homomorph+iso.xml

<?xml version="1.0"?>
<chapter xmlns:xi="http://www.w3.org/2001/XInclude" xml:id="homomorphisms" permid="JXG">
  <title>Homomorphisms</title>
  <introduction permid="aAs">
    <p>
      Basic learning goals
	<ol>
	  <li>
	    Recognizing and checking group homomorphisms. 
	  </li>
	  <li>
	    Basic properties of group homomorphisms. 
	  </li>
	  <li>
	    Recognizing, checking and basic properties of group isomorphisms. 
	  </li>
	  <li>
	    Characterization of cyclic and abelian groups (Fundamental Theorem of Finite Abelian Groups). 
	  </li>
	  <li>
	    Normal subgroups, their interplay with homomorphisms. 
	  </li>
	  <li>
	    Free groups and generated groups. 
	  </li>
	</ol>
      </p>
    
    <p permid="ato">
      One of the most fundamental ideas of algebra is the concept of a homomorphism.
      In the study of groups, we already identified structures of certain operations on sets.
      Now, the next crucial insights come from the study of structure-preserving maps.
      This generalizes the idea of linear maps between vector spaces as considered in linear algebra.
      Recall that linear maps take the sum of two vectors to the sum of their images.
      This idea is relaxed to maps taking composition in one group to the composition in another group. 
    </p>
  </introduction>

  <section xml:id="homomorph-section-group-homomorphisms" permid="uOI">
      <title>Group Homomorphisms</title>
    <p permid="pgN">
      A <term>homomorphism</term><idx><h>Group</h><h>homomorphism of</h></idx><idx><h>Homomorphism</h><h>of groups</h></idx> between groups <m>(G, \cdot)</m> and
      <m>(H, \circ)</m> is a map <m>\phi :G \rightarrow H</m> such that
      <me permid="Stq">
        \phi( g_1 \cdot g_2 ) = \phi( g_1 ) \circ \phi( g_2 )
      </me>
      for <m>g_1, g_2 \in G</m>.
      The range of <m>\phi</m> in <m>H</m> is called the
      <term>homomorphic image</term>
          <idx><h>Homomorphic image</h></idx>
      of <m>\phi</m>.
    </p>
    <example>
      <p>
	First note that a vector space is an abelian group.
	For the special vector space <m>\mathbb{R}^k</m>, one can directly see this from <xref ref="exercise-external-direct-products"/> and the basic insight that <m>(\mathbb{R},+)</m> is an abelian group. 
      </p>
      <p>
	Now, recall the map from <xref ref="matrix-example-linear-transform"/> given by
	 <m>T \colon {\mathbb R}^2 \rightarrow {\mathbb R}^2</m> with
          <me permid="RNa">
            T(x_1, x_2) = (2 x_1 + 5 x_2, - 4 x_1 + 3 x_2) \enspace .
          </me>
	  It maps the abelian group <m>(\mathbb{R}^2,+)</m> to itself while preserving the additive structure, that is
	  <me>
	    T(x_1 + y_1, x_2 + y_2) = T(x_1, x_2) + T(y_1, y_2) \enspace .
	  </me>
      </p>
      <p>
	More generally, recall from the introduction of <xref ref="intro-and-applications"/> that a linear map <m>T : {\mathbb R}^n \rightarrow {\mathbb R}^m</m> fulfills <m>T({\mathbf x}+{\mathbf y}) = T({\mathbf x}) + T({\mathbf y})</m>.
	Hence, a linear map is a homomorphism between the additive structure of two vector spaces. 
	Note that we did not require compatibility with scalar multiplication as we only care about the additive structure of a vector space here. 
      </p>
    </example>

    <p>
      More generally, we use homomorphisms to study relationships between groups. 
    </p>

    <example>
    <p>
      The symmetric group <m>S_n</m> and the group
      <m>{\mathbb Z}_2</m> are related by the fact that <m>S_n</m> can be divided into even and odd permutations that exhibit a group structure like that <m>{\mathbb Z}_2</m>,
      as shown in the following multiplication table.
    </p>
    
    <p permid="umf">
      <me permid="yAz">
        \begin{array}{c|cc}
        &amp; \text{even} &amp; \text{odd} \\ \hline
        \text{even} &amp; \text{even} &amp; \text{odd} \\
        \text{odd}  &amp; \text{odd}  &amp; \text{even}
        \end{array}
      </me>
    </p>
    </example>
    
    <example xml:id="homomorph-example-zn" permid="kZx">
      <p permid="QZC">
        Let <m>G</m> be a group and <m>g \in G</m>.
        Define a map <m>\phi : {\mathbb Z} \rightarrow G</m> by <m>\phi( n ) = g^n</m>.
        Then <m>\phi</m> is a group homomorphism, since
        <me permid="eHI">
          \phi( m + n ) = g^{ m + n} = g^m g^n = \phi( m ) \phi( n )
        </me>.
        This homomorphism maps <m>{\mathbb Z}</m> onto the cyclic subgroup of <m>G</m> generated by <m>g</m>.
      </p>
    </example>

