struct.xml

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<chapter xmlns:xi="http://www.w3.org/2001/XInclude" xml:id="struct">
  <title>The Structure of  Groups</title>
  <introduction>
    <p>
      The ultimate goal of group theory is to classify all groups up to isomorphism;
      that is, given a particular group,
      we should be able to match it up with a known group via an isomorphism.
      For example,
      we have already proved that any finite cyclic group of order <m>n</m> is isomorphic to <m>{\mathbb Z}_n</m>;
      hence, we <q>know</q> all finite cyclic groups.
      It is probably not reasonable to expect that we will ever know all groups;
      however, we can often classify certain types of groups or distinguish between groups in special cases.
    </p>

    <p>
      In this chapter we will characterize all finite abelian groups.
      We shall also investigate groups with sequences of subgroups.
      If a group has a sequence of subgroups, say
      <me>
        G = H_n \supset H_{n - 1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}
      </me>,
      where each subgroup <m>H_i</m> is normal in <m>H_{i+1}</m> and each of the factor groups <m>H_{i+1}/H_i</m> is abelian,
      then <m>G</m> is a solvable group.
      In addition to allowing us to distinguish between certain classes of groups,
      solvable groups turn out to be central to the study of solutions to polynomial equations.
    </p>
  </introduction>

  <section xml:id="struct-section-finite-abelian-groups">
    <title>Finite Abelian Groups</title>
    <p>
      In our investigation of cyclic groups we found that every group of prime order was isomorphic to <m>{\mathbb Z}_p</m>,
      where <m>p</m> was a prime number.
      We also determined that <m>{\mathbb Z}_{mn} \cong {\mathbb Z}_m \times {\mathbb Z}_n</m> when <m>\gcd(m, n) =1</m>.
      In fact, much more is true.
      Every finite abelian group is isomorphic to a direct product of cyclic groups of prime power order;
      that is, every finite abelian group is isomorphic to a group of the type
      <me>
        {\mathbb Z}_{p_1^{\alpha_1}} \times \cdots \times {\mathbb Z}_{p_n^{\alpha_n}}
      </me>,
      where each <m>p_k</m> is prime
      (not necessarily distinct).
    </p>

    <p>
      First, let us examine a slight generalization of finite abelian groups.
      Suppose that <m>G</m> is a group and let
      <m>\{ g_i\}</m> be a set of elements in <m>G</m>,
      where <m>i</m> is in some index set <m>I</m>
      (not necessarily finite).
      The smallest subgroup of <m>G</m> containing all of the <m>g_i</m>'s is the subgroup of <m>G</m>
      <term>generated</term> by the <m>g_i</m>'s.
      If this subgroup of <m>G</m> is in fact all of <m>G</m>,
      then <m>G</m> is generated by the set <m>\{g_i : i \in I \}</m>.
      In this case the <m>g_i</m>'s are said to be the <term>generators</term><idx><h>Generators for a group</h></idx><idx><h>Group</h><h>generators of</h></idx>
      of <m>G</m>.
      If there is a finite set <m>\{ g_i : i \in I \}</m> that generates <m>G</m>,
      then <m>G</m> is <idx><h>Group</h><h>finitely generated</h></idx><idx><h>Finitely generated group</h></idx><term>finitely generated</term>.   
    </p>

    <example xml:id="struct-example-finite-groups">
      <p>
        Obviously, all finite groups are finitely generated.
        For example,
        the group <m>S_3</m> is generated by the permutations <m>(1 \, 2)</m> and <m>(1 \, 2 \,3)</m>.
        The group <m>{\mathbb Z} \times {\mathbb Z}_n</m> is an infinite group but is finitely generated by <m>\{ (1,0), (0,1) \}</m>.
      </p>
    </example>

    <example xml:id="struct-example-infinite-groups">
      <p>
        Not all groups are finitely generated.
        Consider the rational numbers
        <m>{\mathbb Q}</m> under the operation of addition.
        Suppose that <m>{\mathbb Q}</m> is finitely generated with generators <m>p_1/q_1, \ldots,
        p_n/q_n</m>,
        where each <m>p_i/q_i</m> is a fraction expressed in its lowest terms.
        Let <m>p</m> be some prime that does not divide any of the denominators <m>q_1, \ldots, q_n</m>.
        We claim that <m>1/p</m> cannot be in the subgroup of
        <m>{\mathbb Q}</m> that is generated by  <m>p_1/q_1, \ldots,
        p_n/q_n</m>,
        since <m>p</m> does not divide the denominator of any element in this subgroup.
        This fact is easy to see since the sum of any two generators is
        <me>
          p_i / q_i + p_j / q_j = (p_i q_j + p_j q_i)/(q_i q_j)
        </me>.
      </p>
    </example>

    <proposition>
      <statement>
        <p>
          Let <m>H</m> be the subgroup of a group <m>G</m> that is generated by <m>\{ g_i \in G : i \in I \}</m>.
          Then <m>h \in H</m> exactly when it is a product of the form
          <me>
            h = g_{i_1}^{\alpha_1} \cdots g_{i_n}^{\alpha_n}
          </me>,
          where the <m>g_{i_k}</m>s are not necessarily distinct.
        </p>
      </statement>

      <proof>
        <p>
          Let <m>K</m> be the set of all products of the form <m>g_{i_1}^{\alpha_1} \cdots g_{i_n}^{\alpha_n}</m>,
          where the <m>g_{i_k}</m>s are not necessarily distinct.
          Certainly <m>K</m> is a subset of <m>H</m>.
          We need only show that <m>K</m> is a subgroup of <m>G</m>.
          If this is the case, then <m>K=H</m>,
          since <m>H</m> is the smallest subgroup containing all the <m>g_i</m>s.
        </p>

        <p>
          Clearly, the set <m>K</m> is closed under the group operation.
          Since <m>g_i^0 = 1</m>, the identity is in <m>K</m>.
          It remains to show that the inverse of an element 
          <m>g =g_{i_1}^{k_1} \cdots g_{i_n}^{k_n}</m> in <m>K</m> must also be in <m>K</m>.
          However,
          <me>
            g^{-1} = (g_{i_1}^{k_{1}} \cdots g_{i_n}^{k_n})^{-1} = (g_{i_n}^{-k_n} \cdots g_{i_{1}}^{-k_{1}})
          </me>.
        </p>
      </proof>
    </proposition>

    <p>
      The reason that powers of a fixed <m>g_i</m> may occur several times in the product is that we may have a nonabelian group.
      However, if the group is abelian,
      then the <m>g_i</m>s need occur only once.
      For example, a product such as
      <m>a^{-3} b^5 a^7</m> in an abelian group could always be simplified
      (in this case, to <m>a^4 b^5</m>).
    </p>
        <!-- Typo corrected. Suggested by S. Engle. TWJ 11/13/2011 -->
        <!-- Subscript numbering corrected.  TWJ 11/17/2012 -->
        <!-- Subscript numbering corrected<mdash />second try.  TWJ 4/24/2013 -->
    <p>
      Now let us restrict our attention to finite abelian groups.
      We can express any finite abelian group as a finite direct product of cyclic groups.
      More specifically, letting <m>p</m> be prime,
      we define a group <m>G</m> to be a
      <term><m>p</m>-group</term><idx><h>Group</h><h><m>p</m>-group</h></idx>
      if every element in <m>G</m> has as its order a power of <m>p</m>.
      For example,
      both <m>{\mathbb Z}_2 \times {\mathbb Z}_2</m> and <m>{\mathbb Z}_4</m> are <m>2</m>-groups,
      whereas <m>{\mathbb Z}_{27}</m> is a <m>3</m>-group.
      We shall prove the Fundamental Theorem of Finite Abelian Groups which tells us that every finite abelian group is isomorphic to a direct product of cyclic <m>p</m>-groups.
    </p>

    <theorem xml:id="struct-theorem-finite-abelian-groups">
      <title>Fundamental Theorem of Finite Abelian Groups</title>
      <idx>
      <h>Fundamental Theorem</h>
      <h>of Finite Abelian Groups</h>
      </idx>
      <statement>
        <p>
          Every finite abelian group <m>G</m> is isomorphic to a direct product of cyclic groups of the form
          <me>
            {\mathbb Z}_{p_1^{ \alpha_1 }} \times {\mathbb Z}_{p_2^{ \alpha_2 }} \times \cdots \times {\mathbb Z}_{p_n^{ \alpha_n }}
          </me>
          here the <m>p_i</m>'s are primes
          (not necessarily distinct).
        </p>
      </statement>
    </theorem>

    <example xml:id="struct-example-abelian-540">
      <p>
        Suppose that we wish to classify all abelian groups of order <m>540=2^2 \cdot 3^3 \cdot 5</m>.
        The Fundamental Theorem of Finite Abelian Groups tells us that we have the following six possibilities.

