

    <exercise number="5" xml:id="normal-exercise-intersection">
      <statement>
        <p>
          Show that the intersection of two normal subgroups is a normal subgroup.
        </p>
      </statement>
    </exercise>

    <exercise number="6" xml:id="normal-exercise-abelian">
      <statement>
        <p>
          If <m>G</m> is abelian, prove that <m>G/H</m> must also be abelian.
        </p>
      </statement>
    </exercise>

    <exercise number="7" xml:id="normal-exercise-H-GmodH">
      <statement>
        <p>
          Prove or disprove: If <m>H</m> is a normal subgroup of <m>G</m> such that <m>H</m> and <m>G/H</m> are abelian,
          then <m>G</m> is abelian.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand the solution to #8.-->

    <exercise number="8" xml:id="normal-exercise-cyclic">
      <statement>
        <p>
          If <m>G</m> is cyclic, prove that <m>G/H</m> must also be cyclic.
        </p>
      </statement>
      <hint>
        <p>
          If <m>a \in G</m> is a generator for <m>G</m>,
          then <m>aH</m> is a generator for <m>G/H</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="9" xml:id="normal-exercise-H-GmodH-cyclic">
      <statement>
        <p>
          Prove or disprove: If <m>H</m> and <m>G/H</m> are cyclic,
          then <m>G</m> is cyclic.
        </p>
      </statement>
    </exercise>

    <exercise number="10" xml:id="normal-exercise-index-2">
      <statement>
        <p>
          Let <m>H</m> be a subgroup of index <m>2</m> of a group <m>G</m>.
          Prove that <m>H</m> must be a normal subgroup of <m>G</m>.
          Conclude that <m>S_n</m> is not simple for <m>n \geq 3</m>.
        </p>
      </statement>
    </exercise>

    <exercise number="11" xml:id="normal-exercise-one-subgroup">
      <statement>
        <p>
          If a group <m>G</m> has exactly one subgroup <m>H</m> of order <m>k</m>,
          prove that <m>H</m> is normal in <m>G</m>.
        </p>
      </statement>
      <hint>
        <p>
          For any <m>g \in G</m>, show that the map <m>i_g : G \to G</m> defined by
          <m>i_g : x \mapsto gxg^{-1}</m> is an isomorphism of <m>G</m> with itself.
          Then consider <m>i_g(H)</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="12" xml:id="normal-exercise-centralizer">
      <statement>
        <p>
          Define the <term>centralizer</term>
          of an element <m>g</m> in a group <m>G</m> to be the set
          <me>
            C(g) = \{ x \in G : xg = gx \}
          </me>.
          Show that <m>C(g)</m> is a subgroup of <m>G</m>.
          If <m>g</m> generates a normal subgroup of <m>G</m>,
          prove that <m>C(g)</m> is normal in <m>G</m>.
        </p>
      </statement>
      <hint>
        <p>
          Suppose that <m>\langle g \rangle</m> is normal in <m>G</m> and let <m>y</m> be an arbitrary element of <m>G</m>.
          If <m>x \in C(g)</m>, we must show that
          <m>y x y^{-1}</m> is also in <m>C(g)</m>.
          Show that <m>(y x y^{-1}) g = g (y x y^{-1})</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="13" xml:id="normal-exercise-center">
      <statement>
        <p>
          Recall that the <term>center</term>
          of a group <m>G</m> is the set
          <me>
            Z(G) = \{ x \in G : xg = gx \text{ for all } g \in G \}
          </me>.
          <ol>
            <li>
              <p>
                Calculate the center of <m>S_3</m>.
              </p>
            </li>

            <li>
              <p>
                Calculate the center of <m>GL_2 ( {\mathbb R} )</m>.
              </p>
            </li>

            <li>
              <p>
                Show that the center of any group <m>G</m> is a normal subgroup of <m>G</m>.
              </p>
            </li>

            <li>
              <p>
                If <m>G / Z(G)</m> is cyclic,
                show that <m>G</m> is abelian.
              </p>
            </li>
          </ol>
        </p>
      </statement>
    </exercise>

    <exercise number="14" xml:id="normal-exercise-commutator">
      <statement>
        <p>
          Let <m>G</m> be a group and let <m>G' = \langle aba^{- 1} b^{-1} \rangle</m>;
          that is, <m>G'</m> is the subgroup of all finite products of elements in <m>G</m> of the form <m>aba^{-1}b^{-1}</m>.
          The subgroup <m>G'</m> is called the
          <term>commutator subgroup</term> of <m>G</m>.

