A group isomorphic to $Bbb Z_p^*$. [on hold]Proof of Fermat's little theorem using groupsNilpotent action on $p$-groupA finite abelian group that does not contain a subgroup isomorphic to $mathbb Z_poplusmathbb Z_p$, for any prime $p$, is cyclic.Let $G$ be a group of all $2 times 2$ matrices over $Z_p$ with determinant $1$ under matrix multiplication. To find the order of $G$.Matrix group isomorphic to $mathbb Z$.Prove that external direct product is isomorphic to groupIf $p$ is a prime number and $G$ is non-abelian group of order $p^3$, $G/Z(G) cong Z_p times Z_p$.Proper subgroups of $Bbb Z_p$Let G be $mathbb Z_ptimesdotstimes mathbb Z_p$ . Find A(G).Proving $(Bbb R,+)$ has no proper subgroup isomorphic to itself

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A group isomorphic to $Bbb Z_p^*$. [on hold]


Proof of Fermat's little theorem using groupsNilpotent action on $p$-groupA finite abelian group that does not contain a subgroup isomorphic to $mathbb Z_poplusmathbb Z_p$, for any prime $p$, is cyclic.Let $G$ be a group of all $2 times 2$ matrices over $Z_p$ with determinant $1$ under matrix multiplication. To find the order of $G$.Matrix group isomorphic to $mathbb Z$.Prove that external direct product is isomorphic to groupIf $p$ is a prime number and $G$ is non-abelian group of order $p^3$, $G/Z(G) cong Z_p times Z_p$.Proper subgroups of $Bbb Z_p$Let G be $mathbb Z_ptimesdotstimes mathbb Z_p$ . Find A(G).Proving $(Bbb R,+)$ has no proper subgroup isomorphic to itself













0












$begingroup$


Let $p$ be prime. Let $H = 1,2,3,dots ,p-1$ with binary operation $cdot$ defined as $$acdot b := ab pmodp.$$



I have a gut feeling that $H$ is isomorphic to $Bbb Z_p^*$ under multiplication.



I just dont know how to prove it.



Is $H$ a group?



Thank you.










share|cite|improve this question











$endgroup$



put on hold as off-topic by uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u yesterday


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." – uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    $begingroup$
    What's your definition of $mathbbZ_p$, how does it differ from your $H$?
    $endgroup$
    – badjohn
    yesterday







  • 1




    $begingroup$
    You should be able to check for yourself whether $H$ is a group or not.
    $endgroup$
    – uniquesolution
    yesterday















0












$begingroup$


Let $p$ be prime. Let $H = 1,2,3,dots ,p-1$ with binary operation $cdot$ defined as $$acdot b := ab pmodp.$$



I have a gut feeling that $H$ is isomorphic to $Bbb Z_p^*$ under multiplication.



I just dont know how to prove it.



Is $H$ a group?



Thank you.










share|cite|improve this question











$endgroup$



put on hold as off-topic by uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u yesterday


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." – uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    $begingroup$
    What's your definition of $mathbbZ_p$, how does it differ from your $H$?
    $endgroup$
    – badjohn
    yesterday







  • 1




    $begingroup$
    You should be able to check for yourself whether $H$ is a group or not.
    $endgroup$
    – uniquesolution
    yesterday













0












0








0





$begingroup$


Let $p$ be prime. Let $H = 1,2,3,dots ,p-1$ with binary operation $cdot$ defined as $$acdot b := ab pmodp.$$



I have a gut feeling that $H$ is isomorphic to $Bbb Z_p^*$ under multiplication.



I just dont know how to prove it.



Is $H$ a group?



Thank you.










share|cite|improve this question











$endgroup$




Let $p$ be prime. Let $H = 1,2,3,dots ,p-1$ with binary operation $cdot$ defined as $$acdot b := ab pmodp.$$



I have a gut feeling that $H$ is isomorphic to $Bbb Z_p^*$ under multiplication.



I just dont know how to prove it.



Is $H$ a group?