    <example xml:id="homomorph-example-gl2" permid="RgG">
      <p permid="xgL">
        Let <m>G = GL_2( {\mathbb R })</m>.
        If
        <me permid="KOR">
          A =
          \begin{pmatrix}
          a &amp; b \\
          c &amp; d
          \end{pmatrix}
        </me>
        is in <m>G</m>, then the determinant is  nonzero;
        that is, <m>\det(A) = ad - bc \neq 0</m>.
        Also, for any two elements <m>A</m> and <m>B</m> in <m>G</m>,
        <m>\det(AB) = \det(A) \det(B)</m>.
        Using the determinant, we can define a homomorphism
        <m>\phi : GL_2( {\mathbb R }) \rightarrow {\mathbb R}^\ast</m> by <m>A \mapsto \det(A)</m>.
      </p>
    </example>

    <example xml:id="homomorph-example-circle-group" permid="xnP">
      <p permid="dnU">
        Recall that the circle group
        <m>{ \mathbb T}</m> consists of all complex numbers <m>z</m> such that <m>|z|=1</m>.
        We can define a homomorphism <m>\phi</m> from the additive group of real numbers <m>{\mathbb R}</m> to
        <m>{\mathbb T}</m> by <m>\phi : \theta \mapsto \cos \theta + i \sin \theta</m>.
        Indeed,
        <md permid="qWa">
          <mrow>\phi( \alpha + \beta ) &amp; = \cos( \alpha + \beta ) + i \sin( \alpha + \beta )</mrow>
          <mrow>&amp; = (\cos \alpha \cos \beta - \sin \alpha \sin \beta)  + i( \sin \alpha  \cos \beta + \cos \alpha \sin \beta )</mrow>
          <mrow>&amp; = (\cos \alpha + i \sin \alpha )(\cos \beta + i \sin \beta)</mrow>
          <mrow>&amp; = \phi( \alpha ) \phi( \beta )</mrow>
        </md>.
        Geometrically,
        we are simply wrapping the real line around the circle in a group-theoretic fashion.
      </p>
    </example>

    <p>
      The following proposition lists some basic properties of group homomorphisms.
    </p>

    <proposition xml:id="proposition-properties-homomorphism-group">
      <statement>
        <p permid="kSt">
          Let <m>\phi : G_1 \rightarrow G_2</m> be a homomorphism of groups.
          <ol permid="TFN">
            <li permid="Mbo">
              <p permid="rde">
                If <m>e</m> is the identity of <m>G_1</m>,
                then <m>\phi( e)</m> is the identity of <m>G_2</m>.
              </p>
            </li>

            <li permid="six">
              <p permid="Xkn">
                For any element <m>g \in G_1</m>,
                <m>\phi( g^{-1}) = [\phi( g )]^{- 1}</m>. 
              </p>
            </li>

	    <li>
	      <p>
		For any element <m>g \in G_1</m> and <m>n \in \mathbb{Z}</m>, 
                <m>\phi( g^{n}) = [\phi( g )]^{n}</m>.
	      </p>
	    </li>
	    
            <li permid="YpG">
              <p permid="Drw">
                If <m>H_1</m> is a subgroup of <m>G_1</m>,
                then <m>\phi( H_1 )</m> is a subgroup of <m>G_2</m>. 
              </p>
            </li>
          </ol>
        </p>
      </statement>
    </proposition>

    <p permid="NJx">
      Let <m>\phi : G \rightarrow H</m> be a group homomorphism and suppose that <m>e</m> is the identity of <m>H</m>.
      By <xref ref="proposition-properties-homomorphism-group"/>,
      <m>\phi^{-1} ( \{ e \} )</m> is a subgroup of <m>G</m>.
      This subgroup is called the <term>kernel</term><idx><h>Kernel</h><h>of a group homomorphism</h></idx><idx><h>Homomorphism</h><h>kernel of a group</h></idx> of <m>\phi</m> and will be denoted by <m>\ker \phi</m>.

      <notation>
        <usage><m>\ker \phi</m></usage>
        <description>kernel of <m>\phi</m></description>
      </notation>
    </p>
      
    <example xml:id="homomorph-example-g2-to-r" permid="duY">
      <p permid="Jvd">
        Let us examine the homomorphism
        <m>\phi : GL_2( {\mathbb R }) \rightarrow {\mathbb R}^\ast</m> defined by <m>A \mapsto \det( A )</m>.
        Since <m>1</m> is the identity of <m>{\mathbb R}^\ast</m>,
        the kernel of this homomorphism is all
        <m>2 \times 2</m> matrices having determinant one.
        That is, <m>\ker \phi = SL_2( {\mathbb R })</m>.
      </p>
    </example>

    <example xml:id="homomorph-example-kernel" permid="JCh">
      <p permid="pCm">
        The kernel of the group homomorphism <m>\phi : {\mathbb R} \rightarrow {\mathbb C}^\ast</m> defined by
        <m>\phi( \theta ) = \cos \theta + i \sin \theta</m> is <m>\{ 2 \pi n : n \in {\mathbb Z} \}</m>.
        Notice that <m>\ker \phi \cong {\mathbb Z}</m>.
      </p>
    </example>

