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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.

group-theory finite-groups group-isomorphism

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked yesterday

user593805user593805

866

$endgroup$

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
  

  
  
  
  

add a comment | 

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.

group-theory finite-groups group-isomorphism

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked yesterday

user593805user593805

866

$endgroup$

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
  

  
  
  
  

add a comment | 

$begingroup$

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.

group-theory finite-groups group-isomorphism

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked yesterday

user593805user593805

866

$endgroup$

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked yesterday

user593805user593805

866

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked yesterday

user593805user593805

866

edited yesterday

Shaun

9,334113684

edited yesterday

Shaun

9,334113684

asked yesterday

user593805user593805

866

asked yesterday

user593805user593805

866

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

answered yesterday

Chris CusterChris Custer

14.2k3827

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

-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

answered yesterday

vidyarthividyarthi

3,0291833

$endgroup$

add a comment | 

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

answered yesterday

Chris CusterChris Custer

14.2k3827

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered yesterday

Chris CusterChris Custer

14.2k3827

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered yesterday

Chris CusterChris Custer

14.2k3827

$endgroup$

share|cite|improve this answer

answered yesterday

Chris CusterChris Custer

14.2k3827

answered yesterday

Chris CusterChris Custer

14.2k3827

answered yesterday

Chris CusterChris Custer

14.2k3827

-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

answered yesterday

vidyarthividyarthi

3,0291833

$endgroup$

add a comment | 

-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

answered yesterday

vidyarthividyarthi

3,0291833

$endgroup$

add a comment | 

$begingroup$

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

answered yesterday

vidyarthividyarthi

3,0291833

$endgroup$

share|cite|improve this answer

answered yesterday

vidyarthividyarthi

3,0291833

answered yesterday

vidyarthividyarthi

3,0291833

answered yesterday

vidyarthividyarthi

3,0291833