PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

<section>
    <title>Cyclic groups</title>
    <p>
      Often a subgroup will depend entirely on a single element of the group;
      that is, knowing that particular element will allow us to compute any other element in the subgroup.
    </p>

    <example xml:id="example-cyclic-z3">
      <title>An Infinite Cyclic Subgroup, Modular Addition</title>
      <p>
        Suppose that we consider <m>3 \in {\mathbb Z}</m> and look at all multiples
        (both positive and negative)
        of 3.
        As a set, this is
        <me>
          3 {\mathbb Z} = \{ \ldots, -3, 0, 3, 6, \ldots \}
        </me>.
        It is easy to see that <m>3 {\mathbb Z}</m> is a subgroup of the integers.
        This subgroup is completely determined by the element 3 since we can obtain all of the other elements of the group by taking multiples of 3.
        Every element in the subgroup is
        <q>generated</q>
        by 3.
      </p>
    </example>
    <example xml:id="example-cyclic-2n">
      <title>An Infinite Cyclic Subgroup, Multiplication of Rational Numbers</title>
      <p>
        If <m>H = \{ 2^n : n \in {\mathbb Z} \}</m>,
        then <m>H</m> is a subgroup of the multiplicative group of nonzero rational numbers,
        <m>{\mathbb Q}^*</m>.
        If <m>a = 2^m</m> and <m>b = 2^n</m> are in <m>H</m>,
        then <m>ab^{-1} = 2^m 2^{-n} = 2^{m-n}</m> is also in <m>H</m>.
        By <xref ref="proposition-subgroup" />,
        <m>H</m> is a subgroup of <m>{\mathbb Q}^*</m> determined by the element 2.
      </p>
    </example>
    <theorem xml:id="theorem-cyclic-subgroup">
      <statement>
        <p>
          Let <m>G</m> be a group and <m>a</m> be any element in <m>G</m>.
          Then the set
          <me>
            \langle a \rangle  = \{ a^k : k \in {\mathbb Z} \}
          </me>
          is a subgroup of <m>G</m>.
          Furthermore,
          <m>\langle a \rangle</m> is the smallest subgroup of <m>G</m> that contains~<m>a</m>.
          <notation>
            <usage><m>\langle a \rangle</m></usage>
            <description>cyclic group generated by <m>a</m></description>
          </notation>
        </p>
      </statement>
      <proof>
        <p>
          The identity is in <m>\langle a \rangle </m> since <m>a^0 = e</m>.
          If <m>g</m> and <m>h</m> are any two elements in <m>\langle a \rangle </m>,
          then by the definition of <m>\langle a \rangle</m> we can write <m>g = a^m</m> and <m>h = a^n</m> for some integers <m>m</m> and <m>n</m>.
          So <m>gh = a^m a^n = a^{m+n}</m> is again in <m>\langle a \rangle </m>.
          Finally, if <m>g = a^n</m> in <m>\langle a \rangle </m>,
          then the inverse <m>g^{-1} = a^{-n}</m> is also in <m>\langle a \rangle </m>.
          Clearly, any subgroup <m>H</m> of <m>G</m> containing <m>a</m> must contain all the powers of <m>a</m> by closure;
          hence, <m>H</m> contains <m>\langle a \rangle </m>.
          Therefore, <m>\langle a \rangle </m> is the smallest subgroup of <m>G</m> containing <m>a</m>.
        </p>
      </proof>
    </theorem>
    <remark>
      <p>
        If we are using the
        <q>+</q>
        notation, as in the case of the integers under addition,
        we write <m>\langle a \rangle = \{ na : n \in {\mathbb Z} \}</m>.
      </p>
    </remark>
    <p>
      For <m>a \in G</m>, we call <m>\langle a \rangle </m> the
      <term>cyclic subgroup</term>
          <idx><h>Subgroup</h><h>cyclic</h></idx>
      generated by <m>a</m>.
      If <m>G</m> contains some element <m>a</m> such that <m>G = \langle a \rangle </m>,
      then <m>G</m> is a <term>cyclic group</term>.
          <idx><h>Group</h><h>cyclic</h></idx>
      In this case <m>a</m> is a <term>generator</term><idx><h>Generator of a cyclic subgroup</h></idx> of <m>G</m>.
      If <m>a</m> is an element of a group <m>G</m>,
      we define the <term>order</term>
          <idx><h>Element</h><h>order of</h></idx>
      of <m>a</m> to be the smallest positive integer <m>n</m> such that <m>a^n= e</m>,
      and we write <m>|a| = n</m>.
      <notation>
        <usage><m>|a|</m></usage>
        <description>the order of an element <m>a</m></description>
      </notation>
      If there is no such integer <m>n</m>,
      we say that the order of <m>a</m> is infinite and write
      <m>|a| = \infty</m> to denote the order of <m>a</m>.
    </p>
    <example xml:id="example-cyclic-z6">
      <title>Generators of a Finite Cyclic Group</title>
      <p>
        Notice that a cyclic group can have more than a single generator.
        Both 1 and 5 generate <m>{\mathbb Z}_6</m>;
        hence, <m>{\mathbb Z}_6</m> is a cyclic group.
        Not every element in a cyclic group is necessarily a generator of the group.
        The order of <m>2 \in {\mathbb Z}_6</m> is 3.
        The cyclic subgroup generated by 2 is <m>\langle 2 \rangle = \{ 0, 2, 4 \}</m>.
      </p>
    </example>
    <p>
      The groups <m>{\mathbb Z}</m> and <m>{\mathbb Z}_n</m> are cyclic groups.
      The elements 1 and <m>-1</m> are generators for <m>{\mathbb Z}</m>.
      We can certainly generate <m>{\mathbb Z}_n</m> with 1 although there may be other generators of <m>{\mathbb Z}_n</m>,
      as in the case of <m>{\mathbb Z}_6</m>.
    </p>
    <example xml:id="example-cyclic-u9">
      <title>A Cyclic Group of Units</title>
      <p>
        The group of units, <m>U(9)</m>,
        in <m>{\mathbb Z}_9</m> is a cyclic group.
        As a set, <m>U(9)</m> is <m>\{ 1, 2, 4, 5, 7, 8  \}</m>.
        The element 2 is a generator for <m>U(9)</m> since
        <md>
          <mrow>2^1 &amp; = 2 \qquad 2^2 = 4</mrow>
          <mrow>2^3 &amp; = 8 \qquad 2^4 = 7</mrow>
          <mrow>2^5 &amp; =  5 \qquad 2^6 = 1</mrow>
        </md>.
      </p>
    </example>
    <example xml:id="example-cyclic-s3-not-cyclic">
      <title>A Group That is Not Cyclic</title>
      <p>
        Not every group is a cyclic group.
        Consider the symmetry group of an equilateral triangle <m>S_3</m>.
        The subgroups of <m>S_3</m> are shown in <xref ref="figure-subgrps-s3" />.
        Notice that every subgroup is cyclic;
        however, no single element generates the entire group.
      </p>
    </example>
    <figure xml:id="figure-subgrps-s3">
      <caption>Subgroups of <m>S_3</m></caption>