        <ul>
          <li>
            <p>
              <m>{\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li>
            <p>
              <m>{\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_3 \times {\mathbb Z}_9 \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li>
            <p>
              <m>{\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_{27} \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li>
            <p>
              <m>{\mathbb Z}_4 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li>
            <p>
              <m>{\mathbb Z}_4 \times {\mathbb Z}_3 \times {\mathbb Z}_9 \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li>
            <p>
              <m>{\mathbb Z}_4 \times {\mathbb Z}_{27} \times {\mathbb Z}_5</m>.
            </p>
          </li>
        </ul>
      </p>
    </example>

    <p>
      The proof of the Fundamental Theorem of Finite Abelian Groups depends on several lemmas.
    </p>

    <lemma xml:id="struct-lemma-cauchy-finite-abelian">
      <statement>
        <p>
          Let <m>G</m> be a finite abelian group of order <m>n</m>.
          If <m>p</m> is a prime that divides <m>n</m>,
          then <m>G</m> contains an element of order <m>p</m>.
        </p>
      </statement>

      <proof>
        <p>
          We will prove this lemma by induction.
          If <m>n = 1</m>, then there is nothing to show.
          Now suppose that the lemma is true for all groups of order <m>k</m>,
          where <m>k \lt n</m>.
          Furthermore, let <m>p</m> be a prime that divides <m>n</m>.
        </p>

        <p>
          If <m>G</m> has no proper nontrivial subgroups,
          then <m>G = \langle a \rangle</m>,
          where <m>a</m> is any element other than the identity.
          By <xref ref="cyclic-exercise-subgroups-exist"/>, the order of <m>G</m> must be prime.
          Since <m>p</m> divides <m>n</m>,
          we know that <m>p = n</m>,
          and <m>G</m> contains <m>p - 1</m> elements of order <m>p</m>.
        </p>

        <p>
          Now suppose that <m>G</m> contains a nontrivial proper subgroup <m>H</m>.
          Then <m>1 \lt |H| \lt n</m>.
          If <m>p \mid |H|</m>,
          then <m>H</m> contains an element of order <m>p</m> by induction and the lemma is true.
          Suppose that <m>p</m> does not divide the order of <m>H</m>.
          Since <m>G</m> is abelian,
          it must be the case that <m>H</m> is a normal subgroup of <m>G</m>,
          and <m>|G| = |H| \cdot |G/H|</m>.
          Consequently, <m>p</m> must divide <m>|G/H|</m>.
          Since <m>|G/H| \lt |G| = n</m>,
          we know that <m>G/H</m> contains an element <m>aH</m> of order <m>p</m> by the induction hypothesis.
          Thus,
          <me>
            H = (aH)^p = a^pH
          </me>,
          and <m>a^p \in H</m> but <m>a \notin H</m>.
          If <m>|H| = r</m>, then <m>p</m> and <m>r</m> are relatively prime,
          and there exist integers <m>s</m> and <m>t</m> such that <m>sp + tr = 1</m>.
          Furthermore, the order of <m>a^p</m> must divide <m>r</m>,
          and <m>(a^p)^r = (a^r)^p = 1</m>.
        </p>

        <p>
          We claim that <m>a^r</m> has order <m>p</m>.
          We must show that <m>a^r \neq 1</m>.
          Suppose <m>a^r = 1</m>.
          Then
          <md>
            <mrow>a &amp; = a^{sp + tr}</mrow>
            <mrow>&amp; = a^{sp} a^{tr}</mrow>
            <mrow>&amp; = (a^p)^s (a^r)^t</mrow>
            <mrow>&amp; = (a^p)^s 1</mrow>
            <mrow>&amp; = (a^p)^s</mrow>
          </md>.
          Since <m>a^p \in H</m>, it must be the case that
          <m>a= (a^p)^s \in H</m>, which is a contradiction.
          Therefore, <m>a^r \neq 1</m> is an element of order <m>p</m> in <m>G</m>.
        </p>
      </proof>
    </lemma>

    <p>
      <xref ref="struct-lemma-cauchy-finite-abelian"/> is a special case of Cauchy's Theorem (<xref ref="sylow-theorem-cauchy"/>),
      which states that if <m>G</m> is a finite group and <m>p</m> a prime such that <m>p</m> divides the order of <m>G</m>,
      then <m>G</m> contains a subgroup of order <m>p</m>.
      We will prove Cauchy's Theorem in <xref ref="sylow"/>.
    </p>

    <lemma xml:id="struct-lemma-p-group-order">
      <statement>
        <p>
          A finite abelian group is a <m>p</m>-group if and only if its order is a power of <m>p</m>.
        </p>
      </statement>

      <proof>
        <p>
          If <m>|G| = p^n</m> then by Lagrange’s theorem,
          then the order of any <m>g \in G</m> must divide <m>p^n</m>,
          and therefore must be a power of <m>p</m>.
          Conversely, if <m>|G|</m> is not a power of <m>p</m>,
          then it has some other prime divisor <m>q</m>,
          so by <xref ref="struct-lemma-cauchy-finite-abelian"/>,
          <m>G</m> has an element of order <m>q</m> and thus is not a <m>p</m>-group.
        </p>
      </proof>
    </lemma>

    <lemma xml:id="struct-lemma-direct-product-p-groups">
      <statement>
        <p>
          Let <m>G</m> be a finite abelian group of order <m>n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}</m>,
          where where <m>p_1, \ldots, p_k</m> are distinct primes and
          <m>\alpha_1, \alpha_2, \ldots, \alpha_k</m> are positive integers.
          Then <m>G</m> is the internal direct product of subgroups <m>G_1, G_2, \ldots, G_k</m>,
          where <m>G_i</m> is the subgroup of <m>G</m> consisting of all elements of order <m>p_i^r</m>
          for some integer <m>r</m>.
        </p>
      </statement>

      <proof>
        <p>
          Since <m>G</m> is an abelian group,
          we are guaranteed that <m>G_i</m> is a subgroup of <m>G</m> for <m>i = 1, \ldots, k</m>.
          Since the identity has order <m>p_i^0 = 1</m>,
          we know that <m>1 \in G_i</m>.
          If <m>g \in G_i</m> has order <m>p_i^r</m>,
          then <m>g^{-1}</m> must also have order <m>p_i^r</m>.
          Finally, if <m>h \in G_i</m> has order <m>p_i^s</m>, then
          <me>
            (gh)^{p_i^t} = g^{p_i^t} h^{p_i^t} = 1 \cdot 1 = 1
          </me>,
          where <m>t</m> is the maximum of <m>r</m> and <m>s</m>.
        </p>

        <p>
          We must show that
          <me>
            G = G_1 G_2 \cdots  G_k
          </me>
          and <m>G_i \cap G_j = \{1 \}</m> for <m>i \neq j</m>.
          Suppose that <m>g_1 \in G_1</m> is in the subgroup generated by <m>G_2, G_3, \ldots, G_k</m>.
          Then <m>g_1 = g_2 g_3 \cdots g_k</m> for <m>g_i \in G_i</m>.
          Since <m>g_i</m> has order <m>p_i^{\alpha_i}</m>,
          we know that <m>g_i^{p^{\alpha_i}} = 1</m> for <m>i = 2, 3, \ldots, k</m>,
          and <m>g_1^{p_2^{\alpha_2} \cdots p_k^{\alpha_k}} = 1</m>.
          Since the order of <m>g_1</m> is a power of <m>p_1</m> and <m>\gcd(p_1, p_2^{\alpha_2} \cdots p_k^{\alpha_k}) = 1</m>,
          it must be the case that <m>g_1 = 1</m> and the intersection of <m>G_1</m> with any of the subgroups <m>G_2, G_3, \ldots, G_k</m> is the identity.
          A similar argument shows that <m>G_i \cap G_j = \{1 \}</m> for <m>i \neq j</m>.
        </p>

