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4.3 : Homomorphisms

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    132495
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    Definition: Homomorphism

    Let \((G_1,\star_1)\), \((G_2, \star_2)\) be groups.

    Then a function \(h:G_1 \rightarrow G_2\) s.t. \(h(g_1)=g_2, g_1 \in G_1 \text{and } g_2 \in G_2\) is called a homomorphism from the group \(G_1\) to the group \(G_2\) if \(h(g_1 \star_1g_1^{\shortmid})=h(g_1)\star_2 h(g_1^{\shortmid}), \;g_1,g_1^{\shortmid} \in G_1\).

    Further, if \(h\) is a bijection and also a homomorphism, then \(h\) is called an isomorphism.  In this case, we say \(G_1 \cong G_2\), which means that \(G_1\) is congruent to \(G_2\).

    Example \(\PageIndex{1}\)

    Let \(C_n=<a,a^n=e>\) be a cyclic group.

    Define: \(h: C_n \rightarrow \mathbb{Z}_n \) by \(h(a^k)=k \pmod{n}\).  Then

    1. Is \(h\) a homomorphism?

    2. Is \(h\) injective (1-1)?

    3. Is \(h\) surjective (onto)?

    If all three are true, then \(h\) is an isomorphism.

     

    Proof:

    We will show a finite cyclic group is always isomorphic to \((\mathbb{Z}_n, + \pmod{n})\).

     

    Proof of Homomorphism:

    Let \(g_1,g_2 \in C_n\).

    We shall show that \(h(g_1\star_1 g_2)=h(g_1)\star_2h(g_2)\).

    Consider \(g_1=a^{k_1}\) and \(g_2=a^{k_2}\).

    Then \(h(g_1)=k_1 \pmod{n}\) and \(h(g_2)=k_2 \pmod{n}\).

    \(g_1\star_1 g_2=a^{k_1} \star_1 a^{k_2}=a^{k_1+k_2}\).

    Thus \(h(g_1\star_1g_2)=k_1+k_2 \pmod{n}\).

    Now consider \(h(g_1)\star_2 h(g_2)=k_1 \pmod{n} +k_2 \pmod{n}\)

              \(=k_1+k_2 \pmod{n}\).

    Hence \(h(g_1\star_1 g_2)=h(g_1)\star_2h(g_2)\).

    Hence \(h\) is a homomorphism.◻

     

    Proof of Injection (1-1):Screen Shot 2023-07-06 at 4.34.19 PM.png

    Let \(g_1, g_2 \in G\) s.t. \(h(g_1)=h(g_2)\).

    Since \(g_1, g_2 \in C_n\), \(g_1=a^{k_1}\) and \(g_2=a^{k_2}, \; k_1,k_2 \in \mathbb{Z}\).

    Now, \(h(g_1)=k_1 \pmod{n}=k_2 \pmod{n}=h(g_2)\).
    Therefore, \(k_1=k_2 \in \mathbb{Z}_n\).

    Hence \(h\) is injective (1-1).◻

     

    Proof of Surjective (onto):

    Let \(k \in \mathbb{Z}_n\).Screen Shot 2023-07-06 at 4.35.09 PM.png

    Since \(a^k \in C_n\) by definition \(h(a^k)=k \pmod{n}\).

    Hence \(h\) is surjective.◻

     

    Since a finite cyclic group is homomorphic, injective and surjective to \((\mathbb{Z}_n, + \pmod{n})\), it is isomorphic to \((\mathbb{Z}_n, + \pmod{n})\). ◻  

     

    Example \(\PageIndex{2}\)

     Define  \( h: (\mathbb{Z}, +) \to ( \mathbb{Z}_n, \cdot (mod n) \) defined by

    \( h(k)= a^k, \) where \( \mathbb{Z}_n = <a>, k \in \mathbb{Z}. \) Then 

    For \( k, m \in \mathbb{Z},  h(k+m)= a^{k+m}= a^k a^m= h(k)h(m). \) Thus \(h\) is a homomorphism.

    Notice that 

     

    \( 0 \)

    \( \pm n \)

    \( \pm 2n \)

    \( \pm 3n \)

    .

    .

    .

    .

    \( 1 \)

    \( 1 \pm n \)

    \( 1 \pm 2n \)

    \( 1 \pm 3n \)

    .