    <example xml:id="homomorph-example-z7" permid="pJq">
      <p permid="VJv">
        Suppose that we wish to determine all possible homomorphisms <m>\phi</m> from
        <m>{\mathbb Z}_7</m> to <m>{\mathbb Z}_{12}</m>.
        Since the kernel of <m>\phi</m> must be a subgroup of <m>{\mathbb Z}_7</m>,
        there are only two possible kernels,
        <m>\{ 0 \}</m> and all of <m>{\mathbb Z}_7</m>.
        The image of a subgroup of
        <m>{\mathbb Z}_7</m> must be a subgroup of <m>{\mathbb Z}_{12}</m>.
        Hence, there is no injective homomorphism;
        otherwise, <m>{\mathbb Z}_{12}</m> would have a subgroup of order <m>7</m>,
        which is impossible.
        Consequently,
        the only possible homomorphism from <m>{\mathbb Z}_7</m> to
        <m>{\mathbb Z}_{12}</m> is the one mapping all elements to zero.
      </p>
    </example>

    <example xml:id="homomorph-example-gn" permid="VQz">
      <p permid="BQE">
        Let <m>G</m> be a group.
        Suppose that <m>g \in G</m> and <m>\phi</m> is the homomorphism from
        <m>{\mathbb Z}</m> to <m>G</m> given by <m>\phi( n ) = g^n</m>.
        If the order of <m>g</m> is infinite,
        then the kernel of this homomorphism is <m>\{ 0 \}</m> since <m>\phi</m> maps
        <m>{\mathbb Z}</m> onto the cyclic subgroup of <m>G</m> generated by <m>g</m>.
        However, if the order of <m>g</m> is finite,
        say <m>n</m>,
        then the kernel of <m>\phi</m> is <m>n {\mathbb Z}</m>.
      </p>
    </example>
  </section>

  <section xml:id="isomorph-section-definitions" permid="Evp">
    <title>Isomorphisms</title>
    <introduction>
      <p>
      Many groups may appear to be different at first glance,
      but can be shown to be the same by a simple renaming of the group elements.
      For example,
      <m>{\mathbb Z}_4</m> and the subgroup of the circle group
      <m>{\mathbb T}</m> generated by <m>i</m> can be shown to be the same by demonstrating a one-to-one correspondence between the elements of the two groups and between the group operations.
      In such a case we say that the groups are isomorphic. 
      This is exactly the case when there is a bijective homomorphism between the groups, i.e.,  whose kernel is trivial. 
    </p>
  </introduction>

    <introduction>
      <p>
        Two groups <m>(G, \cdot)</m> and
        <m>(H, \circ)</m> are <term>isomorphic</term>
        <idx><h>Group</h><h>isomorphic</h></idx>
        if there exists a group homomorphism
	<m>\phi : G \rightarrow H</m>
	that is a one-to-one and onto map. 
        If <m>G</m> is isomorphic to <m>H</m>,
        we write <m>G \cong H</m>.
        <notation>
          <usage><m>G \cong H</m></usage>
          <description><m>G</m> is isomorphic to a group <m>H</m></description>
        </notation>

        The map <m>\phi</m> is called an <term>isomorphism</term>.
            <idx><h>Group</h><h>isomorphism of</h></idx>
            <idx><h>Isomorphism</h><h>of groups</h></idx>
      </p>
    </introduction>

      <example xml:id="isomorph-example-z4" permid="gha">
        <p permid="Qaa">
          To show that <m>{\mathbb Z}_4 \cong \langle i \rangle</m>,
          define a map <m>\phi: {\mathbb Z}_4 \rightarrow \langle i \rangle</m> by <m>\phi(n) = i^n</m>.
          We must show that <m>\phi</m> is bijective and preserves the group operation.
          The map <m>\phi</m> is one-to-one and onto because
          <md permid="EJX">
            <mrow>\phi(0) &amp; = 1</mrow>
            <mrow>\phi(1) &amp; = i</mrow>
            <mrow>\phi(2) &amp; = -1</mrow>
            <mrow>\phi(3) &amp; = -i</mrow>
          </md>.
          Since
          <me permid="YCO">
            \phi(m + n) = i^{m+n} = i^m i^n = \phi(m) \phi( n)
          </me>,
          the group operation is preserved.
        </p>
      </example>

      <example xml:id="isomorph-example-real" permid="Moj">
        <p permid="whj">
          We can define an isomorphism <m>\phi</m> from the additive group of real numbers
          <m>( {\mathbb R}, + )</m> to the multiplicative group of positive real numbers 
          <m>( {\mathbb R^+}, \cdot )</m> with the exponential map; that is,
          <me permid="kRg">
            \phi( x + y) = e^{x + y} = e^x e^y = \phi( x ) \phi( y)
          </me>.
          Of course, we must still show that <m>\phi</m> is one-to-one and onto,
          but this can be determined using calculus.
        </p>
      </example>