Thank you.







group-theory finite-groups group-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Shaun

9,334113684




9,334113684










asked yesterday









user593805user593805

866




866




put on hold as off-topic by uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u yesterday


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." – uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u yesterday


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." – uniquesolution, Derek Holt, Shaun, Lee David Chung Lin, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 3




    $begingroup$
    What's your definition of $mathbbZ_p$, how does it differ from your $H$?
    $endgroup$
    – badjohn
    yesterday







  • 1




    $begingroup$
    You should be able to check for yourself whether $H$ is a group or not.
    $endgroup$
    – uniquesolution
    yesterday












  • 3




    $begingroup$
    What's your definition of $mathbbZ_p$, how does it differ from your $H$?
    $endgroup$
    – badjohn
    yesterday







  • 1




    $begingroup$
    You should be able to check for yourself whether $H$ is a group or not.
    $endgroup$
    – uniquesolution
    yesterday







3




3




$begingroup$
What's your definition of $mathbbZ_p$, how does it differ from your $H$?
$endgroup$
– badjohn
yesterday





$begingroup$
What's your definition of $mathbbZ_p$, how does it differ from your $H$?
$endgroup$
– badjohn
yesterday





1




1




$begingroup$
You should be able to check for yourself whether $H$ is a group or not.
$endgroup$
– uniquesolution
yesterday




$begingroup$
You should be able to check for yourself whether $H$ is a group or not.
$endgroup$
– uniquesolution
yesterday










2 Answers
2






active

oldest

votes


















1












$begingroup$

The group you refer to is not so much isomorphic to $Bbb Z_p^*$, as the definition of it. That accounts for your gut feeling.



$H$ is a group, as the nonzero elements of a field always form a multiplicative group by definition.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    In the definition of $Bbb Z_n^*$ I learnt first of all, the underlying set was $$[a]_nmid gcd(a,n)=1,$$ where $[a]_n:=bmid aequiv bpmodn$; thus it is not (strictly) the group the OP describes.
    $endgroup$
    – Shaun
    yesterday











  • $begingroup$
    @Shaun but when $n$ is prime it is.
    $endgroup$
    – Chris Custer
    yesterday










  • $begingroup$
    No, it's not: $1, dots, p-1$ is not a set of equivalence classes described in my previous comment; it's a set of representatives.
    $endgroup$
    – Shaun
    yesterday


















-1












$begingroup$

We have, that $acdot b$ is closed under the operation, the operation $cdot$ being associative, having an identity($1$), and each element having an inverse, by Fermat theorem, the inverse is $a^p-2$ for any $p-2$. Hence the operation defines a group. This is the multiplicative group of a finite field of order $p$.






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












    $begingroup$

    The group you refer to is not so much isomorphic to $Bbb Z_p^*$, as the definition of it. That accounts for your gut feeling.



    $H$ is a group, as the nonzero elements of a field always form a multiplicative group by definition.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      In the definition of $Bbb Z_n^*$ I learnt first of all, the underlying set was $$[a]_nmid gcd(a,n)=1,$$ where $[a]_n:=bmid aequiv bpmodn$; thus it is not (strictly) the group the OP describes.
      $endgroup$
      – Shaun
      yesterday











    • $begingroup$
      @Shaun but when $n$ is prime it is.
      $endgroup$
      – Chris Custer
      yesterday










    • $begingroup$
      No, it's not: $1, dots, p-1$ is not a set of equivalence classes described in my previous comment; it's a set of representatives.
      $endgroup$
      – Shaun
      yesterday















    1












    $begingroup$

    The group you refer to is not so much isomorphic to $Bbb Z_p^*$, as the definition of it. That accounts for your gut feeling.



    $H$ is a group, as the nonzero elements of a field always form a multiplicative group by definition.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      In the definition of $Bbb Z_n^*$ I learnt first of all, the underlying set was $$[a]_nmid gcd(a,n)=1,$$ where $[a]_n:=bmid aequiv bpmodn$; thus it is not (strictly) the group the OP describes.
      $endgroup$
      – Shaun
      yesterday











    • $begingroup$
      @Shaun but when $n$ is prime it is.
      $endgroup$
      – Chris Custer
      yesterday










    • $begingroup$
      No, it's not: $1, dots, p-1$ is not a set of equivalence classes described in my previous comment; it's a set of representatives.
      $endgroup$
      – Shaun
      yesterday













    1












    1








    1





    $begingroup$

    The group you refer to is not so much isomorphic to $Bbb Z_p^*$, as the definition of it. That accounts for your gut feeling.