        <p>
          Next, we must show that it possible to write every <m>g \in G</m> as a product
          <m>g_1 \cdots g_k</m>, where <m>g_i \in G_i</m>.
          Since the order of <m>g</m> divides the order of <m>G</m>,
          we know that
          <me>
            |g| = p_1^{\beta_1}  p_2^{\beta_2} \cdots p_k^{\beta_k}
          </me>
          for some integers <m>\beta_1, \ldots, \beta_k</m>.
          Letting <m>a_i = |g| / p_i^{\beta_i}</m>,
          the <m>a_i</m>'s are relatively prime;
          hence, there exist integers
          <m>b_1, \ldots,
          b_k</m> such that <m>a_1 b_1 + \cdots + a_k b_k = 1</m>.
          Consequently,
          <me>
            g = g^{a_1 b_1 + \cdots + a_k b_k} = g^{a_1 b_1} \cdots  g^{a_k b_k}
          </me>.
          Since
          <me>
            g^{(a_i b_i ) p_i^{\beta_i}} = g^{b_i |g|} = e
          </me>,
          it follows that <m>g^{a_i b_i}</m> must be in <m>G_{i}</m>.
          Let <m>g_i = g^{a_i b_i}</m>.
          Then <m>g = g_1 \cdots g_k \in G_1 G_2 \cdots G_k</m>.
          Therefore, <m>G = G_1 G_2 \cdots G_k</m> is an internal direct product of subgroups.
        </p>
      </proof>
    </lemma>
<!--
 		<lemma>
			<statement>
				<p>Every finite abelian group <m>G</m> is the internal direct product of <m>p</m>-groups.</p>
			</statement>
			<proof>
				<p>If <m>|G|= 1</m>, then the theorem is trivial. Suppose that the order of <m>G</m> is greater than 1, say 
					<me>|G| = p_1^{\alpha_1} \cdots p_n^{\alpha_n},</me>
				where <m>p_1, \ldots, p_n</m> are all prime, and define <m>G_i</m> to be the set of elements in <m>G</m> of order <m>p_i^k</m> for some integer <m>k</m>. Since <m>G</m> is an abelian group, we are guaranteed that <m>G_i</m> is a subgroup of <m>G</m> for <m>i = 1, \ldots, n</m>. We must show that
					<me>G = G_1 G_2 \cdots  G_n.</me>
				That is, we must be able to write every <m>g \in G</m> as a unique product <m>g_{p_1} \cdots g_{p_n}</m> where <m>g_{p_i}</m> is of the order of some power of <m>p_i</m>. Since the order of <m>g</m> divides the order of <m>G</m>, we know that 
					<me>|g| = p_1^{\beta_1}  p_2^{\beta_2} \cdots p_n^{\beta_n}</me>
				for some integers <m>\beta_1, \ldots, \beta_n</m>. Letting <m>a_i = |g| / p_i^{\beta_i}</m>, the <m>a_i</m>'s are relatively prime; hence, there exist integers <m>b_1, \ldots, b_n</m> such that <m>a_1 b_1 + \cdots + a_n b_n = 1</m>. Consequently, 
					<me>g = g^{a_1 b_1 + \cdots + a_n b_n} = g^{a_1 b_1} \cdots  g^{a_n b_n}.</me>
				Since
					<me>g^{(a_i b_i ) p_i^{\beta_i}} = g^{b_i |g|} = e,</me>
				it follows that <m>g^{a_i b_i}</m> must be in <m>G_{i}</m>. Let <m>g_i = g^{a_i b_i}</m>. Then <m>g = g_1 \cdots g_n</m> and <m>G_i \cap G_j = \{ e \}</m> for <m>i \neq j</m>.</p>

				<p>To show uniqueness, suppose that <m>g = g_1 \cdots g_n = h_1 \cdots h_n</m>, with <m>h_i \in G_i</m>. Then
					<me>e = (g_1 \cdots g_n)(h_1 \cdots h_n)^{-1} = g_1 h_1^{-1} \cdots g_n h_n^{-1}.</me>
				The order of <m>g_i h_i^{-1}</m> is a power of <m>p_i</m>; hence, the order of <m>g_1 h_1^{-1} \cdots g_n h_n^{-1}</m> is the least common multiple of the orders of the <m>g_i h_i^{-1}</m>.  This must be 1, since the order of the identity is 1. Therefore, <m>|g_i h_i^{-1}| =1</m> or <m>g_i =h_i</m> for <m>i = 1, \ldots, n</m>.</p>
			</proof>

		</lemma>
-->
        <!-- Changed the statement and proof of the theorem to -->
        <!-- reflect that we are dealing with internal direct products.  TWJ 4/24/2013 -->
        <!--

		<p>The proof of the Fundamental Theorem relies on the following two lemmas.</p>

		 <lemma>
			<statement>
				<p>Let <m>G</m> be a finite abelian group and <m>p</m> a prime such that <m>p</m> divides the order of <m>G</m>. Then <m>G</m> contains an element of order <m>p</m>.</p>
			</statement>
			<proof>
				<p>We will use mathematical induction and the fact that if <m>G</m> is a cyclic group and <m>m \mid |G|</m>, then <m>G</m> must have an element of order <m>m</m> (<xref ref="cyclic-exercise-subgroups-exist" /> in <xref ref="cyclic" />).  Suppose that <m>|G| = n</m>.  By the Fundamental Theorem of Arithmetic <xref ref="theorem-fund-theorem-arithmetic" />), we can write
					<me>n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},</me>
				where <m>p_1, p_2, \ldots, p_k</m> are distinct primes.  We will induct on <m>S(n) = e_1 + e_2 + \cdots + e_k</m>.  If <m>S(n) = 1</m>, then the order of <m>G</m> is <m>p</m> and there is nothing to prove.  Now assume that the lemma is true for all groups <m>G'</m>, where <m>S(|G'|) \lt S(n)</m>.  Let <m>a \in G</m> such that <m>|a| \gt 1</m>.  If <m>p \mid |a|</m>, then we are done by the comments above.  If  <m>p \nmid |a|</m>, let <m>H = \langle a \rangle</m>.  Since <m>G</m> is an abelian group, we know that <m>H</m> is normal in <m>G</m>, and <m>S(G/H) \lt S(n)</m>. Moreover, <m>p</m> divides <m>|G/H| = |G|/|H|</m>, since <m>p \mid |G|</m> but <m>p \nmid |H|</m>.  By the induction hypothesis there exists an element <m>bH \in G/H</m> of order <m>p</m>.  Suppose that the order of <m>b</m> in <m>G</m> is <m>m</m>.  Then <m>(bH)^m = eH</m> in <m>G/H</m>, and so <m>p</m> must divide <m>m</m>.  Therefore, <m>b^{m/p}</m> is an element of order <m>p</m> in <m>G</m>.</p>
			</proof>

		</lemma>
-->
    <p>
      If remains for us to determine the possible structure of each <m>p_i</m>-group <m>G_i</m> in <xref ref="struct-lemma-direct-product-p-groups"/>.
    </p>

    <lemma xml:id="struct-lemma-struct-finite-abelian">
      <statement>
        <p>
          Let <m>G</m> be a finite abelian <m>p</m>-group and suppose that <m>g \in G</m> has maximal order.
          Then <m>G</m> is isomorphic to
          <m>\langle g \rangle \times H</m> for some subgroup <m>H</m> of <m>G</m>.
        </p>
      </statement>

      <proof>
        <p>
          By <xref ref="struct-lemma-p-group-order"/>,
          we may assume that the order of <m>G</m> is <m>p^n</m>.
          We shall induct on <m>n</m>.
          If <m>n= 1</m>,
          then <m>G</m> is cyclic of order <m>p</m> and must be generated by <m>g</m>.
          Suppose now that the statement of the lemma holds for all integers <m>k</m> with
          <m>1 \leq k \lt n</m> and let <m>g</m> be of maximal order in <m>G</m>,
          say <m>|g| = p^{m}</m>.
          Then <m>a^{p^m} = e</m> for all <m>a \in G</m>.
          Now choose <m>h</m> in <m>G</m> such that <m>h \notin \langle g \rangle</m>,
          where <m>h</m> has the smallest possible order.
          Certainly such an <m>h</m> exists;
          otherwise, <m>G = \langle g \rangle</m> and we are done.
          Let <m>H = \langle h \rangle</m>.
        </p>