      <example xml:id="isomorph-example-rational" permid="svs">
        <p permid="cos">
          The integers are isomorphic to the subgroup of
          <m>{\mathbb Q}^\ast</m> consisting of elements of the form <m>2^n</m>.
          Define a map <m>\phi: {\mathbb Z} \rightarrow {\mathbb Q}^\ast</m> by <m>\phi( n ) = 2^n</m>.
          Then
          <me permid="QYp">
            \phi( m + n ) = 2^{m + n} = 2^m 2^n = \phi( m ) \phi( n )
          </me>.
          By definition the map <m>\phi</m> is onto the subset
          <m>\{2^n :n \in {\mathbb Z} \}</m> of  <m>{\mathbb Q}^\ast</m>.
          To show that the map is injective,
          assume that <m>m \neq n</m>.
          If we can show that <m>\phi(m) \neq \phi(n)</m>, then we are done.
          Suppose that <m>m \gt n</m> and assume that <m>\phi(m) = \phi(n)</m>.
          Then <m>2^m = 2^n</m> or <m>2^{m - n} = 1</m>,
          which is impossible since <m>m - n \gt 0</m>.
        </p>
      </example>

      <example xml:id="isomorph-example-units" permid="YCB">
        <p permid="IvB">
          The groups <m>{\mathbb Z}_8</m> and
          <m>{\mathbb Z}_{12}</m> cannot be isomorphic since they have different orders;
          however, it is true that <m>U(8) \cong U(12)</m>.
          We know that
          <md permid="xfy">
            <mrow>U(8) &amp; = \{1, 3, 5, 7 \}</mrow>
            <mrow>U(12) &amp; = \{1, 5, 7, 11 \}</mrow>
          </md>.
          An isomorphism <m>\phi : U(8) \rightarrow U(12)</m> is then given by
          <md permid="dmH">
            <mrow>1 &amp; \mapsto  1</mrow>
            <mrow>3 &amp; \mapsto  5</mrow>
            <mrow>5 &amp; \mapsto  7</mrow>
            <mrow>7 &amp; \mapsto  11</mrow>
          </md>.
          The map <m>\phi</m> is not the only possible isomorphism between these two groups.
          We could define another isomorphism <m>\psi</m> by <m>\psi(1) = 1</m>,
          <m>\psi(3) = 11</m>, <m>\psi(5) = 5</m>, <m>\psi(7) = 7</m>.
          In fact, both of these groups are isomorphic to <m>{\mathbb Z}_2 \times {\mathbb Z}_2</m>
          (see <xref ref="groups-example-z2xz2"/> in <xref ref="groups"/>).
        </p>
      </example>

      <example xml:id="isomorph-example-not-isomorph" permid="EJK">
        <p permid="oCK">
          Even though <m>S_3</m> and
          <m>{\mathbb Z}_6</m> possess the same number of elements,
          we would suspect that they are not isomorphic,
          because <m>{\mathbb Z}_6</m> is abelian and <m>S_3</m> is nonabelian.
          To demonstrate that this is indeed the case,
          suppose that <m>\phi : {\mathbb Z}_6 \rightarrow  S_3</m> is an isomorphism.
          Let <m>a , b \in S_3</m> be two elements such that <m>ab \neq ba</m>.
          Since <m>\phi</m> is an isomorphism,
          there exist elements <m>m</m> and <m>n</m> in <m>{\mathbb Z}_6</m> such that
          <me permid="JtQ">
            \phi( m ) = a \quad \text{and} \quad \phi( n ) = b
          </me>.
          However,
          <me permid="pAZ">
            ab = \phi(m ) \phi(n) = \phi(m + n) = \phi(n + m) = \phi(n ) \phi(m) = ba
          </me>,
          which contradicts the fact that <m>a</m> and <m>b</m> do not commute.
        </p>
      </example>

      <theorem xml:id="isomorph-theorem-1" permid="bxh">
        <statement>
          <p permid="idE">
            Let <m>\phi : G \rightarrow H</m> be an isomorphism of two groups.
            Then the following statements are true.

            <ol permid="gvv">
              <li permid="Cdz">
                <p permid="UQX">
                  <m>\phi^{-1} : H \rightarrow G</m> is an isomorphism.
                </p>
              </li>

              <li permid="ikI">
                <p permid="AYg">
                  <m>|G| = |H|</m>.
                </p>
              </li>

              <li permid="OrR">
                <p permid="hfp">
                  If <m>G</m> is abelian, then <m>H</m> is abelian.
                </p>
              </li>

              <li permid="uza">
                <p permid="Nmy">
                  If <m>G</m> is cyclic, then <m>H</m> is cyclic.
                </p>
              </li>

              <li permid="aGj">
                <p permid="ttH">
                  If <m>G</m> has a subgroup of order <m>n</m>,
                  then <m>H</m> has a subgroup of order <m>n</m>.
                </p>
              </li>
            </ol>
          </p>
        </statement>

