Epimorphism between free Abelian groups












0












$begingroup$


I have found a statement (without proof) that there is no epimorphism in Group category $mathbb{Z}times mathbb{Z} to mathbb{Z}times mathbb{Z}times mathbb{Z} $ existing.
Could you please help, why it's true?






abstract-algebra group-theory abelian-groups







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$endgroup$












  • $begingroup$
    You need to use capital Z in the mathbb font. Lower case z doesn't work. And no epimorphism between those as what? Sets? Groups? Modules? Rings?
    $endgroup$
    – Arthur
    Dec 11 '18 at 14:29










  • $begingroup$
    In Groups category
    $endgroup$
    – Igor Kozyrev
    Dec 11 '18 at 14:31










  • $begingroup$
    Tensor with $Bbb{Q}$.
    $endgroup$
    – jgon
    Dec 11 '18 at 14:33










  • $begingroup$
    Look at the matrix of your morphism.
    $endgroup$
    – Pierre-Yves Gaillard
    Dec 11 '18 at 15:08
















0












$begingroup$


I have found a statement (without proof) that there is no epimorphism in Group category $mathbb{Z}times mathbb{Z} to mathbb{Z}times mathbb{Z}times mathbb{Z} $ existing.
Could you please help, why it's true?






abstract-algebra group-theory abelian-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    You need to use capital Z in the mathbb font. Lower case z doesn't work. And no epimorphism between those as what? Sets? Groups? Modules? Rings?
    $endgroup$
    – Arthur
    Dec 11 '18 at 14:29










  • $begingroup$
    In Groups category
    $endgroup$
    – Igor Kozyrev
    Dec 11 '18 at 14:31










  • $begingroup$
    Tensor with $Bbb{Q}$.
    $endgroup$
    – jgon
    Dec 11 '18 at 14:33










  • $begingroup$
    Look at the matrix of your morphism.
    $endgroup$
    – Pierre-Yves Gaillard
    Dec 11 '18 at 15:08














0












0








0





$begingroup$


I have found a statement (without proof) that there is no epimorphism in Group category $mathbb{Z}times mathbb{Z} to mathbb{Z}times mathbb{Z}times mathbb{Z} $ existing.
Could you please help, why it's true?






abstract-algebra group-theory abelian-groups







share|cite|improve this question











$endgroup$




I have found a statement (without proof) that there is no epimorphism in Group category $mathbb{Z}times mathbb{Z} to mathbb{Z}times mathbb{Z}times mathbb{Z} $ existing.
Could you please help, why it's true?






abstract-algebra group-theory abelian-groups




abstract-algebra group-theory abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 21:18









user26857

39.4k124183




39.4k124183










asked Dec 11 '18 at 14:26









Igor KozyrevIgor Kozyrev

464




464












  • $begingroup$
    You need to use capital Z in the mathbb font. Lower case z doesn't work. And no epimorphism between those as what? Sets? Groups? Modules? Rings?
    $endgroup$
    – Arthur
    Dec 11 '18 at 14:29










  • $begingroup$
    In Groups category
    $endgroup$
    – Igor Kozyrev
    Dec 11 '18 at 14:31










  • $begingroup$
    Tensor with $Bbb{Q}$.
    $endgroup$
    – jgon
    Dec 11 '18 at 14:33










  • $begingroup$
    Look at the matrix of your morphism.
    $endgroup$
    – Pierre-Yves Gaillard
    Dec 11 '18 at 15:08


















  • $begingroup$
    You need to use capital Z in the mathbb font. Lower case z doesn't work. And no epimorphism between those as what? Sets? Groups? Modules? Rings?
    $endgroup$
    – Arthur
    Dec 11 '18 at 14:29










  • $begingroup$
    In Groups category
    $endgroup$
    – Igor Kozyrev
    Dec 11 '18 at 14:31










  • $begingroup$
    Tensor with $Bbb{Q}$.
    $endgroup$
    – jgon
    Dec 11 '18 at 14:33










  • $begingroup$
    Look at the matrix of your morphism.
    $endgroup$
    – Pierre-Yves Gaillard
    Dec 11 '18 at 15:08
















$begingroup$
You need to use capital Z in the mathbb font. Lower case z doesn't work. And no epimorphism between those as what? Sets? Groups? Modules? Rings?
$endgroup$
– Arthur
Dec 11 '18 at 14:29




$begingroup$
You need to use capital Z in the mathbb font. Lower case z doesn't work. And no epimorphism between those as what? Sets? Groups? Modules? Rings?
$endgroup$
– Arthur
Dec 11 '18 at 14:29