          <notation>
            <usage><m>G'</m></usage>
            <description>commutator subgroup of <m>G</m></description>
          </notation>

          <ol>
            <li>
              <p>
                Show that <m>G'</m> is a normal subgroup of <m>G</m>.
              </p>
            </li>

            <li>
              <p>
                Let <m>N</m> be a normal subgroup of <m>G</m>.
                Prove that <m>G/N</m> is abelian if and only if <m>N</m> contains the commutator subgroup of <m>G</m>.
              </p>
            </li>
          </ol>
        </p>
      </statement>
      <hint>
        <p>
          (a) Let <m>g \in G</m> and <m>h \in G'</m>.
          If <m>h = aba^{-1}b^{-1}</m>, then
          <md>
            <mrow>ghg^{-1} &amp; = gaba^{-1}b^{-1}g^{-1}</mrow>
            <mrow>&amp; = (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})</mrow>
            <mrow>&amp; = (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}</mrow>
          </md>.
          We also need to show that if
          <m>h = h_1 \cdots h_n</m> with <m>h_i = a_i b_i a_i^{-1} b_i^{-1}</m>,
          then <m>ghg^{-1}</m> is a product of elements of the same type.
          However, <m>ghg^{-1} = g h_1 \cdots h_n g^{-1} = (gh_1g^{-1})(gh_2g^{-1}) \cdots (gh_ng^{-1})</m>.
        </p>
      </hint>
    </exercise>

  </exercises>

  <section xml:id="normal-sage" component="sage-info">
    <title>Sage</title>
    <introduction>
      <p>
        Sage has several convenient functions that will allow us to investigate quickly if a subgroup is normal,
        and if so, the nature of the resulting quotient group.
        But for an initial understanding,
        we can also work with the raw cosets.
        Let us get our hands dirty first,
        then learn about the easy way.
      </p>
    </introduction>

    <subsection>
      <title>Multiplying Cosets</title>
      <p>
        The definiton of a factor group requires a normal subgroup,
        and then we <em>define</em> a way to <q>multiply</q>
        two cosets of the subgroup to produce another coset.
        It is important to realize that we can interpret the definition of a normal subgroup to be <em>exactly</em>
        the condition we need for our new multiplication to be workable.
        We will do two examples <mdash/> first with a normal subgroup,
        then with a subgroup that is not normal.
      </p>

      <p>
        Consider the dihedral group <m>D_{8}</m> that is the symmetry group of an <m>8</m>-gon.
        If we take the element that creates a quarter-turn,
        we can use it generate a cyclic subgroup of order 4.
        This will be a normal subgroup
        (trust us for the moment on this).
        First, build the (right) cosets (notice there is no output):
      </p>

            <sage>
                <input>
        G = DihedralGroup(8)
        quarter_turn = G('(1,3,5,7)(2,4,6,8)')
        S = G.subgroup([quarter_turn])
        C = G.cosets(S)
        </input>
            </sage>

      <p>
        So <c>C</c> is a list of lists,
        with every element of the group <c>G</c> occuring exactly once somewhere.
        You could ask Sage to print out <c>C</c> for you if you like,
        but we will try to avoid that here.
        We want to multiply two cosets (lists) together.
        How do we do this?
        Take <em>any</em> element out of the first list,
        and <em>any</em> element out of the second list and multiply them together
        (which we know how to do since they are elements of <c>G</c>).
        Now we have an element of <c>G</c>.
        What do we do with this element,
        since we really want a coset as the result of the product of two cosets?
        Simple <mdash/> we see which coset the product is in.
        Let us give it a try.
        We will multiply coset <m>1</m> with coset <m>3</m>
        (there are <m>4</m> cosets by Lagrange's Theorem).
        Study the following code carefully to see if you can understand what it is doing,
        and <em>then</em> read the explanation that follows.
      </p>

            <sage>
                <input>
        p = C[1][0]*C[3][0]
        [i for i in srange(len(C)) if p in C[i]]
        </input>
                <output>
        [2]
        </output>
            </sage>

