Isomorphism between two groups












1












$begingroup$



Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.




I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:16












  • $begingroup$
    You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:20










  • $begingroup$
    What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:22










  • $begingroup$
    $Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:27












  • $begingroup$
    No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:29


















1












$begingroup$



Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.




I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:16












  • $begingroup$
    You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:20










  • $begingroup$
    What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:22










  • $begingroup$
    $Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:27












  • $begingroup$
    No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:29
















1












1








1


0



$begingroup$



Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.




I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.






abstract-algebra







share|cite|improve this question











$endgroup$





Prove that $langle Bbb Z/nBbb Z, +rangle$ and $langle U_{n}, {}cdot{}rangle$ are isomorphic binary structure where $U_n$ is roots of unity and $Bbb Z/nBbb Z$ is integers modulo $n$.




I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function. Please help me, if you have any good idea.






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 9:50









Chinnapparaj R

5,4942928




5,4942928










asked Dec 4 '18 at 9:07









mathsstudentmathsstudent

386




386












  • $begingroup$
    That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:16












  • $begingroup$
    You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:20










  • $begingroup$
    What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:22










  • $begingroup$
    $Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:27












  • $begingroup$
    No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:29




















  • $begingroup$
    That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:16












  • $begingroup$
    You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:20










  • $begingroup$
    What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:22










  • $begingroup$
    $Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
    $endgroup$
    – mathsstudent
    Dec 4 '18 at 9:27












  • $begingroup$
    No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
    $endgroup$
    – Arthur
    Dec 4 '18 at 9:29


















$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16






$begingroup$
That's better. Now, if you're stuck on a general problem (like here, "show that for all $n$, something something"), it usually helps to check a few small examples and see if you can get some insight. What about $Bbb Z/Bbb Z_2$ and $U_2$? What about $Bbb Z/3Bbb Z$ and $U_3$? Can you find isomorhisms there? What about $4$ and $5$? Does that generalize in any way?
$endgroup$
– Arthur
Dec 4 '18 at 9:16














$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20




$begingroup$
You are right. Thank you. I can define a map between $Z/3Z$ and $U3$ for example : $f(3k)$ = $(3k)^i$ or can not ?
$endgroup$
– mathsstudent
Dec 4 '18 at 9:20












$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22




$begingroup$
What are the elements of $Bbb Z/3Bbb Z$, and what are the elements of $U_3$?
$endgroup$
– Arthur
Dec 4 '18 at 9:22












$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27






$begingroup$
$Z/3Z= {0, 3, 6, 9, 12,....}$ and $U3={1,t, t^{2}}$ where $t=e^{2pi(i)/3}$
$endgroup$
– mathsstudent
Dec 4 '18 at 9:27














$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29






$begingroup$
No, $0, 3, 6, 9, ldots$ is $3Bbb Z$, not $Bbb Z/3Bbb Z$. The three elements of $Bbb Z/3Bbb Z$ are ${ldots, -3, 0, 3, 6, ldots}$ and ${ldots,-2, 1, 4, 7, ldots}$ and ${ldots,-1, 2, 5, 8, ldots}$, usually called $[0], [1]$ and $[2]$.
$endgroup$
– Arthur
Dec 4 '18 at 9:29












1 Answer
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$begingroup$

Outline:



$$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$



For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$






share|cite|improve this answer









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    1 Answer
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    active

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    $begingroup$

    Outline:



    $$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$



    For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Outline:



      $$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$



      For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Outline:



        $$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$



        For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$






        share|cite|improve this answer









        $endgroup$



        Outline:



        $$U_n={1, zeta,zeta^2,cdots, zeta^{n-1}}=langle zeta rangle$$ where $zeta=e^frac{2 pi i}{n}$



        For a sake of simplicity, identify $Bbb Z/ n Bbb Z$ with $Bbb Z_n$. Here $Bbb Z_n =langle 1 rangle$. Then the map $$ f:Bbb Z_n ni 1^i mapsto zeta^i in U_n$$ is an isomorphism (!). Here $1^n$ means $underbrace{1+1+cdots+1}_{n ;times}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 9:25









        Chinnapparaj RChinnapparaj R

        5,4942928




        5,4942928






























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