$begingroup$
In Groups category
$endgroup$
– Igor Kozyrev
Dec 11 '18 at 14:31




$begingroup$
In Groups category
$endgroup$
– Igor Kozyrev
Dec 11 '18 at 14:31












$begingroup$
Tensor with $Bbb{Q}$.
$endgroup$
– jgon
Dec 11 '18 at 14:33




$begingroup$
Tensor with $Bbb{Q}$.
$endgroup$
– jgon
Dec 11 '18 at 14:33












$begingroup$
Look at the matrix of your morphism.
$endgroup$
– Pierre-Yves Gaillard
Dec 11 '18 at 15:08




$begingroup$
Look at the matrix of your morphism.
$endgroup$
– Pierre-Yves Gaillard
Dec 11 '18 at 15:08










1 Answer
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$begingroup$

Assume that there is an epimorphism $f:mathbb{Z}timesmathbb{Z}tomathbb{Z}timesmathbb{Z}timesmathbb{Z}$, then by quotient $mathbb{Ztimesmathbb{Z}}timesmathbb{Z}$ with $2mathbb{Z}times2mathbb{Z}times2mathbb{Z}$, we also get an epimorphism $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus3}$. Since $(mathbb{Z}/2mathbb{Z})^{oplus3}$ is 2-torsion, $hat{f}$ factors as $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus2}to(mathbb{Z}/2mathbb{Z})^{oplus3}$, in which each morphism is an epimorphism, this is impossible if you compute the order of the two finite groups (4 is less than 8).






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    $begingroup$

    Assume that there is an epimorphism $f:mathbb{Z}timesmathbb{Z}tomathbb{Z}timesmathbb{Z}timesmathbb{Z}$, then by quotient $mathbb{Ztimesmathbb{Z}}timesmathbb{Z}$ with $2mathbb{Z}times2mathbb{Z}times2mathbb{Z}$, we also get an epimorphism $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus3}$. Since $(mathbb{Z}/2mathbb{Z})^{oplus3}$ is 2-torsion, $hat{f}$ factors as $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus2}to(mathbb{Z}/2mathbb{Z})^{oplus3}$, in which each morphism is an epimorphism, this is impossible if you compute the order of the two finite groups (4 is less than 8).






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Assume that there is an epimorphism $f:mathbb{Z}timesmathbb{Z}tomathbb{Z}timesmathbb{Z}timesmathbb{Z}$, then by quotient $mathbb{Ztimesmathbb{Z}}timesmathbb{Z}$ with $2mathbb{Z}times2mathbb{Z}times2mathbb{Z}$, we also get an epimorphism $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus3}$. Since $(mathbb{Z}/2mathbb{Z})^{oplus3}$ is 2-torsion, $hat{f}$ factors as $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus2}to(mathbb{Z}/2mathbb{Z})^{oplus3}$, in which each morphism is an epimorphism, this is impossible if you compute the order of the two finite groups (4 is less than 8).






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Assume that there is an epimorphism $f:mathbb{Z}timesmathbb{Z}tomathbb{Z}timesmathbb{Z}timesmathbb{Z}$, then by quotient $mathbb{Ztimesmathbb{Z}}timesmathbb{Z}$ with $2mathbb{Z}times2mathbb{Z}times2mathbb{Z}$, we also get an epimorphism $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus3}$. Since $(mathbb{Z}/2mathbb{Z})^{oplus3}$ is 2-torsion, $hat{f}$ factors as $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus2}to(mathbb{Z}/2mathbb{Z})^{oplus3}$, in which each morphism is an epimorphism, this is impossible if you compute the order of the two finite groups (4 is less than 8).






        share|cite|improve this answer









        $endgroup$



        Assume that there is an epimorphism $f:mathbb{Z}timesmathbb{Z}tomathbb{Z}timesmathbb{Z}timesmathbb{Z}$, then by quotient $mathbb{Ztimesmathbb{Z}}timesmathbb{Z}$ with $2mathbb{Z}times2mathbb{Z}times2mathbb{Z}$, we also get an epimorphism $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus3}$. Since $(mathbb{Z}/2mathbb{Z})^{oplus3}$ is 2-torsion, $hat{f}$ factors as $hat{f}:mathbb{Z}timesmathbb{Z}to(mathbb{Z}/2mathbb{Z})^{oplus2}to(mathbb{Z}/2mathbb{Z})^{oplus3}$, in which each morphism is an epimorphism, this is impossible if you compute the order of the two finite groups (4 is less than 8).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 14:45









        BonbonBonbon

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