        <p>
          We claim that <m>\langle g \rangle \cap H = \{ e \}</m>.
          It suffices to show that <m>|H|=p</m>.
          Since <m>|h^p| = |h| / p</m>,
          the order of <m>h^p</m> is smaller than the order of <m>h</m> and must be in
          <m>\langle g \rangle</m> by the minimality of <m>h</m>;
          that is, <m>h^p = g^r</m> for some number <m>r</m>.
          Hence,
          <me>
            (g^r)^{p^{m - 1}} = (h^p)^{p^{m - 1}} = h^{p^{m}} = e
          </me>,
          and the order of <m>g^r</m> must be less than or equal to <m>p^{m-1}</m>.
          Therefore, <m>g^r</m> cannot generate <m>\langle g \rangle</m>.
          Notice that <m>p</m> must occur as a factor of <m>r</m>,
          say <m>r = ps</m>, and <m>h^p = g^r = g^{ps}</m>.
          Define <m>a</m> to be <m>g^{-s}h</m>.
          Then <m>a</m> cannot be in <m>\langle g \rangle</m>;
          otherwise, <m>h</m> would also have to be in <m>\langle g \rangle</m>.
          Also,
          <me>
            a^p = g^{-sp} h^p = g^{-r} h^p = h^{-p} h^p = e
          </me>.
          We have now formed an element <m>a</m> with order <m>p</m> such that <m>a \notin \langle g \rangle</m>.
          Since <m>h</m> was chosen to have the smallest order of all of the elements that are not in <m>\langle g\rangle</m>,
          <m>|H|  = p</m>.
        </p>

        <p>
          Now we will show that the order of <m>gH</m> in the factor group <m>G/H</m> must be the same as the order of <m>g</m> in <m>G</m>.
          If <m>|gH| \lt |g| =  p^m</m>, then
          <me>
            H = (gH)^{p^{m-1}} =  g^{p^{m-1}} H;
          </me>
          hence, <m>g^{p^{m-1}}</m> must be in <m>\langle g \rangle \cap H = \{ e \}</m>,
          which contradicts the fact that the order of <m>g</m> is <m>p^m</m>.
          Therefore, <m>gH</m> must have maximal order in <m>G/H</m>.
          By the Correspondence Theorem and our induction hypothesis,
          <me>
            G/H \cong \langle gH \rangle \times K/H
          </me>
          for some subgroup <m>K</m> of <m>G</m> containing <m>H</m>.
          We claim that <m>\langle g \rangle \cap K = \{ e \}</m>.
          If <m>b \in \langle g \rangle \cap K</m>,
          then <m>bH \in \langle gH \rangle \cap K/H = \{ H \}</m> and <m>b \in \langle g \rangle \cap H = \{ e \}</m>.
          It follows that <m>G = \langle g \rangle K</m> implies that <m>G \cong \langle g \rangle \times K</m>.
        </p>
      </proof>
    </lemma>
        <!-- Changed the statement and proof of the theorem to -->
        <!-- reflect that we are dealing with direct products.  Suggested by P. Diethelm. -->
        <!-- TWJ 4/24/2013 -->
    <p>
      The proof of the Fundamental Theorem of Finite Abelian Groups follows very quickly from <xref ref="struct-lemma-direct-product-p-groups"/> and <xref ref="struct-lemma-struct-finite-abelian"/>. By <xref ref="struct-lemma-direct-product-p-groups"/>, <m>G</m> is a product of <m>p</m>-groups.
      Suppose <m>G</m> is a <m>p</m>-group and let <m>g</m> be an element of maximal order in <m>G</m>.
      If <m>\langle g \rangle = G</m>, then we are done;
      otherwise, <m>G \cong {\mathbb Z}_{|g|} \times H</m> for some subgroup <m>H</m> contained in <m>G</m> by the <xref ref="struct-lemma-struct-finite-abelian"/>.
      Since <m>|H| \lt |G|</m>, we can apply mathematical induction.
    </p>

    <p>
      We now state the more general theorem for all finitely generated abelian groups.
      The proof of this theorem can be found in any of the references at the end of this chapter.
    </p>

    <theorem>
      <title>The Fundamental Theorem of Finitely Generated Abelian Groups</title>
      <statement>
        <p>
          Every finitely generated abelian group <m>G</m> is isomorphic to a direct product of cyclic groups of the form
          <me>
            {\mathbb Z}_{p_1^{ \alpha_1 }} \times {\mathbb Z}_{p_2^{ \alpha_2 }} \times \cdots \times {\mathbb Z}_{p_n^{ \alpha_n }} \times {\mathbb Z} \times \cdots \times {\mathbb Z}
          </me>,
          where the <m>p_i</m>'s are primes
          (not necessarily distinct).
        </p>
      </statement>
    </theorem>
  </section>

  <section xml:id="struct-section-solvable-groups">
    <title>Solvable Groups</title>
    <p>
      A <term>subnormal series</term><idx><h>Subnormal series of a group</h></idx>  of a group <m>G</m> is a finite sequence of subgroups
      <me>
        G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}
      </me>,
      where <m>H_i</m> is a normal subgroup of <m>H_{i+1}</m>.
      If each subgroup <m>H_i</m> is normal in <m>G</m>,
      then the series is called a <idx><h>Normal series of a group</h></idx><term>normal series</term>.
      The <term>length</term> of a  subnormal or normal series is the number of proper inclusions.
    </p>

    <example xml:id="struct-example-normal-series">
      <p>
        Any series of subgroups of an abelian group is a normal series.
        Consider the following  series of groups:
        <md>
          <mrow>{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\},</mrow>
          <mrow>{\mathbb Z}_{24} \supset \langle 2 \rangle \supset \langle 6 \rangle \supset \langle 12 \rangle \supset \{0\}</mrow>
        </md>.
      </p>
    </example>

    <example xml:id="struct-example-subnormal-series">
      <p>
        A subnormal series need not be a normal series.
        Consider the following subnormal series of the group <m>D_4</m>:
        <me>
          D_4 \supset \{ (1), (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \} \supset \{  (1), (1 \, 2)(3 \, 4) \} \supset \{ (1) \}
        </me>.
        The subgroup <m>\{  (1), (1 \, 2)(3 \, 4) \}</m> is not normal in <m>D_4</m>;
        consequently, this series is not a normal series.
      </p>
    </example>

    <p>
      A subnormal (normal) series <m>\{ K_j \}</m> is a
      <term>refinement of a subnormal (normal) series</term>
      <m>\{ H_i \}</m> if <m>\{ H_i \} \subset \{ K_j \}</m>.
      That is, each <m>H_i</m> is one of the <m>K_j</m>.
    </p>

    <example xml:id="struct-example-refinement">
      <p>
        The series
        <me>
          {\mathbb Z} \supset 3{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 90{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\}
        </me>
        is a refinement of the series
        <me>
          {\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\}
        </me>.
      </p>
    </example>

    <p>
      The best way to study a subnormal or normal series of subgroups,
      <m>\{ H_i \}</m> of <m>G</m>,
      is actually to study the factor groups <m>H_{i+1}/H_i</m>.
      We say that two subnormal (normal) series <m>\{H_i \}</m> and
      <m>\{ K_j \}</m> of a group <m>G</m> are <term>isomorphic</term>
      if there is a one-to-one correspondence between the collections of factor groups
      <m>\{H_{i+1}/H_i \}</m> and <m>\{ K_{j+1}/ K_j \}</m>.
    </p>