        <proof>
          <p>
            Assertions (1) and (2) follow from the fact that <m>\phi</m> is a bijection.
          </p>
          <p permid="TZZ">
            (3) Suppose that <m>h_1</m> and <m>h_2</m> are elements of <m>H</m>.
            Since <m>\phi</m> is onto,
            there exist elements <m>g_1, g_2 \in G</m> such that
            <m>\phi(g_1) = h_1</m> and <m>\phi(g_2) = h_2</m>.
            Therefore,
            <me permid="VIi">
              h_1 h_2 = \phi(g_1) \phi(g_2) =  \phi(g_1 g_2) = \phi(g_2 g_1) = \phi(g_2) \phi(g_1) = h_2 h_1
            </me>.
          </p>
	  <p>
	    (4)
	    Let <m>g</m> be a generator of <m>G</m>.
	    Then
	    <me>
	      \phi\left(G\right) = \phi\left(\{g^k \mid k \in \mathbb{Z}\}\right) = \{\phi(g^k) \mid k \in \mathbb{Z}\} \enspace .
	    </me>
	    Since <m>\phi(G) = H</m>, we get from <xref ref="proposition-properties-homomorphism-group"/> that
	    <me>
	      H = \{\phi(g^k) \mid k \in \mathbb{Z}\} = \{\phi(g)^k \mid k \in \mathbb{Z}\} \enspace .
	    </me>
	    Therefore, also <m>H</m> is cyclic and the generators of <m>G</m> and <m>H</m> are in bijection. 
	  </p>
	  <p>
	  (5)
	  Let <m>G_1</m> be a subgroup of <m>G</m> of order <m>n</m>.
	  Then <m>\phi(G_1)</m> is a subgroup of <m>H</m> by <xref ref="proposition-properties-homomorphism-group"/>.
	  Since <m>\phi</m> is a bijection, this subgroup has cardinality <m>|\phi(G_1)| = |G_1| = n</m>. 
	  </p>
        </proof>
      </theorem>

      <p permid="CGp">
        We are now in a position to characterize all cyclic groups.
      </p>

      <theorem xml:id="isomorph-theorem-2" permid="HEq">
        <statement>
          <p permid="OkN">
            All cyclic groups of infinite order are isomorphic to <m>{\mathbb Z}</m>.
          </p>
        </statement>

        <proof permid="pAM">
          <p permid="Ahi">
            Let <m>G</m> be a cyclic group with infinite order and suppose that <m>a</m> is a generator of <m>G</m>.
            Define a map <m>\phi : {\mathbb Z} \rightarrow  G</m> by <m>\phi : n \mapsto a^n</m>.
            Then
            <me permid="BPr">
              \phi( m+n ) = a^{m+n} = a^m a^n = \phi( m ) \phi( n )
            </me>.
            To show that <m>\phi</m> is injective,
            suppose that <m>m</m> and <m>n</m> are two elements in
            <m>{\mathbb Z}</m>, where <m>m \neq n</m>.
            We can assume that <m>m \gt n</m>.
            We must show that <m>a^m \neq a^n</m>.
            Let us suppose the contrary;
            that is, <m>a^m = a^n</m>.
            In this case <m>a^{m - n} = e</m>, where <m>m - n \gt 0</m>,
            which contradicts the fact that <m>a</m> has infinite order.
            Our map is onto since any element in <m>G</m> can be written as <m>a^n</m> for some integer <m>n</m> and <m>\phi(n) = a^n</m>.
          </p>
        </proof>
      </theorem>

      <theorem xml:id="isomorph-theorem-3" permid="nLz">
        <statement>
          <p permid="urW">
            If <m>G</m> is a cyclic group of order <m>n</m>,
            then <m>G</m> is isomorphic to <m>{\mathbb Z}_n</m>.
          </p>
        </statement>

        <proof permid="VHV">
          <p permid="gor">
            Let <m>G</m> be a cyclic group of order <m>n</m> generated by <m>a</m> and define a map
            <m>\phi : {\mathbb Z}_n \rightarrow G</m> by <m>\phi : k \mapsto a^k</m>,
            where <m>0 \leq k \lt n</m>.
            One can check right from the definitions that <m>\phi</m> is an isomorphism. 
          </p>
        </proof>
      </theorem>
            <!-- RAB: 2014/08/18 Should ref be "corollary-isomorph-4"?   -->
            <!-- If so, search for refs (or see them fail with xsltproc) -->

      <p permid="iNy">
        The main goal in group theory is to classify all groups;
        however, it makes sense to consider two groups to be the same if they are isomorphic.
        We state this result in the following theorem. 
      </p>

      <theorem xml:id="isomorph-theorem-5" permid="TSI">
        <statement>
          <p permid="azf">
            The isomorphism of groups determines an equivalence relation on the class of all groups.
          </p>
        </statement>
      </theorem>

      <p permid="OUH">
        Hence, we can modify our goal of classifying all groups to classifying all groups
        <term>up to isomorphism</term>;
        that is, we will consider two groups to be the same if they are isomorphic.
      </p>

      <p>
	We finish the discussion of isomorphisms with an important class of groups which has a nice representative up to isomorphism.
	The Fundamental Theorem of Finite Abelian Groups tells us that every finite abelian group is isomorphic to a direct product of cyclic groups whose order is a prime power.
      </p>

      <theorem xml:id="struct-theorem-finite-abelian-groups" permid="lUM">
      <title>Fundamental Theorem of Finite Abelian Groups</title>
      <idx>
      <h>Fundamental Theorem</h>
      <h>of Finite Abelian Groups</h>
      </idx>
      <statement>
        <p permid="ddQ">
          Every finite abelian group <m>G</m> is isomorphic to a direct product of cyclic groups of the form
          <me permid="HKh">
            {\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" permid="voy">
      <p permid="mxC">
        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 permid="KbX">
          <li permid="iEH">
            <p permid="eEI">
              <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 permid="OLQ">
            <p permid="KLR">
              <m>{\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_3 \times {\mathbb Z}_9 \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li permid="uSZ">
            <p permid="qTa">
              <m>{\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_{27} \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li permid="bai">
            <p permid="Xaj">
              <m>{\mathbb Z}_4 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li permid="Hhr">
            <p permid="Dhs">
              <m>{\mathbb Z}_4 \times {\mathbb Z}_3 \times {\mathbb Z}_9 \times {\mathbb Z}_5</m>;
            </p>
          </li>

