Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x],...











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Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.




I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?






abstract-algebra







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  • There is always a trivial homomorphism between groups.
    – CyclotomicField
    Nov 19 at 5:22










  • That isn't well-defined.
    – Lord Shark the Unknown
    Nov 19 at 5:35















up vote
0
down vote

favorite













Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.




I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?






abstract-algebra







share|cite|improve this question
























  • There is always a trivial homomorphism between groups.
    – CyclotomicField
    Nov 19 at 5:22










  • That isn't well-defined.
    – Lord Shark the Unknown
    Nov 19 at 5:35













up vote
0
down vote

favorite









up vote
0
down vote

favorite












Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.




I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?






abstract-algebra







share|cite|improve this question
















Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.




I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?






abstract-algebra




abstract-algebra






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edited Nov 19 at 5:52









Tianlalu

2,854832




2,854832










asked Nov 19 at 4:56









david D

825




825












  • There is always a trivial homomorphism between groups.
    – CyclotomicField
    Nov 19 at 5:22










  • That isn't well-defined.
    – Lord Shark the Unknown
    Nov 19 at 5:35


















  • There is always a trivial homomorphism between groups.
    – CyclotomicField
    Nov 19 at 5:22










  • That isn't well-defined.
    – Lord Shark the Unknown
    Nov 19 at 5:35
















There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22




There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22












That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35




That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35










1 Answer
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It's enough to prove it for each component.
begin{align}
&varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
&x+4Bbb Zmapsto x+4Bbb Z\
&varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
&y+6Bbb Zmapsto 2y+4Bbb Z\
&varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
&x+4Bbb Zmapsto 0+3Bbb Z\
&varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
&y+6Bbb Zmapsto y+3Bbb Z\
end{align}
In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
begin{CD}
Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
@VVV @|\
Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
end{CD}






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    It's enough to prove it for each component.
    begin{align}
    &varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
    &x+4Bbb Zmapsto x+4Bbb Z\
    &varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
    &y+6Bbb Zmapsto 2y+4Bbb Z\
    &varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
    &x+4Bbb Zmapsto 0+3Bbb Z\
    &varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
    &y+6Bbb Zmapsto y+3Bbb Z\
    end{align}
    In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
    begin{CD}
    Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
    @VVV @|\
    Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
    end{CD}






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      It's enough to prove it for each component.
      begin{align}
      &varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
      &x+4Bbb Zmapsto x+4Bbb Z\
      &varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
      &y+6Bbb Zmapsto 2y+4Bbb Z\
      &varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
      &x+4Bbb Zmapsto 0+3Bbb Z\
      &varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
      &y+6Bbb Zmapsto y+3Bbb Z\
      end{align}
      In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
      begin{CD}
      Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
      @VVV @|\
      Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
      end{CD}






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
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        down vote



        accepted






        It's enough to prove it for each component.
        begin{align}
        &varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
        &x+4Bbb Zmapsto x+4Bbb Z\
        &varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
        &y+6Bbb Zmapsto 2y+4Bbb Z\
        &varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
        &x+4Bbb Zmapsto 0+3Bbb Z\
        &varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
        &y+6Bbb Zmapsto y+3Bbb Z\
        end{align}
        In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
        begin{CD}
        Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
        @VVV @|\
        Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
        end{CD}






        share|cite|improve this answer












        It's enough to prove it for each component.
        begin{align}
        &varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
        &x+4Bbb Zmapsto x+4Bbb Z\
        &varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
        &y+6Bbb Zmapsto 2y+4Bbb Z\
        &varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
        &x+4Bbb Zmapsto 0+3Bbb Z\
        &varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
        &y+6Bbb Zmapsto y+3Bbb Z\
        end{align}
        In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
        begin{CD}
        Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
        @VVV @|\
        Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
        end{CD}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 8:18









        Fabio Lucchini

        7,79311326




        7,79311326






























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