    $H$ is a group, as the nonzero elements of a field always form a multiplicative group by definition.






    share|cite|improve this answer









    $endgroup$



    The group you refer to is not so much isomorphic to $Bbb Z_p^*$, as the definition of it. That accounts for your gut feeling.



    $H$ is a group, as the nonzero elements of a field always form a multiplicative group by definition.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    Chris CusterChris Custer

    14.2k3827




    14.2k3827











    • $begingroup$
      In the definition of $Bbb Z_n^*$ I learnt first of all, the underlying set was $$[a]_nmid gcd(a,n)=1,$$ where $[a]_n:=bmid aequiv bpmodn$; thus it is not (strictly) the group the OP describes.
      $endgroup$
      – Shaun
      yesterday











    • $begingroup$
      @Shaun but when $n$ is prime it is.
      $endgroup$
      – Chris Custer
      yesterday










    • $begingroup$
      No, it's not: $1, dots, p-1$ is not a set of equivalence classes described in my previous comment; it's a set of representatives.
      $endgroup$
      – Shaun
      yesterday
















    • $begingroup$
      In the definition of $Bbb Z_n^*$ I learnt first of all, the underlying set was $$[a]_nmid gcd(a,n)=1,$$ where $[a]_n:=bmid aequiv bpmodn$; thus it is not (strictly) the group the OP describes.
      $endgroup$
      – Shaun
      yesterday











    • $begingroup$
      @Shaun but when $n$ is prime it is.
      $endgroup$
      – Chris Custer
      yesterday










    • $begingroup$
      No, it's not: $1, dots, p-1$ is not a set of equivalence classes described in my previous comment; it's a set of representatives.
      $endgroup$
      – Shaun
      yesterday















    $begingroup$
    In the definition of $Bbb Z_n^*$ I learnt first of all, the underlying set was $$[a]_nmid gcd(a,n)=1,$$ where $[a]_n:=bmid aequiv bpmodn$; thus it is not (strictly) the group the OP describes.
    $endgroup$
    – Shaun
    yesterday





    $begingroup$
    In the definition of $Bbb Z_n^*$ I learnt first of all, the underlying set was $$[a]_nmid gcd(a,n)=1,$$ where $[a]_n:=bmid aequiv bpmodn$; thus it is not (strictly) the group the OP describes.
    $endgroup$
    – Shaun
    yesterday













    $begingroup$
    @Shaun but when $n$ is prime it is.
    $endgroup$
    – Chris Custer
    yesterday




    $begingroup$
    @Shaun but when $n$ is prime it is.
    $endgroup$
    – Chris Custer
    yesterday












    $begingroup$
    No, it's not: $1, dots, p-1$ is not a set of equivalence classes described in my previous comment; it's a set of representatives.
    $endgroup$
    – Shaun
    yesterday




    $begingroup$
    No, it's not: $1, dots, p-1$ is not a set of equivalence classes described in my previous comment; it's a set of representatives.
    $endgroup$
    – Shaun
    yesterday











    -1












    $begingroup$

    We have, that $acdot b$ is closed under the operation, the operation $cdot$ being associative, having an identity($1$), and each element having an inverse, by Fermat theorem, the inverse is $a^p-2$ for any $p-2$. Hence the operation defines a group. This is the multiplicative group of a finite field of order $p$.






    share|cite|improve this answer









    $endgroup$

















      -1












      $begingroup$

      We have, that $acdot b$ is closed under the operation, the operation $cdot$ being associative, having an identity($1$), and each element having an inverse, by Fermat theorem, the inverse is $a^p-2$ for any $p-2$. Hence the operation defines a group. This is the multiplicative group of a finite field of order $p$.






      share|cite|improve this answer









      $endgroup$















        -1












        -1








        -1





        $begingroup$

        We have, that $acdot b$ is closed under the operation, the operation $cdot$ being associative, having an identity($1$), and each element having an inverse, by Fermat theorem, the inverse is $a^p-2$ for any $p-2$. Hence the operation defines a group. This is the multiplicative group of a finite field of order $p$.






        share|cite|improve this answer









        $endgroup$



        We have, that $acdot b$ is closed under the operation, the operation $cdot$ being associative, having an identity($1$), and each element having an inverse, by Fermat theorem, the inverse is $a^p-2$ for any $p-2$. Hence the operation defines a group. This is the multiplicative group of a finite field of order $p$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        vidyarthividyarthi

        3,0291833




        3,0291833













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