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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$endgroup$

















    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

      9,065113683




      9,065113683










      asked Jan 4 at 17:23









      FailousaFailousa

      184




      184






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          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
















          3












          $begingroup$

          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














          3












          3








          3





          $begingroup$

          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

          4,256919




          4,256919












          • $begingroup$
            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

          181k1485203




          181k1485203












          • $begingroup$
            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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          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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          File:DeusFollowingSea.jpg

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          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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