    <example xml:id="struct-example-isomorph-series">
      <p>
        The two normal series
        <md>
          <mrow>{\mathbb Z}_{60} \supset \langle 3 \rangle \supset  \langle 15 \rangle \supset \{ 0 \}</mrow>
          <mrow>{\mathbb Z}_{60} \supset \langle 4 \rangle \supset  \langle 20 \rangle \supset \{ 0 \}</mrow>
        </md>
        of the group <m>{\mathbb Z}_{60}</m> are isomorphic since
        <md>
          <mrow>{\mathbb Z}_{60} / \langle 3 \rangle \cong \langle 20 \rangle / \{ 0 \} \cong {\mathbb Z}_{3}</mrow>
          <mrow>\langle 3 \rangle / \langle 15 \rangle \cong \langle 4 \rangle /  \langle 20 \rangle \cong {\mathbb Z}_{5}</mrow>
          <mrow>\langle 15 \rangle / \{ 0 \} \cong {\mathbb Z}_{60} / \langle 4 \rangle \cong {\mathbb Z}_4</mrow>
        </md>.
      </p>
    </example>

    <p>
      A subnormal series <m>\{ H_i \}</m> of a group <m>G</m> is a
      <term>composition series</term><idx><h>Composition series</h></idx>
      if all the factor groups are simple;
      that is, if none of the factor groups of the series contains a normal subgroup.
      A normal series <m>\{ H_i \}</m> of <m>G</m> is a
      <term>principal series</term><idx><h>Principal series</h></idx>
      if all the factor groups are simple.
    </p>

    <example xml:id="struct-example-composition-series">
      <p>
        The group <m>{\mathbb Z}_{60}</m> has  a composition series
        <me>
          {\mathbb Z}_{60} \supset \langle 3 \rangle \supset  \langle 15 \rangle \supset \langle 30 \rangle  \supset \{ 0 \}
        </me>
        with factor groups
        <md>
          <mrow>{\mathbb Z}_{60} / \langle 3 \rangle &amp; \cong  {\mathbb Z}_{3}</mrow>
          <mrow>\langle 3 \rangle / \langle 15 \rangle &amp; \cong  {\mathbb Z}_{5}</mrow>
          <mrow>\langle 15 \rangle / \langle 30 \rangle &amp; \cong  {\mathbb Z}_{2}</mrow>
          <mrow>\langle 30 \rangle / \{ 0 \} &amp; \cong  {\mathbb Z}_2</mrow>
        </md>.
        Since <m>{\mathbb Z}_{60}</m> is an abelian group,
        this series is automatically a principal series.
        Notice that a composition series need not be unique.
        The series
        <me>
          {\mathbb Z}_{60} \supset \langle 2 \rangle \supset \langle 4 \rangle \supset  \langle 20 \rangle \supset \{ 0 \}
        </me>
        is also a composition series.
      </p>
    </example>

    <example xml:id="struct-example-sn_series">
      <p>
        For <m>n \geq 5</m>, the series
        <me>
          S_n \supset A_n \supset \{ (1) \}
        </me>
        is a composition series for <m>S_n</m> since
        <m>S_n / A_n \cong {\mathbb Z}_2</m> and <m>A_n</m> is simple.
      </p>
    </example>
        <!-- typo corrected.  Suggested by L. Franklin. -->
        <!-- TWJ - 12/19/2011 -->
    <example xml:id="struct-example-z-series">
      <p>
        Not every group has a composition series or a principal series.
        Suppose that
        <me>
          \{ 0 \} = H_0 \subset H_1 \subset \cdots \subset H_{n-1} \subset H_n = {\mathbb Z}
        </me>
        is a subnormal series for the integers under addition.
        Then <m>H_1</m> must be of the form
        <m>k {\mathbb Z}</m> for some <m>k \in {\mathbb N}</m>.
        In this case <m>H_1 / H_0 \cong k {\mathbb Z}</m> is an infinite cyclic group with many nontrivial proper normal subgroups.
      </p>
    </example>
        <!-- changed n to k in the example.  Suggested by P. Diethelm. -->
        <!-- TWJ 4/24/2013 -->
    <p>
      Although composition series need not be unique as in the case of <m>{\mathbb Z}_{60}</m>,
      it turns out that any two composition series are related.
      The factor groups of the two composition series for
      <m>{\mathbb Z}_{60}</m> are <m>{\mathbb Z}_2</m>,
      <m>{\mathbb Z}_2</m>, <m>{\mathbb Z}_3</m>, and <m>{\mathbb Z}_5</m>;
      that is, the two composition series are isomorphic.
      The Jordan-Hölder Theorem says that this is always the case.
    </p>

    <theorem>
      <title>Jordan-Hölder</title>
      <idx>
      <h>Jordan-Hölder Theorem</h>
      </idx>
      <statement>
        <p>
          Any two composition series of <m>G</m> are isomorphic.
        </p>
      </statement>

      <proof>
        <p>
          We shall employ mathematical induction on the length of the composition series.
          If the length of a composition series is 1, then <m>G</m> must be a simple group.
          In this case any two composition series are isomorphic.
        </p>

        <p>
          Suppose now that the theorem is true for all groups having a composition series of length <m>k</m>,
          where <m>1 \leq k \lt n</m>.
          Let
          <md>
            <mrow>G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}</mrow>
            <mrow>G = K_m \supset K_{m-1} \supset \cdots \supset K_1 \supset K_0 = \{ e \}</mrow>
          </md>
          be two composition series for <m>G</m>.
          We can form two new subnormal series for <m>G</m> since
          <m>H_i \cap K_{m-1}</m> is normal in <m>H_{i+1} \cap K_{m-1}</m> and
          <m>K_j \cap H_{n-1}</m> is normal in <m>K_{j+1} \cap H_{n-1}</m>:
          <md>
            <mrow>G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}</mrow>
            <mrow>G = K_m \supset K_{m-1} \supset K_{m-1} \cap H_{n-1} \supset \cdots \supset K_0 \cap H_{n-1} = \{ e \}</mrow>
          </md>.
          Since <m>H_i \cap K_{m-1}</m> is normal in <m>H_{i+1} \cap K_{m-1}</m>,
          the Second Isomorphism Theorem (<xref ref="homomorph-theorem-second-isomorphism"/>) implies that
          <md>
            <mrow>(H_{i+1} \cap K_{m-1}) / (H_i \cap K_{m-1}) &amp; =  (H_{i+1} \cap K_{m-1}) / (H_i \cap ( H_{i+1} \cap K_{m-1} ))</mrow>
            <mrow>&amp; \cong  H_i (H_{i+1} \cap K_{m-1})/ H_i</mrow>
          </md>,
          where <m>H_i</m> is normal in <m>H_i (H_{i+1} \cap K_{m-1})</m>.
          Since <m>\{ H_i \}</m> is a composition series,
          <m>H_{i+1} / H_i</m> must be simple;
          consequently,
          <m>H_i (H_{i+1} \cap K_{m-1})/ H_i</m> is either <m>H_{i+1}/H_i</m> or <m>H_i/H_i</m>.
          That is, <m>H_i (H_{i+1} \cap K_{m-1})</m> must be either <m>H_i</m> or <m>H_{i+1}</m>.
          Removing any nonproper inclusions from the series
          <me>
            H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}
          </me>,
          we have a composition series for <m>H_{n-1}</m>.
          Our induction hypothesis says that this series must be equivalent to the composition series
          <me>
            H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}
          </me>.
          Hence, the composition series
          <me>
            G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}
          </me>
          and
          <me>
            G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}
          </me>
          are equivalent.
          If <m>H_{n-1} = K_{m-1}</m>,
          then the composition series <m>\{H_i \}</m> and
          <m>\{ K_j \}</m> are equivalent and we are done;
          otherwise, <m>H_{n-1} K_{m-1}</m>  is a normal subgroup of <m>G</m> properly containing <m>H_{n-1}</m>.
          In this case <m>H_{n-1} K_{m-1} = G</m> and we can apply the Second Isomorphism Theorem once again; that is,
          <me>
            K_{m-1} / (K_{m-1} \cap H_{n-1}) \cong (H_{n-1} K_{m-1}) / H_{n-1} = G/H_{n-1}
          </me>.
          Therefore,
          <me>
            G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}
          </me>
          and
          <me>
            G = K_m \supset K_{m-1} \supset K_{m-1} \cap H_{n-1} \supset \cdots \supset K_0 \cap H_{n-1} = \{ e \}
          </me>
          are equivalent and the proof of the theorem is complete.
        </p>
      </proof>
    </theorem>