    .

    .

    .

    \( \cdots\)

    \( \cdots\)

    \( n-1 \)

    \(n-1  \pm n \)

    \( n-1 \pm 2n \)

    \( n-1 \pm 3n \)

    .

    .

    .

    .




     

    \(\mathbb{Z} \)

      \( \downarrow\)

    \( \downarrow\)

    \( \downarrow\)

    \( \downarrow\)

    \( \downarrow\)

     

    \( a^0\)

    \(a^1\)

       

    \(a^{n-1}\)

    \(\mathbb{Z}_n \)

     

    Hence \( h \) is onto but not one to one.

    Properties of homomorphisms:

    Theorem \(\PageIndex{1}\)

    Let \(h: G_1 \rightarrow G_2\) be a homomorphism.  Then:

    1. \( h(e_1)=e_2 \), where \( e_1, e_2 \) are identities of \( G_1, G_2 \) respectively.Screen Shot 2023-07-06 at 4.39.50 PM.png

    2. \(h(g_1^{-1})=h(g_1)^{-1}, g_1 \in G_1\).

    3. \(h(g^n)=h(g)^n\), \( n \in \mathbb{Z} \)

     

    Proof:

    1. Since \( e_1 \) is the identity of \( G_1 \), \( e_1 e_1=e_1. \)

    Since \( h \) is a homomorphism, \( h(e_1e_1)= h(e_1) h(e_1)=h(e_1)=e_2 h(e_1). \)

    By cancellation law, \(e_2 = h(e_1). \)

     

    1. Let \( g_1 \in G_1.\) Then \( g_1g_1^{-1}=e_1.\) 

    Since \( h \) is a homomorphism, \( h(g_1g_1^{-1})=h(g_1) h(g_1^{-1}) = h(e_1)=e_2.\)

    Hence \( (h(g_1))^{-1} = h(g_1^{-1}). \)

     

    1. Use induction on \( \mathbb{Z}_+ \).

    Screen Shot 2023-07-06 at 4.41.46 PM.pngDefinition: 

    Let \( h: G_1 \to G_2\) be a homomorphism. Then 

    1. Screen Shot 2023-07-06 at 4.43.00 PM.pngThe kernel of \(h\) is defined as \( ker(h)=\{g_1 \in G_1| h(g_1)=e_2 \}. \) 

    2. The image  of \(h\) is defined as \( im(h)= \{ g_2 \in G_2| h(g_1)=g_2, g_1 \in G_1\}.\)

     

     

     

    Theorem \(\PageIndex{1}\)

     Let \( h: G_1 \to G_2\) be a homomorphism. Then \( ker(h)  \leq  G_1 \) and \( im(h)  \leq  G_2. \) Also, \(\ker (h) \) is a normal subgroup of \(G _1\).

    Proof:

    Let \( h: G_1 \to G_2\) be a homomorphism. 

    Since \(  h(e_1)=e_2, e_1 \in ker(h).\) Let \( g_1, g_1^{'} \in ker(h).\)  Then \( h(g_1)=e_2 \) and \( h( g_1^{'}) =e_2. \)

    Similarly, we can show that \( im(h) \leq G_2. \)

    We shall show that \( g_1^{-1} g_1^{'} \in ker(h).\) 

    Consider \( h(g_1^{-1} g_1^{'})= h(g_1^{-1})h( g_1^{'}) = h(g_1)^{-1} h( g_1^{'}) = e_2^{-1} e_2=e_2. \) Hence  \( g_1^{-1} g_1^{'} \in ker(h).\) Thus  \( ker(h) \leq G_1. \)

    We shall show that \(\ker (h) \) is a normal subgroup of \(G_1 \).

    Let \(h \in \ker(h) \) s.t. \(h(h)=e_H \).

    We will show \(ghg^{-1}\in \ker(h), \forall \; g \in G_1 \).

    Consider \(h(ghg^{-1})=h(g) h(h) h(g^{-1}) \)

                                   \(= h(g) h(h) (h(g))^{-1} \)

                                   \(=h(g) e_H (h(g))^{-1} \)

                                   \(=h(g) (h(g))^{-1} \)

                                   \(=e_H \).

    So \(\ker(h) \) is a normal subgroup of \(G_1 \).

     

     


    4.3 : Homomorphisms is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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