      <p>
        What have we accomplished?
        In the first line we create <c>p</c> as the product of two group elements,
        one from coset <m>1</m> and one from coset <m>3</m>
        (<c>C[1]</c>, <c>C[3]</c>).
        Since we can choose <em>any</em> element from each coset,
        we choose the first element of each
        (<c>C[ ][0]</c>).
        Then we count our way through all the cosets,
        selecting only cosets that contain <c>p</c>.
        Since <c>p</c> will only be in one coset,
        we expect a list with just one element.
        Here, our one-element list contains only <c>2</c>.
        So we say the product of coset <m>1</m> and coset <m>3</m> is coset <m>2</m>.
      </p>

      <p>
        The point here is that this result
        (coset <m>1</m> times coset <m>3</m> is coset <m>2</m>)
        should always be the same,
        <em>no matter which elements we pick from the two cosets to form</em> <c>p</c>.
        So let us do it again,
        but this time we will not simply choose the first element from each of coset <m>1</m> and coset <m>3</m>,
        instead we will choose the third element of coset <m>1</m> and the second element of coset <m>3</m>
        (remember, we are counting from zero!).
      </p>

            <sage>
                <input>
        p = C[1][2]*C[3][1]
        [i for i in srange(len(C)) if p in C[i]]
        </input>
                <output>
        [2]
        </output>
            </sage>

      <p>
        Good.
        We have the same result.
        If you are still trusting us on <c>S</c> being a normal subgroup of <c>G</c>,
        then this is the result that the theory predicts.
        Make a copy of the above compute cell and try other choices for the representatives of each coset.
        Then try the product of other cosets,
        with varying representatives.
      </p>

      <sage type="practice"/>

      <p>
        Now is a good time to introduce a way to extend Sage and add new functions.
        We will design a coset-multiplication function.
        Read the following carefully and then see the subsequent explanation.
      </p>

            <sage>
                <input>
        def coset_product(i, j, C):
           p = C[i][0]*C[j][0]
           c = [k for k in srange(len(C)) if p in C[k]]
           return c[0]
        </input>
            </sage>

      <p>
        The first line creates a new Sage function named <c>coset_product</c>.
        This is accomplished with the word <c>def</c>,
        and note the colon ending the line.
        The inputs to the function are the numbers of the cosets we want to multiply and the complete list of the cosets.
        The middle two lines should look familiar from above.
        We know <c>c</c> is a one-element list,
        so <c>c[0]</c> will extract this one coset number,
        and <c>return</c> is what determines that this is the output of the function.
        Notice that the indentation above must be exactly as shown.
        We could have written all this computation on a single line without making a new function,
        but that begins to get unwieldly.
        You need to execute the code block above to actually
        <em>define</em> the function,
        and there will be no output if successful.
        Now we can use our new function to repeat our work above:
      </p>

            <sage>
                <input>
        coset_product(1, 3, C)
        </input>
                <output>
        2
        </output>
            </sage>

      <p>
        Now you know the basics of how to add onto Sage and do much more than it was designed for.
        And with some practice,
        you could suggest and contribute new functions to Sage,
        since it is an open source project.
        Nice.
      </p>

      <p>
        Now let us examine a situation where the subgroup is not normal.
        So we will see that our definition of coset multiplication is insufficient in this case.
        And realize that our new <c>coset_product</c> function is also useless since it assumes the cosets come from a normal subgroup.
      </p>

      <p>
        Consider the alternating group <m>A_4</m> which we can interpet as the symmetry group of a tetrahedron.
        For a subgroup,
        take an element that fixes one vertex and rotates the opposite face <mdash/> this will generate a cyclic subgroup of order 3, and by Lagrange's Theorem we will get four cosets.
        We compute them here.
        (Again, no output is requested.)
      </p>

            <sage>
                <input>
        G = AlternatingGroup(4)
        face_turn = G("(1,2,3)")
        S = G.subgroup([face_turn])
        C = G.cosets(S)
        </input>
            </sage>

      <p>
        Again, let's consider the product of coset <m>1</m> and coset <m>3</m>:
      </p>

            <sage>
                <input>
        p = C[1][0]*C[3][0]
        [i for i in srange(len(C)) if p in C[i]]
        </input>
                <output>
        [0]
        </output>
            </sage>

      <p>
        Again, but now for coset <m>3</m>,
        choose the second element of the coset to produce the product <c>p</c>:
      </p>