    <p>
      A group <m>G</m> is <term>solvable</term><idx><h>Group</h><h>solvable</h></idx>
      if it has a subnormal series
      <m>\{ H_i \}</m> such that all of the factor groups <m>H_{i+1} / H_i</m> are abelian.
      Solvable groups will play a fundamental role when we study Galois theory and the solution of polynomial equations.
    </p>
        <!-- Corrected the definition of a solvable group.  Suggested by K. Halasz. -->
        <!-- TWJ 1/10/2014 -->
    <example xml:id="solvable">
      <p>
        The group <m>S_4</m> is solvable since
        <me>
          S_4 \supset A_4 \supset \{ (1), (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \} \supset \{ (1) \}
        </me>
        has abelian factor groups; however,
        for <m>n \geq 5</m> the series
        <me>
          S_n \supset A_n \supset \{ (1) \}
        </me>
        is a composition series for <m>S_n</m> with a nonabelian factor group.
        Therefore, <m>S_n</m> is not a solvable group for <m>n \geq 5</m>.
      </p>
    </example>

    <paragraphs component="sage-blurb">
      <title>Sage</title>
      <p>
        Sage is able to create direct products of cyclic groups,
        though they are realized as permutation groups.
        This is a situation that should improve.
        However, with a classification of finite abelian groups,
        we can describe how to construct in Sage every group of order less than <m>16</m>.
      </p>
    </paragraphs>

  </section>

  <reading-questions>

    <exercise label="rq-struct-200">
      <statement>
        <p>
          How many abelian groups are there of order <m>200 = 2^3 5^2</m>?
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-struct-729">
      <statement>
        <p>
          How many abelian groups are there of order <m>729=3^6</m>?
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-struct-subgroup-order6">
      <statement>
        <p>
          Find a subgroup of order 6 in <m>\mathbb Z_8\times\mathbb Z_3\times\mathbb Z_3</m>.
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-struct-72">
      <statement>
        <p>
          It can be shown that an abelian group of order <m>72</m> contains a subgroup of order <m>8</m>.
          What
          are the possibilities for this subgroup?
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-struct-principal-series">
      <statement>
        <p>
          What is a principal series of the group <m>G</m>?
          Your answer should not use new terms defined in this chapter.
        </p>
      </statement>
      <response/>
    </exercise>

  </reading-questions>
    <!-- Exercises with Solutions  -->
    <!-- File: struct.xml  -->
    <!-- Title: The Structure of Groups -->
  <exercises xml:id="struct-exercises" filenamebase="struct">
    <title>Exercises</title>

    <exercise number="1" xml:id="struct-exercise-order-40">
      <statement>
        <p>
          Find all of the abelian groups of order less than or equal to <m>40</m> up to isomorphism.
        </p>
      </statement>
      <hint>
        <p>
          There are three possible groups.
        </p>
      </hint>
    </exercise>

    <exercise number="2" xml:id="struct-exercise-order-200">
      <statement>
        <p>
          Find all of the abelian groups of order <m>200</m> up to isomorphism.
        </p>
      </statement>
    </exercise>

    <exercise number="3" xml:id="struct-exercise-order-729">
      <statement>
        <p>
          Find all of the abelian groups of order <m>720</m> up to isomorphism.
        </p>
      </statement>
    </exercise>

    <exercise number="4" xml:id="struct-exercise-composition-series">
      <statement>
        <p>
          Find all of the composition series for each of the following groups.

          <ol cols="2">
            <li>
              <p>
                <m>{\mathbb Z}_{12}</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Z}_{48}</m>
              </p>
            </li>

            <li>
              <p>
                The quaternions, <m>Q_8</m>
              </p>
            </li>

            <li>
              <p>
                <m>D_4</m>
              </p>
            </li>

            <li>
              <p>
                <m>S_3 \times {\mathbb Z}_4</m>
              </p>
            </li>

            <li>
              <p>
                <m>S_4</m>
              </p>
            </li>

            <li>
              <p>
                <m>S_n</m>, <m>n \geq 5</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}</m>
              </p>
            </li>
          </ol>
        </p>
      </statement>
      <hint>
        <p>
          (a) <m>\{ 0 \} \subset \langle 6 \rangle \subset \langle 3 \rangle \subset {\mathbb Z}_{12}</m>; (e) <m>\{ (1) \} \times \{ 0 \} \subset \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \} \times \{ 0 \} \subset S_3 \times \{ 0 \} \subset S_3 \times \langle 2 \rangle\subset S_3 \times {\mathbb Z}_4</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="5" xml:id="struct-exercise-infinite-direct-product">
      <statement>
        <p>
          Show that the infinite direct product
          <m>G = {\mathbb Z}_2 \times {\mathbb Z}_2 \times \cdots</m> is not finitely generated.
        </p>
      </statement>
    </exercise>

    <exercise number="6" xml:id="struct-exercise-abelian-group-subgroup">
      <statement>
        <p>
          Let <m>G</m> be an abelian group of order <m>m</m>.
          If <m>n</m> divides <m>m</m>,
          prove that <m>G</m> has a subgroup of order <m>n</m>.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="7" xml:id="struct-exercise-torsion-group">
      <statement>
        <p>
          A group <m>G</m> is a <term>torsion group</term>
          if every element of <m>G</m> has finite order.
          Prove that a finitely generated abelian torsion group must be finite.
        </p>
      </statement>
      <hint>
        <p>
          Use the Fundamental Theorem of Finitely Generated Abelian Groups.
        </p>
      </hint>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="8" xml:id="struct-exercise-finitely-generated-abelian-group">
      <statement>
        <p>
          Let <m>G</m>,
          <m>H</m>, and <m>K</m> be finitely generated abelian groups.
          Show that if <m>G \times H \cong G \times K</m>, then <m>H \cong K</m>.
          Give a counterexample to show that this cannot be true in general.
        </p>
      </statement>
    </exercise>

    <exercise number="9" xml:id="struct-exercise-GxH-solvable">
      <statement>
        <p>
          Let <m>G</m> and <m>H</m> be solvable groups.
          Show that <m>G \times H</m> is also solvable.
        </p>
      </statement>
    </exercise>

    <exercise number="10" xml:id="struct-exercise-composition-principal-series">
      <statement>
        <p>
          If <m>G</m> has a composition (principal) series and if <m>N</m> is a proper normal subgroup of <m>G</m>,
          show there exists a composition (principal) series containing <m>N</m>.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="11" xml:id="struct-exercise-N-GmodN-comp-series">
      <statement>
        <p>
          Prove or disprove: Let <m>N</m> be a normal subgroup of <m>G</m>.
          If <m>N</m> and <m>G/N</m> have composition series,
          then <m>G</m> must also have a composition series.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="12" xml:id="struct-exercise-N-normal-solvable">
      <statement>
        <p>
          Let <m>N</m> be a normal subgroup of <m>G</m>.
          If <m>N</m> and <m>G/N</m> are solvable groups,
          show that <m>G</m> is also a solvable group.
        </p>
      </statement>
      <hint>
        <p>
          If <m>N</m> and <m>G/N</m> are solvable,
          then they have solvable series
          <md>
            <mrow>N = N_n \supset N_{n - 1} \supset \cdots \supset N_1 \supset N_0 = \{ e \}</mrow>
            <mrow>G/N = G_n/N \supset G_{n - 1}/N \supset \cdots G_1/N \supset G_0/N = \{ N \}</mrow>
          </md>.
        </p>
      </hint>
    </exercise>

    <exercise number="13" xml:id="struct-exercise-solvable-series">
      <statement>
        <p>
          Prove that <m>G</m> is a solvable group if and only if <m>G</m> has a series of subgroups
          <me>
            G = P_n \supset P_{n - 1} \supset \cdots \supset P_1 \supset P_0 = \{ e \}
          </me>
          where <m>P_i</m> is normal in
          <m>P_{i + 1}</m> and the order of <m>P_{i + 1} / P_i</m> is prime.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="14" xml:id="struct-exercise-subgroup-solvable">
      <statement>
        <p>
          Let <m>G</m> be a solvable group.
          Prove that any subgroup of <m>G</m> is also solvable.
        </p>
      </statement>
    </exercise>

