If $g:mathbb{Z_{10}}rightarrow U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or...












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Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?



Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.










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closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.












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    $begingroup$
    $1$ has order $10$ in $mathbb{Z}_{10}$.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:18


















-1












$begingroup$


Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?



Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.










share|cite|improve this question











$endgroup$



closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    $1$ has order $10$ in $mathbb{Z}_{10}$.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:18
















-1












-1








-1





$begingroup$


Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?



Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.










share|cite|improve this question











$endgroup$




Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?



Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.







abstract-algebra group-theory functions group-homomorphism






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edited Dec 10 '18 at 5:42









Shaun

9,241113684




9,241113684










asked Dec 10 '18 at 5:15









AMN52AMN52

326




326




closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    $1$ has order $10$ in $mathbb{Z}_{10}$.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:18
















  • 1




    $begingroup$
    $1$ has order $10$ in $mathbb{Z}_{10}$.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:18










1




1




$begingroup$
$1$ has order $10$ in $mathbb{Z}_{10}$.
$endgroup$
– Randall
Dec 10 '18 at 5:18






$begingroup$
$1$ has order $10$ in $mathbb{Z}_{10}$.
$endgroup$
– Randall
Dec 10 '18 at 5:18












2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.






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$endgroup$





















    0












    $begingroup$

    In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.






          share|cite|improve this answer









          $endgroup$



          Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 5:24









          ShaunShaun

          9,241113684




          9,241113684























              0












              $begingroup$

              In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.






                  share|cite|improve this answer









                  $endgroup$



                  In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 6:27









                  Chris CusterChris Custer

                  13.6k3827




                  13.6k3827















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