            <sage>
                <input>
        p = C[1][0]*C[3][1]
        [i for i in srange(len(C)) if p in C[i]]
        </input>
                <output>
        [2]
        </output>
            </sage>

      <p>
        So, is the product of coset <m>1</m> and coset <m>3</m> equal to coset <m>0</m> or coset <m>2</m>?
        We cannot say!
        So there is <em>no way</em> to construct a quotient group for this subgroup.
        You can experiment some more with this subgroup, but in some sense,
        we are done with this example <mdash/> there is nothing left to say.
      </p>
    </subsection>

    <subsection>
      <title>Sage Methods for Normal Subgroups</title>
      <p>
        You can easily ask Sage if a subgroup is normal or not.
        This is viewed as a property of the subgroup,
        but you must tell Sage what the <q>supergroup</q> is,
        since the answer can change depending on this value. (For example <c>H.is_normal(H)</c> will always be <c>True</c>.) Here are our two examples from above.
      </p>

            <sage>
                <input>
        G = DihedralGroup(8)
        quarter_turn = G('(1,3,5,7)(2,4,6,8)')
        S = G.subgroup([quarter_turn])
        S.is_normal(G)
        </input>
                <output>
        True
        </output>
            </sage>
            <sage>
                <input>
        G = AlternatingGroup(4)
        face_turn = G("(1,2,3)")
        S = G.subgroup([face_turn])
        S.is_normal(G)
        </input>
                <output>
        False
        </output>
            </sage>

      <p>
        The text proves in <xref ref="normal-section-simplicity-of-an"/> that <m>A_5</m> is simple,
        <ie/> <m>A_5</m> has no normal subgroups.
        We could build every subgroup of <m>A_5</m> and ask if it is normal in <m>A_5</m> using the <c>.is_normal()</c> method.
        But Sage has this covered for us already.
      </p>

            <sage>
                <input>
        G = AlternatingGroup(5)
        G.is_simple()
        </input>
                <output>
        True
        </output>
            </sage>

      <p>
        We can also build a quotient group when we have a normal subgroup.
      </p>

            <sage>
                <input>
        G = DihedralGroup(8)
        quarter_turn = G('(1,3,5,7)(2,4,6,8)')
        S = G.subgroup([quarter_turn])
        Q = G.quotient(S)
        Q
        </input>
                <output>
        Permutation Group with generators [(1,2)(3,4), (1,3)(2,4)]
        </output>
            </sage>

      <p>
        This is useful, but also a bit unsettling.
        We have the quotient group, but any notion of cosets has been lost,
        since <c>Q</c> is returned as a new permutation group on a different set of symbols.
        We cannot presume that the numbers used for the new permutation group <c>Q</c> bear any resemblance to the cosets we get from the <c>.cosets()</c> method.
        But we can see that the quotient group is described as a group generated by two elements of order two.
        We could ask for the order of the group,
        or by Lagrange's Theorem we know the quotient has order <m>4</m>.
        We can say now that there are only two groups of order four,
        the cyclic group of order <m>4</m> and a non-cyclic group of order <m>4</m>,
        known to us as the Klein <m>4</m>-group or <m>{\mathbb Z}_2\times{\mathbb Z}_2</m>.
        This quotient group looks like the non-cyclic one since the cyclic group of order 4 has just one element of order 2.
        Let us see what Sage says.
      </p>

            <sage>
                <input>
        Q.is_isomorphic(KleinFourGroup())
        </input>
                <output>
        True
        </output>
            </sage>

      <p>
        Yes, that's it.
      </p>

      <p>
        Finally, Sage can build us a list of all of the normal subgroups of a group.
        The list of groups themselves, as we have seen before,
        is sometimes an overwhelming amount of information.
        We will demonstrate by just listing the orders of the normal subgroups produced.
      </p>

            <sage>
                <input>
        G = DihedralGroup(8)
        N = G.normal_subgroups()
        [H.order() for H in N]
        </input>
                <output>
        [16, 8, 8, 8, 4, 2, 1]
        </output>
            </sage>

      <p>
        So, in particular, we see that our
        <q>quarter-turn</q> subgroup is the <em>only</em>
        normal subgroup of order <m>4</m> in this group.
      </p>
    </subsection>