    <exercise number="15" xml:id="struct-exercise-Gmod-solvable">
      <statement>
        <p>
          Let <m>G</m> be a solvable group and <m>N</m> a normal subgroup of <m>G</m>.
          Prove that <m>G/N</m> is solvable.
        </p>
      </statement>
    </exercise>

    <exercise number="16" xml:id="struct-exercise-Dn-solvable">
      <statement>
        <p>
          Prove that <m>D_n</m> is solvable for all integers <m>n</m>.
        </p>
      </statement>
      <hint>
        <p>
          Use the fact that <m>D_n</m> has a cyclic subgroup of index <m>2</m>.
        </p>
      </hint>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="17" xml:id="struct-exercise-G-compostion-series">
      <statement>
        <p>
          Suppose that <m>G</m> has a composition series.
          If <m>N</m> is a normal subgroup of <m>G</m>,
          show that <m>N</m> and <m>G/N</m> also have composition series.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="18" xml:id="struct-exercise-cyclic">
      <statement>
        <p>
          Let <m>G</m> be a cyclic <m>p</m>-group with subgroups <m>H</m> and <m>K</m>.
          Prove that either <m>H</m> is contained in <m>K</m> or <m>K</m> is contained in <m>H</m>.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="19" xml:id="struct-exercise-G-solvable-normal-abelian-subgroup">
      <statement>
        <p>
          Suppose that <m>G</m> is a solvable group with order <m>n \geq 2</m>.
          Show that <m>G</m> contains a normal nontrivial abelian subgroup.
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="20" xml:id="struct-exercise-commutator-subgroup">
      <statement>
        <p>
          Recall that the <term>commutator subgroup</term>
          <m>G'</m> of a group <m>G</m> is defined as the subgroup of <m>G</m> generated by elements of the form
          <m>a^{-1} b ^{-1} ab</m> for <m>a, b \in G</m>.
          We can define a series of subgroups of <m>G</m> by <m>G^{(0)} = G</m>,
          <m>G^{(1)} = G'</m>, and <m>G^{(i + 1)} = (G^{(i)})'</m>.

          <ol>
            <li>
              <p>
                Prove that <m>G^{(i+1)}</m> is normal in <m>(G^{(i)})'</m>.
                The series of subgroups
                <me>
                  G^{(0)} = G \supset G^{(1)} \supset G^{(2)} \supset \cdots
                </me>
                is called the <term>derived series</term> of <m>G</m>.
              </p>
            </li>

            <li>
              <p>
                Show that <m>G</m> is solvable if and only if
                <m>G^{(n)} = \{ e \}</m> for some integer <m>n</m>.
              </p>
            </li>
          </ol>
        </p>
      </statement>
    </exercise>

    <!--Todo Expand this solution.-->

    <exercise number="21" xml:id="struct-exercise-abelian-factor-group">
      <statement>
        <p>
          Suppose that <m>G</m> is a solvable group with order <m>n \geq 2</m>.
          Show that <m>G</m> contains a normal nontrivial abelian factor group.
        </p>
      </statement>
      <hint>
        <p>
          <m>G/G'</m> is abelian.
        </p>
      </hint>
    </exercise>

    <!--Todo Expand solution.-->

    <exercise number="22" xml:id="struct-exercise-zassenhaus">
      <title>Zassenhaus Lemma</title>
      <statement>
        <p>
          Let <m>H</m> and <m>K</m> be subgroups of a group <m>G</m>.
          Suppose also that <m>H^*</m> and <m>K^*</m> are normal subgroups of <m>H</m> and <m>K</m> respectively.
          Then

          <ol>
            <li>
              <p>
                <m>H^* ( H \cap K^*)</m> is a normal subgroup of <m>H^* ( H \cap K)</m>.
              </p>
            </li>

            <li>
              <p>
                <m>K^* ( H^* \cap K)</m> is a normal subgroup of <m>K^* ( H \cap K)</m>.
              </p>
            </li>

            <li>
              <p>
                <m>H^* ( H \cap K) / H^* ( H \cap K^*) \cong K^* ( H \cap K) / K^* ( H^* \cap K) \cong (H \cap K) / (H^* \cap K)(H \cap K^*)</m>.
              </p>
            </li>
          </ol>
        </p>
      </statement>
    </exercise>

    <exercise number="23" xml:id="struct-exercise-schreir">
      <title>Schreier's Theorem</title>
      <statement>
        <p>
          Use the Zassenhaus Lemma to prove that two subnormal (normal) series of a group <m>G</m> have isomorphic refinements.
        </p>
      </statement>
    </exercise>

    <exercise number="24" xml:id="struct-exercise-jordan-holder">
      <statement>
        <p>
          Use Schreier's Theorem to prove the Jordan-Hölder Theorem.
        </p>
      </statement>
    </exercise>

  </exercises>

  <exercises xml:id="struct-exercises-programming">
    <title>Programming Exercises</title>
    <exercise xml:id="struct-programming-exercise-abelian-groups">
      <statement>
        <p>
          Write a program that will compute all possible abelian groups of order <m>n</m>.
          What is the largest <m>n</m> for which your program will work?
        </p>
      </statement>
    </exercise>

  </exercises>

  <references xml:id="struct-references">
        <!-- %References checked and updated - TWJ 6/22/2010 -->
    <title>References and Suggested Readings</title>
        <!--Fix INTRO
		<p>Each of the following references contains a proof of the Fundamental Theorem of Finitely Generated Abelian Groups.</p>

	-->
    <biblio type="raw">
<!-- was [1] -->
      Hungerford, T. W.
      <title>Algebra</title>. Springer, New York, 1974.
    </biblio>
        <!-- %Reference updated 6/22/2010 - TWJ -->
    <biblio type="raw">
<!-- was [2] -->
      Lang, S.
      <title>Algebra</title>. 3rd ed. Springer, New York, 2002.
    </biblio>
        <!-- %Reference updated 6/22/2010 - TWJ -->
    <biblio type="raw">
<!-- was [3] --><!-- %Reference updated 6/22/2010 - TWJ -->
      Rotman, J. J.
      <title>An Introduction to the Theory of Groups</title>. 4th ed. Springer, New York, 1995.
    </biblio>
        <!-- %Reference updated 6/22/2010 - TWJ -->
  </references>



  <section xml:id="struct-sage" component="sage-info">
    <title>Sage</title>
    <introduction>
      <p>
        Cyclic groups, and direct products of cyclic groups,
        are implemented in Sage as permutation groups.
        However, these groups quickly become very unwieldly representations and it should be easier to work with finite abelian groups in Sage.
        So we will postpone any specifics for this chapter until that happens.
        However, now that we understand the notion of isomorphic groups and the structure of finite abelian groups,
        we can return to our quest to classify all of the groups with order less than <m>16</m>.
      </p>
    </introduction>

    <subsection>
      <title>Classification of Finite Groups</title>
      <p>
        It does not take any sophisticated tools to understand groups of order <m>2p</m>,
        where <m>p</m> is an odd prime.
        There are two possibilities <mdash/> a cyclic group of order <m>2p</m> and the dihedral group of order <m>2p</m> that is the set of symmetries of a regular <m>p</m>-gon.
        The proof requires some close, tight reasoning,
        but the required theorems are generally just concern orders of elements, Lagrange's Theorem and cosets.
        See <xref ref="isomorph-exercise-classify-2p"/>.
        This takes care of orders <m>n=6,\,10,\,14</m>.
      </p>

      <p>
        For <m>n=9</m>,
        the upcoming <xref ref="actions-corollary-p2abelian"/> will tell us that any group of order <m>p^2</m>
        (where <m>p</m> is a prime)
        is abelian.
        So we know from this section that the only two possibilities are
        <m>{\mathbb Z}_9</m> and <m>{\mathbb Z}_3\times{\mathbb Z}_3</m>.
        Similarly, the upcoming <xref ref="sylow-theorem-pq-cyclic"/> will tell us that every group of order <m>n=15</m> is abelian.
        Now this leaves just one possibility for this order:
        <m>{\mathbb Z}_3\times{\mathbb Z}_5\cong{\mathbb Z}_{15}</m>.
      </p>