      <image xml:id="cyclic-s3-subgroups">
<latex-image>
\begin{tikzpicture}[scale=1]
\draw  (0,0.3) -- (2.6,1.2);
\draw  (2,0.3) -- (2.8,1.2);
\draw  (4,0.3) -- (3.2,1.2);
\draw  (6,0.3) -- (3.4,1.2);
\draw  (0,-0.3) -- (2.6,-1.2);
\draw  (2,-0.3) -- (2.8,-1.2);
\draw  (4,-0.3) -- (3.2,-1.2);
\draw  (6,-0.3) -- (3.4,-1.2);
\node at (0, 0) {$\{  \identity, \rho_1, \rho_2\}$};
\node at (2, 0) {$\{  \identity, \mu_1\}$};
\node at (4, 0) {$\{  \identity, \mu_2 \}$};
\node at (6, 0) {$\{  \identity, \mu_3 \}$};
\node at (3, 1.5) {$S_3$};
\node at (3,-1.5) {$\{ \identity \}$};
\end{tikzpicture}
</latex-image>
      </image>
    </figure>
    <theorem xml:id="theorem-cyclic-abelian">
      <statement>
        <p>
          Every cyclic group is abelian.
        </p>
      </statement>
      <proof>
        <p>
          Let <m>G</m> be a cyclic group and <m>a \in G</m> be a generator for <m>G</m>.
          If <m>g</m> and <m>h</m> are in <m>G</m>,
          then they can be written as powers of <m>a</m>,
          say <m>g = a^r</m> and <m>h = a^s</m>.
          Since
          <me>
            g  h = a^r a^s = a^{r+s} = a^{s+r} = a^s a^r = h g
          </me>,
          <m>G</m> is abelian.
        </p>
      </proof>
    </theorem>
  </section>