          <li permid="noA">
            <p permid="joB">
              <m>{\mathbb Z}_4 \times {\mathbb Z}_{27} \times {\mathbb Z}_5</m>.
            </p>
          </li>
        </ul>
      </p>
    </example>
      
  </section>
  
  <section xml:id="kernels-group-homomorphisms" permid="kgh">
    <title>Normal subgroups and kernels</title>
    <p permid="fmy">
      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
      <me>
	gHg^{-1} := \{ghg^{-1} \mid h \in H\}  = H \text{ for all } g \in G .
      </me>
    </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>ghg^{-1} = h</m> for all <m>g \in G</m> and <m>h \in H</m> by commutativity,
          it will always be the case that <m>gHg^{-1} = H</m>.
        </p>
      </example>


      <proposition xml:id="homomorph-proposition-preimage-subgroup" permid="preimage-subgroup">
	<statement>
          <p permid="EwP">
            Let <m>\phi : G_1 \rightarrow G_2</m> be a homomorphism of groups and <m>H_2</m> is a subgroup of <m>G_2</m>. 
            <ol>
	      <li>
                <m>\phi^{-1}(H_2) = \{ g \in G _1: \phi(g) \in H_2 \}</m> is a subgroup of <m>G_1</m>.
	      </li>
	      <li>
		If <m>H_2</m> is normal in <m>G_2</m>, then <m>\phi^{-1}(H_2)</m> is normal in <m>G_1</m>.
	      </li>
	    </ol>
	  </p>
	</statement>
	<proof>
          <p>
	    Let <m>H_2</m> be a subgroup of <m>G_2</m> and define <m>H_1</m> to be <m>\phi^{-1}(H_2)</m>;
            that is, <m>H_1</m> is the set of all
            <m>g \in G_1</m> such that <m>\phi(g) \in H_2</m>.
            The identity is in <m>H_1</m> since <m>\phi(e) = e'</m>.
            If <m>a</m> and <m>b</m> are in <m>H_1</m>,
            then <m>\phi(ab^{-1}) = \phi(a)[ \phi(b) ]^{-1}</m> is in <m>H_2</m> since <m>H_2</m> is a subgroup of <m>G_2</m>.
            Therefore, <m>ab^{-1} \in H_1</m> and <m>H_1</m> is a subgroup of <m>G_1</m>.
	  </p>
	  <p>
            If <m>H_2</m> is normal in <m>G_2</m>,
            we must show that <m>g^{-1} h g \in H_1</m> for
            <m>h \in H_1</m> and <m>g \in G_1</m>.
            But
            <me permid="PyK">
              \phi( g^{-1} h g) = [ \phi(g) ]^{-1} \phi( h ) \phi( g ) \in H_2
              </me>,
              since <m>H_2</m> is a normal subgroup of <m>G_2</m>.
              Therefore, <m>g^{-1}hg \in H_1</m>.
          </p>
	</proof>
      </proposition>

      <p>
	Note that the trivial subgroup consisting only of the identity element is normal.
	Together with <xref ref="homomorph-proposition-preimage-subgroup"/> this yields the following theorem,
	which says that with every homomorphism of groups we can naturally associate a normal subgroup.
      </p>

    <theorem permid="Aiv">
      <statement>
        <p permid="giA">
          Let <m>\phi : G \rightarrow H</m> be a group homomorphism.
          Then the kernel of <m>\phi</m> is a normal subgroup of <m>G</m>.
        </p>
      </statement>
    </theorem>

    <p>
      Let <m>G</m> be a group and <m>T</m> be an arbitrary subset.
      Then the <term>normal closure</term> of <m>T</m> is the smallest normal subgroup of <m>G</m> that contains <m>T</m>.
      Note that this is well-defined as there is always a normal subgroup containing <m>T</m>, at least the group <m>G</m> itself. 
    </p>
    <p>
      We list a few insights about the normal closure.
      The proof is omitted as this statement is not so central for the course but it gives background on the construction which we will exhibit in <xref ref="generated-groups-and-word-problem"/>. 
    </p>
    
    <proposition xml:id="proposition-properties-normal-closure">
      <statement>
	<ol>
	  <li>
	    <p>
	      The normal closure is the intersection of all normal subgroups of <m>G</m> containing <m>T</m>.
	    </p>
	  </li>
	  <li>
	    <p>
	      The normal closure is generated by <m>\{g^{-1}tg \mid g \in G, t \in T\}</m> that means that it equals
	      <me>
		\left\{g_1^{-1}t_1^{\sigma_1}g_1g_2^{-1}\dots t_n^{\sigma_n}g_n \mid n \geq 0, g_i \in G, t_i \in T, \sigma_i \in \{1,-1\} \forall i \in \{1,\dots,n\}\right\}. 
	      </me>
	    </p>
	  </li>
<!--	  <li>
	    <p>
	      If <m>T</m> is a subgroup, then the smallest normal subgroup of <m>G</m> containing <m>T</m> is <m>\bigcup_{g \in G} g^{-1}Tg</m>.  
	    </p>
	  </li> -->
	</ol>
      </statement>
    </proposition>
    
