Construct a non trivial homomorphism from $mathbb{Z}_2timesmathbb{Z}_4$ to $mathbb{Z}_8$












0












$begingroup$


I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:



$(0,0)$: $1$
$(0,1)$: $4$
$(1,0)$: $2$
$(1,1)$: $4$
$(0,2)$: $2$
$(1,2)$: $2$
$(0,3)$: $4$
$(1,3)$: $4$



Can anyone help me?






abstract-algebra group-theory cyclic-groups group-homomorphism direct-product







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    0












    $begingroup$


    I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:



    $(0,0)$: $1$
    $(0,1)$: $4$
    $(1,0)$: $2$
    $(1,1)$: $4$
    $(0,2)$: $2$
    $(1,2)$: $2$
    $(0,3)$: $4$
    $(1,3)$: $4$



    Can anyone help me?






    abstract-algebra group-theory cyclic-groups group-homomorphism direct-product







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:



      $(0,0)$: $1$
      $(0,1)$: $4$
      $(1,0)$: $2$
      $(1,1)$: $4$
      $(0,2)$: $2$
      $(1,2)$: $2$
      $(0,3)$: $4$
      $(1,3)$: $4$



      Can anyone help me?






      abstract-algebra group-theory cyclic-groups group-homomorphism direct-product







      share|cite|improve this question











      $endgroup$




      I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:



      $(0,0)$: $1$
      $(0,1)$: $4$
      $(1,0)$: $2$
      $(1,1)$: $4$
      $(0,2)$: $2$
      $(1,2)$: $2$
      $(0,3)$: $4$
      $(1,3)$: $4$



      Can anyone help me?






      abstract-algebra group-theory cyclic-groups group-homomorphism direct-product




      abstract-algebra group-theory cyclic-groups group-homomorphism direct-product






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 18:16









      Shaun

      9,442113684




      9,442113684










      asked Jan 4 at 17:23









      FailousaFailousa

      185




      185






















          2 Answers
          2






          active

          oldest

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          4












          $begingroup$

          A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
            $endgroup$
            – Failousa
            Jan 4 at 19:43












          • $begingroup$
            $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
            $endgroup$
            – user3482749
            Jan 4 at 19:46










          • $begingroup$
            Ok.Thank you very much!
            $endgroup$
            – Failousa
            Jan 4 at 19:53



















          2












          $begingroup$

          Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
          $$
          (x,y)mapsto alpha(x)+beta(y)
          $$
          where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.



          Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
            $endgroup$
            – Failousa
            Jan 4 at 19:58










          • $begingroup$
            @Failousa You can prove it directly.
            $endgroup$
            – egreg
            Jan 4 at 20:25











          Your Answer





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          2 Answers
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          $begingroup$

          A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
            $endgroup$
            – Failousa
            Jan 4 at 19:43












          • $begingroup$
            $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
            $endgroup$
            – user3482749
            Jan 4 at 19:46










          • $begingroup$
            Ok.Thank you very much!
            $endgroup$
            – Failousa
            Jan 4 at 19:53
















          4












          $begingroup$

          A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
            $endgroup$
            – Failousa
            Jan 4 at 19:43












          • $begingroup$
            $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
            $endgroup$
            – user3482749
            Jan 4 at 19:46










          • $begingroup$
            Ok.Thank you very much!
            $endgroup$
            – Failousa
            Jan 4 at 19:53














          4












          4








          4





          $begingroup$

          A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.






          share|cite|improve this answer









          $endgroup$



          A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 17:26









          user3482749user3482749

          4,296919




          4,296919












          • $begingroup$
            Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
            $endgroup$
            – Failousa
            Jan 4 at 19:43












          • $begingroup$
            $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
            $endgroup$
            – user3482749
            Jan 4 at 19:46










          • $begingroup$
            Ok.Thank you very much!
            $endgroup$
            – Failousa
            Jan 4 at 19:53


















          • $begingroup$
            Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
            $endgroup$
            – Failousa
            Jan 4 at 19:43












          • $begingroup$
            $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
            $endgroup$
            – user3482749
            Jan 4 at 19:46










          • $begingroup$
            Ok.Thank you very much!
            $endgroup$
            – Failousa
            Jan 4 at 19:53
















          $begingroup$
          Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
          $endgroup$
          – Failousa
          Jan 4 at 19:43






          $begingroup$
          Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
          $endgroup$
          – Failousa
          Jan 4 at 19:43














          $begingroup$
          $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
          $endgroup$
          – user3482749
          Jan 4 at 19:46




          $begingroup$
          $f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
          $endgroup$
          – user3482749
          Jan 4 at 19:46












          $begingroup$
          Ok.Thank you very much!
          $endgroup$
          – Failousa
          Jan 4 at 19:53




          $begingroup$
          Ok.Thank you very much!
          $endgroup$
          – Failousa
          Jan 4 at 19:53











          2












          $begingroup$

          Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
          $$
          (x,y)mapsto alpha(x)+beta(y)
          $$
          where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.



          Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
            $endgroup$
            – Failousa
            Jan 4 at 19:58










          • $begingroup$
            @Failousa You can prove it directly.
            $endgroup$
            – egreg
            Jan 4 at 20:25
















          2












          $begingroup$

          Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
          $$
          (x,y)mapsto alpha(x)+beta(y)
          $$
          where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.



          Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
            $endgroup$
            – Failousa
            Jan 4 at 19:58










          • $begingroup$
            @Failousa You can prove it directly.
            $endgroup$
            – egreg
            Jan 4 at 20:25














          2












          2








          2





          $begingroup$

          Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
          $$
          (x,y)mapsto alpha(x)+beta(y)
          $$
          where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.



          Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?






          share|cite|improve this answer









          $endgroup$



          Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
          $$
          (x,y)mapsto alpha(x)+beta(y)
          $$
          where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.



          Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 17:38









          egregegreg

          184k1486205




          184k1486205












          • $begingroup$
            I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
            $endgroup$
            – Failousa
            Jan 4 at 19:58










          • $begingroup$
            @Failousa You can prove it directly.
            $endgroup$
            – egreg
            Jan 4 at 20:25


















          • $begingroup$
            I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
            $endgroup$
            – Failousa
            Jan 4 at 19:58










          • $begingroup$
            @Failousa You can prove it directly.
            $endgroup$
            – egreg
            Jan 4 at 20:25
















          $begingroup$
          I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
          $endgroup$
          – Failousa
          Jan 4 at 19:58




          $begingroup$
          I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
          $endgroup$
          – Failousa
          Jan 4 at 19:58












          $begingroup$
          @Failousa You can prove it directly.
          $endgroup$
          – egreg
          Jan 4 at 20:25




          $begingroup$
          @Failousa You can prove it directly.
          $endgroup$
          – egreg
          Jan 4 at 20:25


















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          Cabo Verde

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            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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          Gyllenstierna

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