(For any group $H$) Bijection between $H$ and the group consists of all the homomorphisms between $(mathbb Z,...












1












$begingroup$


My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience



I just started learning abstract algebra and soon I encountered this problem.




Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.





I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.



Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.



So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$



$$
b(h) = i_h
$$



And finally the last two tasks.



a) injective



$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$



b) surjective



But I don't know how to prove the surjective part.






abstract-algebra group-theory functions group-homomorphism







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$endgroup$












  • $begingroup$
    Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
    $endgroup$
    – Joppy
    Dec 28 '18 at 9:15
















1












$begingroup$


My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience



I just started learning abstract algebra and soon I encountered this problem.




Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.





I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.



Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.



So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$



$$
b(h) = i_h
$$



And finally the last two tasks.



a) injective



$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$



b) surjective



But I don't know how to prove the surjective part.






abstract-algebra group-theory functions group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
    $endgroup$
    – Joppy
    Dec 28 '18 at 9:15














1












1








1


1



$begingroup$


My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience



I just started learning abstract algebra and soon I encountered this problem.




Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.





I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.



Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.



So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$



$$
b(h) = i_h
$$



And finally the last two tasks.



a) injective



$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$



b) surjective



But I don't know how to prove the surjective part.






abstract-algebra group-theory functions group-homomorphism







share|cite|improve this question











$endgroup$




My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience



I just started learning abstract algebra and soon I encountered this problem.




Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $phi : G rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.





I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.



Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.



So I can construct a function named $b$ which is from $H$ to $IS$. Thence $mathrm { b } : mathrm { H } rightarrow mathrm { IS }$



$$
b(h) = i_h
$$



And finally the last two tasks.



a) injective



$forall h _ { 1 } , h _ { 2 } in H text { if } b left( h _ { 1 } right) = b left( h _ { 2 } right) text { we will have } b left( h _ { 1 } right) ( 1 ) = b left( h _ { 2 } right) ( 1 ) text { which will result in the fact } h _ { 1 } = h _ { 2 }$



b) surjective



But I don't know how to prove the surjective part.






abstract-algebra group-theory functions group-homomorphism




abstract-algebra group-theory functions group-homomorphism






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edited Dec 28 '18 at 9:56









Shaun

10.5k113687




10.5k113687










asked Dec 28 '18 at 9:04









ShuumatsuShuumatsu

637




637












  • $begingroup$
    Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
    $endgroup$
    – Joppy
    Dec 28 '18 at 9:15


















  • $begingroup$
    Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
    $endgroup$
    – Joppy
    Dec 28 '18 at 9:15
















$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15




$begingroup$
Define a function $c: IS to H$, by $c(f) = f(1)$. Can you show that $b$ and $c$ are mutually inverse?
$endgroup$
– Joppy
Dec 28 '18 at 9:15










3 Answers
3






active

oldest

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1












$begingroup$

For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:



$$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:



        $$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
        for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:



          $$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
          for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:



            $$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
            for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.






            share|cite|improve this answer









            $endgroup$



            For any $phiin IS$, we have $h:=phi(1)in H$. We will claim that $$b(h) = i_h=phi$$ as follows:



            $$i_h(n)=h^n=(phi(1))^n=phi(n*1)=phi(n)$$
            for all $nin mathbb{Z}$. Thus $i_h=phi$. This means that $phi=b(h)$ is in the image of $b$. So $b$ is surjective.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 28 '18 at 9:16









            yoyoyoyo

            756410




            756410























                0












                $begingroup$

                You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.






                    share|cite|improve this answer









                    $endgroup$



                    You have to prove that $IS={i_{h}, hin H}$, $i_{h}$ defined above. $mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $mathbb{Z}to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 28 '18 at 9:14









                    ecrinecrin

                    3477




                    3477























                        0












                        $begingroup$

                        For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.






                            share|cite|improve this answer









                            $endgroup$



                            For surjectivity, let $f in text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 28 '18 at 9:17









                            Anurag AAnurag A

                            26.4k12351




                            26.4k12351






























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