  </section>

  <exercises xml:id="normal-sage-exercises" component="sage-exercises">
    <title>Sage Exercises</title>

    <exercise number="1" xml:id="normal-sage-exercise-A5">
      <statement>
        <p>
          Build every subgroup of the alternating group on 5 symbols,
          <m>A_5</m>, and check that each is not a normal subgroup
          (except for the two trivial cases).
          This command might take a couple seconds to run.
          Compare this with the time needed to run the <c>.is_simple()</c> method and realize that there is a significant amount of theory and cleverness brought to bear in speeding up commands like this.
          (It is possible that your Sage installation lacks <acro>GAP</acro>'s <q>Table of Marks</q>
          library and you will be unable to compute the list of subgroups.)
        </p>
      </statement>
    </exercise>

    <exercise number="2" xml:id="normal-sage-exercise-D8-quotient">
      <statement>
        <p>
          Consider the quotient group of the group of symmetries of an <m>8</m>-gon,
          formed with the cyclic subgroup of order <m>4</m> generated by a quarter-turn.
          Use the <c>coset_product</c> function to determine the Cayley table for this quotient group.
          Use the number of each coset,
          as produced by the <c>.cosets()</c> method as names for the elements of the quotient group.
          You will need to build the table <q>by hand</q>
          as there is no easy way to have Sage's Cayley table command do this one for you.
          You can build a table in the Sage Notebook pop-up editor (shift-click on a blue line) or you might read the documentation of the <c>html.table()</c> method.
        </p>
      </statement>
    </exercise>

    <exercise number="3" xml:id="normal-sage-exercise-D8-cyclic-subgroup">
      <statement>
        <p>
          Consider the cyclic subgroup of order <m>4</m> in the symmetries of an <m>8</m>-gon.
          Verify that the subgroup is normal by first building the raw left and right cosets (without using the <c>.cosets()</c> method) and then checking their equality in Sage,
          all with a single command that employs sorting with the <c>sorted()</c> command.
        </p>
      </statement>
    </exercise>

    <exercise number="4" xml:id="normal-sage-exercise-D8-normal-subgroup">
      <statement>
        <p>
          Again, use the same cyclic subgroup of order <m>4</m> in the group of symmetries of an <m>8</m>-gon.
          Check that the subgroup is normal by using part (2) of <xref ref="normal-theorem-normal-equivalents"/>.
          Construct a one-line command that does the complete check and returns <c>True</c>.
          Maybe sort the elements of the subgroup <c>S</c> first,
          then slowly build up the necessary lists,
          commands, and conditions in steps.
          Notice that this check does not require ever building the cosets.
        </p>
      </statement>
    </exercise>

    <exercise number="5" xml:id="normal-sage-exercise-tetrahedron">
      <statement>
        <p>
          Repeat the demonstration from the previous subsection that for the symmetries of a tetrahedron,
          a cyclic subgroup of order <m>3</m> results in an undefined coset multiplication.
          Above, the default setting for the <c>.cosets()</c> method builds right cosets <mdash/> but in this problem,
          work instead with left cosets.
          You need to choose two cosets to multiply,
          and then demonstrate two choices for representatives that lead to different results for the product of the cosets.
        </p>
      </statement>
    </exercise>

    <exercise number="6" xml:id="normal-sage-exercise-n-gon">
      <statement>
        <p>
          Construct some dihedral groups of order <m>2n</m> (<ie/> symmetries of an <m>n</m>-gon,
          <m>D_{n}</m> in the text, <c>DihedralGroup(n)</c> in Sage).
          Maybe all of them for <m>3\leq n \leq 100</m>.
          For each dihedral group,
          construct a list of the orders of each of the normal subgroups (so use <c>.normal_subgroups()</c>).
          You may need to wait ten or twenty seconds for this to finish - be patient.
          Observe enough examples to hypothesize a pattern to your observations,
          check your hypothesis against each of your examples and then state your hypothesis clearly.
        </p>

        <p>
          Can you predict how many normal subgroups there are in the dihedral group
          <m>D_{470448}</m> without using Sage to build all the normal subgroups?
          Can you <em>describe</em> all of the normal subgroups of a dihedral group in a way that would let us predict all of the normal subgroups of <m>D_{470448}</m> without using Sage?
        </p>
      </statement>
    </exercise>

  </exercises>
</chapter>