  </section>

  <section xml:id="free-and-generated-groups">
    <title>Free groups</title>
    <p>
      The <term>free group</term> on a given set <m>S</m> is formed by all words over the alphabet <m>S</m>, that is all expressions of the form
      <me>
	s_1^{\sigma_1} \dots s_n^{\sigma_n}
      </me>
      where <m>s_1,\dots,s_n \in S</m> and <m>\sigma_1,\dots,\sigma_n \in \{-1,1\}</m>.
      The binary operation of the group is concatenation of words.
      More precisely, the elements of the group are the equivalence classes of words under the identification of subwords <m>x x^{-1}</m> and <m>x^{-1} x</m> for <m>x \in S</m> with the empty word <m>\epsilon</m>.
      We will denote the free group by <m>\mathcal{F}(S)</m>. 
      The neutral element of this group is the empty word <m>\epsilon</m>.
      For example, for <m>S = \{a,b\}</m>, the words <m>aa^{-1}</m> or <m>b^{-1}b</m> are in the same equivalence class as <m>\epsilon</m> and the words <m>ab^{-1}aa^{-1}b</m> and <m>aa^{-1}a^{-1}aa</m> are in the same equivalence class as <m>a</m> (among many others). 
      The proof of the following statement implies that the free group is actually a group. 
    </p>

    <p>
      Recall from <xref ref="groups-subsection-subgroup-examples"/> what it means for a group to be generated.
      With this, the free group on <m>S</m> is generated by <m>S</m>.
      The number of generators of a free group is the <term>rank</term> of the free group. 
    </p>
    
    <proposition permid="eqn">
      <statement>
        <p permid="Vzr">
          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 permid="vvP">
            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 permid="QWC">
        <p permid="Pal">
          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 permid="vhu">
          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 permid="bCY">
            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>

    <example>
      <p>
	The free group on one element <m>a</m> denoted by <m>\mathcal{F}(\{a\})</m> is isomorphic with <m>\mathbb{Z}</m>. 
      </p>
    </example>
    
  <p>
      Let <m>G</m> be a group and <m>\{g_s \colon s \in S\} \subseteq G</m> be a subset generating it, e.g., the set of all elements of <m>G</m>.
      Then there is a <term>canonical homomorphism</term> such that
      <men xml:id="homomorphism-finitely-generated">
	s \mapsto g_s
      </men>
      for all <m>s \in S</m>.
      Taking a suitable subset of the elements in the kernel yields a so-called <term>presentation</term> of <m>G</m>.
      We will discuss this further in <xref ref="generated-groups-and-word-problem"/>. 

  </p>
  </section>

  
<!--    <section xml:id="additional-homomorphisms">
      <title> Additional insights </title>

<p>
universal property of free group?!?
</p>

<p>
fundamental group, algebraic topology
</p>
    </section>
-->

  <exercises xml:id="exercises-homomorphism-core" filenamebase="homomorphism">
    <title>Core Exercises</title>

     <exercise xml:id="exercise-properties-homomorphism-group">
	<statement>
	  <p>
	    Prove <xref ref="proposition-properties-homomorphism-group"/>. 
	  </p>
	</statement>
	<solution>
	  <p permid="etn">
          (1) Suppose that <m>e</m> and <m>e'</m> are the identities of <m>G_1</m> and <m>G_2</m>,
          respectively; then
          <me permid="Xdj">
            e' \phi(e) = \phi(e) = \phi(e e) = \phi(e) \phi(e)
          </me>.
          By cancellation, <m>\phi(e) = e'</m>.
        </p>

        <p permid="KAw">
          (2) This statement follows from the fact that
          <me permid="Dks">
            \phi( g^{-1}) \phi(g) = \phi(g^{-1} g) = \phi(e) = e'
          </me>.
        </p>
	<p>
	  (3) This follows by induction from the homomorphism property <m>\phi(g^n)=\phi(g^{n-1}g) = \phi(g^{n-1})\phi(g)=\phi(g)^{n-1}\phi(g)</m>. 
	</p>
        <!--  Corrected notation error.  Suggested by P. Diethelm.  TWJ 16/8/2013. -->
        <p permid="qHF">
          (4) The set <m>\phi(H_1)</m> is nonempty since the identity of <m>G_2</m> is in <m>\phi(H_1)</m>.
          Suppose that <m>H_1</m> is a subgroup of <m>G_1</m> and let <m>x</m> and <m>y</m> be in <m>\phi(H_1)</m>.
          There exist elements <m>a,
          b \in H_1</m> such that <m>\phi(a) = x</m> and <m>\phi(b)=y</m>.
          Since
          <me permid="jrB">
            xy^{-1} = \phi(a)[ \phi(b)]^{-1} = \phi(a b^{-1} ) \in \phi(H_1)
          </me>,
          <m>\phi(H_1)</m> is a subgroup of <m>G_2</m> by <xref ref="groups-proposition-subgroup"/>.
        </p>
	</solution>
      </exercise>