      <p>
        We have just two orders left to analyze:
        <m>n=8</m> and <m>n=12</m>.
        The possibilities are groups we already know, with one exception.
        However, the analysis that these are the <em>only</em>
        possibilities is more complicated,
        and will not be pursued now, nor in the next few chapters.
        Notice that <m>n=16</m> is more complicated still,
        with <m>14</m> different possibilities
        (which explains why we stopped here).
      </p>

      <p>
        For <m>n=8</m> there are <m>3</m> abelian groups,
        and the two non-abelian groups are the dihedral group
        (symmetries of a square)
        and the quaternions.
      </p>

      <p>
        For <m>n=12</m> there are <m>2</m> abelian groups,
        and <m>3</m> non-abelian groups.
        We know two of the non-abelian groups as a dihedral group,
        and the alternating group on <m>4</m> symbols
        (which is also the symmetries of a tetrahedron).
        The third non-abelian group is an example of a <q>dicyclic</q> group,
        which is an infinite family of groups,
        each with order divisible by <m>4</m>.
        The order <m>12</m> dicyclic group can also be constructed as a
        <q>semi-direct product</q>
        of two cyclic groups <mdash/> this is a construction worth knowing as you pursue further study of group theory.
        The order <m>8</m> dicyclic group is also the quaternions and more generally,
        the dicyclic groups of order <m>2^k</m>,
        <m>k&gt;2</m> are known as <q>generalized quaternion groups.</q>
      </p>

      <p>
        The following examples will show you how to construct some of these groups,
        while also exercising a few of the commands and allowing us to be more certain the following table is accurate.
      </p>

            <sage>
                <input>
        S = SymmetricGroup(3)
        D = DihedralGroup(3)
        S.is_isomorphic(D)
        </input>
                <output>
        True
        </output>
            </sage>
            <sage>
                <input>
        C3 = CyclicPermutationGroup(3)
        C5 = CyclicPermutationGroup(5)
        DP = direct_product_permgroups([C3, C5])
        C  = CyclicPermutationGroup(15)
        DP.is_isomorphic(C)
        </input>
                <output>
        True
        </output>
            </sage>
            <sage>
                <input>
        Q  = QuaternionGroup()
        DI = DiCyclicGroup(2)
        Q.is_isomorphic(DI)
        </input>
                <output>
        True
        </output>
            </sage>
    </subsection>

    <subsection>
      <title>Groups of Small Order as Permutation Groups</title>
      <p>
        We list here constructions, as permutation groups in Sage,
        for all of the groups of order less than <m>16</m>.
      </p>
            <!-- comments are the number of dividing rules following the row in LaTeX original -->
      <table>
        <title>The Groups of Order <m>\mathbf{15}</m> or Less in Sage</title>
        <tabular halign="left" top="medium">
          <row bottom="medium">
            <cell>Order</cell>
            <cell>Construction</cell>
            <cell>Notes, Alternatives</cell>
          </row>
          <row>
            <cell>1</cell>
            <cell><c>CyclicPermutationGroup(1)</c></cell>
            <cell>Trivial</cell>
          </row>
            <!--double-->
          <row>
            <cell>2</cell>
            <cell><c>CyclicPermutationGroup(2)</c></cell>
            <cell><c>SymmetricGroup(2)</c></cell>
          </row>
            <!--double-->
          <row>
            <cell>3</cell>
            <cell><c>CyclicPermutationGroup(3)</c></cell>
            <cell>Prime order</cell>
          </row>
            <!--double-->
          <row>
            <cell>4</cell>
            <cell><c>CyclicPermutationGroup(4)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>4</cell>
            <cell><c>KleinFourGroup()</c></cell>
            <cell>Abelian, non-cyclic</cell>
          </row>
            <!--double-->
          <row>
            <cell>5</cell>
            <cell><c>CyclicPermutationGroup(5)</c></cell>
            <cell>Prime order</cell>
          </row>
            <!--double-->
          <row>
            <cell>6</cell>
            <cell><c>CyclicPermutationGroup(6)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>6</cell>
            <cell><c>SymmetricGroup(3)</c></cell>
            <cell>Non-abelian</cell>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell/>
            <cell><c>DihedralGroup(3)</c></cell>
          </row>
            <!--double-->
          <row>
            <cell>7</cell>
            <cell><c>CyclicPermutationGroup(7)</c></cell>
            <cell>Prime order</cell>
          </row>
            <!--double-->
          <row>
            <cell>8</cell>
            <cell><c>CyclicPermutationGroup(8)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>8</cell>
            <cell><c>C2=CyclicPermutationGroup(2)</c></cell>
            <cell/>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell><c>C4=CyclicPermutationGroup(4)</c></cell>
            <cell/>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell><c>G=direct_product_permgroups([C2,C4])</c></cell>
            <cell>Abelian, non-cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>8</cell>
            <cell><c>C2=CyclicPermutationGroup(2)</c></cell>
            <cell/>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell><c>G=direct_product_permgroups([C2,C2,C2])</c></cell>
            <cell>Abelian, non-cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>8</cell>
            <cell><c>DihedralGroup(4)</c></cell>
            <cell>Non-abelian</cell>
          </row>
            <!--single-->
          <row>
            <cell>8</cell>
            <cell><c>QuaternionGroup()</c></cell>
            <cell>Quaternions</cell>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell/>
            <cell><c>DiCyclicGroup(2)</c></cell>
          </row>
            <!--double-->
          <row>
            <cell>9</cell>
            <cell><c>CyclicPermutationGroup(9)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>9</cell>
            <cell><c>C3=CyclicPermutationGroup(3)</c></cell>
            <cell/>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell><c>G=direct_product_permgroups([C3,C3])</c></cell>
            <cell>Abelian, non-cyclic</cell>
          </row>
            <!--double-->
          <row>
            <cell>10</cell>
            <cell><c>CyclicPermutationGroup(10)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>10</cell>
            <cell><c>DihedralGroup(5)</c></cell>
            <cell>Non-abelian</cell>
          </row>
            <!--double-->
          <row>
            <cell>11</cell>
            <cell><c>CyclicPermutationGroup(11)</c></cell>
            <cell>Prime order</cell>
          </row>
            <!--double-->
          <row>
            <cell>12</cell>
            <cell><c>CyclicPermutationGroup(12)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>12</cell>
            <cell><c>C2=CyclicPermutationGroup(2)</c></cell>
            <cell/>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell><c>C6=CyclicPermutationGroup(6)</c></cell>
            <cell/>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell><c>G=direct_product_permgroups([C2,C6])</c></cell>
            <cell>Abelian, non-cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>12</cell>
            <cell><c>DihedralGroup(6)</c></cell>
            <cell>Non-abelian</cell>
          </row>
            <!--single-->
          <row>
            <cell>12</cell>
            <cell><c>AlternatingGroup(4)</c></cell>
            <cell>Non-abelian</cell>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell/>
            <cell>Symmetries of tetrahedron</cell>
          </row>
            <!--single-->
          <row>
            <cell>12</cell>
            <cell><c>DiCyclicGroup(3)</c></cell>
            <cell>Non-abelian</cell>
          </row>
            <!--none-->
          <row>
            <cell/>
            <cell/>
            <cell>Semi-direct product <m>Z_3\rtimes Z_4</m></cell>
          </row>
            <!--double-->
          <row>
            <cell>13</cell>
            <cell><c>CyclicPermutationGroup(13)</c></cell>
            <cell>Prime order</cell>
          </row>
            <!--double-->
          <row>
            <cell>14</cell>
            <cell><c>CyclicPermutationGroup(14)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--single-->
          <row>
            <cell>14</cell>
            <cell><c>DihedralGroup(7)</c></cell>
            <cell>Non-abelian</cell>
          </row>
            <!--double-->
          <row bottom="medium">
            <cell>15</cell>
            <cell><c>CyclicPermutationGroup(15)</c></cell>
            <cell>Cyclic</cell>
          </row>
            <!--double-->
        </tabular>

      </table>
    </subsection>

  </section>

  <exercises xml:id="struct-sage-exercises" component="sage-exercises">
    <title>Sage Exercises</title>
    <introduction>
      <p>
        There are no Sage exercises for this chapter.
      </p>
    </introduction>

  </exercises>
</chapter>