Section 4.1 Cyclic groups

<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>

    <!-- RAB 2014/08/18 Maybe this id should have 3z, not z3? -->
    <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 <md>3 {\mathbb Z} = \{ \ldots, -3, 0, 3, 6, \ldots \}</md>. 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 <md>\langle a \rangle  = \{ a^k : k \in {\mathbb Z} \}</md> 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>
        <!-- Replaced figure with tikz figure - TWJ 5/6/2010 -->
        <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 <md>g  h = a^r a^s = a^{r+s} = a^{s+r} = a^s a^r = h g</md>, <m>G</m> is abelian.</p>
        </proof>
    </theorem>
</section>