    <exercise xml:id="exercise-permutation-matrices">
	<statement>
	  <p>
	    Consider the map <m>\psi \colon S_3 \to GL_3({\mathbb R})</m> given by
	    <m>\pi \mapsto M</m>
	    where
	    <me>
	      M_{ij} =
	      \begin{cases}
	      1 &amp; \text{ for } i = \pi(j), \\
	      0 &amp; \text{ otherwise} .
	      \end{cases}
	    </me>
	    For example, one obtains
	    <me>
	      \psi\left( (2\,1\,3) \right) =
	      \begin{pmatrix}
	      0 &amp; 1 &amp; 0 \\
	      0 &amp; 0 &amp; 1 \\
	      1 &amp; 0 &amp; 0
	      \end{pmatrix} \enspace .
	    </me>
	    <ol>
	      <li>Determine explicitly the image of <m>\psi</m>. </li>
	      <li>Show that <m>\psi</m> is a group homomorphism. </li>
	      <li>What is the kernel of <m>\psi</m>? Is it an isomorphism? </li>
	      <li>Give a generalization of the statement for <m>S_n</m> for arbirary natural numbers <m>n</m>. </li>
	      <li>What does the determinant of the image of a permutation say about the permutation?</li>
	    </ol>
	  </p>
	</statement>
	<solution>
	  <p>
	    (1) The image has six elements.
	    Additional to the example above, one gets
	    <me>
	      \psi\left( (1) \right) =
	      \begin{pmatrix}
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 1 &amp; 0 \\
	      0 &amp; 0 &amp; 1
	      \end{pmatrix}
	    </me>
	    and 
	    <me>
	      \psi\left( (12) \right) =
	      \begin{pmatrix}
	      0 &amp; 1 &amp; 0 \\
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 0 &amp; 1
	      \end{pmatrix} \enspace .
	    </me>
	    The remaining three matrices are obtained analogously. 
	  </p>
	  <p> 
	    (2) There are essentially two ways to approach this.
	    Either one can check the homomorphism property for all pairs explicitly or one shows it more abstractly.
	    The first approach goes through all pairs of elements in <m>S_3</m> like
	    <me>
	      \begin{pmatrix}
	      0 &amp; 1 &amp; 0 \\
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 0 &amp; 1
	      \end{pmatrix} \cdot
	      \begin{pmatrix}
	      0 &amp; 0 &amp; 1 \\
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 1 &amp; 0
	      \end{pmatrix} =
	      \psi\left( (12) (123) \right) \stackrel{\text{?}}{=} 
	      \psi\left( (1)(23) \right) =
	      \begin{pmatrix}
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 0 &amp; 1 \\
	      0 &amp; 1 &amp; 0
	      \end{pmatrix}
	    </me>
	    and indeed
	    <me>
	      \begin{pmatrix}
	      0 &amp; 1 &amp; 0 \\
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 0 &amp; 1
	      \end{pmatrix} \cdot
	      \begin{pmatrix}
	      0 &amp; 0 &amp; 1 \\
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 1 &amp; 0
	      \end{pmatrix} =
	      \begin{pmatrix}
	      1 &amp; 0 &amp; 0 \\
	      0 &amp; 0 &amp; 1 \\
	      0 &amp; 1 &amp; 0
	      \end{pmatrix}\enspace .
	    </me>
	    Note that one has to check actually <m>6 \cdot 6 = 36</m> pairs then. 
	    For the second approach, let <m>\alpha, \beta \in S_3</m> and <m>j \in \{1,2,3\}</m>.
	    Then the <m>(i,j)</m>-entry of the matrix <m>\psi(\alpha)\cdot \psi(\beta)</m> is
	    <me>
	      \sum_{k=1}^{3} \psi(\alpha)_{ik} \cdot \psi(\beta)_{kj} \enspace .
	    </me>
	    Properly checking this expression with the definition of <m>\psi</m> yields the claim. 
	  </p>
	  <p>
	    (3) The kernel is <m>\{id\}</m>, it is an isomorphism.
	  </p>
	  <p>
	    (4) Indeed, there is such an isomorphism from <m>S_n</m> to the group of <term>permutation matrices</term> for all positive integers <m>n</m>. 
	  </p>
	  <p>
	    Even or odd. 
	  </p>
	  </solution>
    </exercise>

    <exercise permid="Trs"  xml:id="homomorph-exercise-which-are-homomorph">
      <statement>
        <p permid="SOP">
          Which of the following maps are homomorphisms?
          If the map is a homomorphism, what is the kernel?

          <ol permid="zMW">
            <li permid="kDY">
              <p permid="PFO">
                <m>\phi : {\mathbb R}^\ast \rightarrow GL_2 ( {\mathbb R})</m> defined by
                <me permid="CeZ">
                  \phi( a ) =
                  \begin{pmatrix}
                  1 &amp; 0 \\
                  0 &amp; a
                  \end{pmatrix}
                </me>
              </p>
            </li>

            <li permid="QLh">
              <p permid="vMX">
                <m>\phi : {\mathbb R} \rightarrow GL_2 ( {\mathbb R})</m> defined by
                <me permid="imi">
                  \phi( a ) =
                  \begin{pmatrix}
                  1 &amp; 0 \\
                  a &amp; 1
                  \end{pmatrix}
                </me>
              </p>
            </li>