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Identity element of a group, help [closed]

-1

$begingroup$

So far I have done the RHS to $(x-3)(y-3)+12$ and I have done this to find the identity element, $(e-3)(x-3)+12=x$, then re-arranged to $(e-3)=(x-12)/(x-3)$, now I am stuck.

abstract-algebra group-theory

share|cite|improve this question

edited Mar 20 at 16:13

Javi

3,1242932

asked Mar 20 at 10:15

nathannathan

334

$endgroup$

closed as off-topic by Dietrich Burde, Thomas Shelby, Riccardo.Alestra, jgon, Parcly Taxel Mar 21 at 0:42

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." – Dietrich Burde, Thomas Shelby, Riccardo.Alestra, jgon, Parcly Taxel

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Apply definition of identity element. Put $y=0$ in $x*y=y$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 20 at 10:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Again one of the various variations of this question.
  $endgroup$
  – Dietrich Burde
  Mar 20 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $e=4$ will do the job for the identity.
  $endgroup$
  – Dbchatto67
  Mar 20 at 10:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You factored wrong. It factors as $(x-3)(y-3)+3$.
  $endgroup$
  – jgon
  Mar 20 at 15:49
  

  
  
  
  

add a comment | 

-1

$begingroup$

So far I have done the RHS to $(x-3)(y-3)+12$ and I have done this to find the identity element, $(e-3)(x-3)+12=x$, then re-arranged to $(e-3)=(x-12)/(x-3)$, now I am stuck.

abstract-algebra group-theory

share|cite|improve this question

edited Mar 20 at 16:13

Javi

3,1242932

asked Mar 20 at 10:15

nathannathan

334

$endgroup$

closed as off-topic by Dietrich Burde, Thomas Shelby, Riccardo.Alestra, jgon, Parcly Taxel Mar 21 at 0:42

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." – Dietrich Burde, Thomas Shelby, Riccardo.Alestra, jgon, Parcly Taxel

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Apply definition of identity element. Put $y=0$ in $x*y=y$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 20 at 10:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Again one of the various variations of this question.
  $endgroup$
  – Dietrich Burde
  Mar 20 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $e=4$ will do the job for the identity.
  $endgroup$
  – Dbchatto67
  Mar 20 at 10:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You factored wrong. It factors as $(x-3)(y-3)+3$.
  $endgroup$
  – jgon
  Mar 20 at 15:49
  

  
  
  
  

add a comment | 

$begingroup$

So far I have done the RHS to $(x-3)(y-3)+12$ and I have done this to find the identity element, $(e-3)(x-3)+12=x$, then re-arranged to $(e-3)=(x-12)/(x-3)$, now I am stuck.

abstract-algebra group-theory

share|cite|improve this question

edited Mar 20 at 16:13

Javi

3,1242932

asked Mar 20 at 10:15

nathannathan

334

$endgroup$

share|cite|improve this question

edited Mar 20 at 16:13

Javi

3,1242932

asked Mar 20 at 10:15

nathannathan

334

share|cite|improve this question

edited Mar 20 at 16:13

Javi

3,1242932

asked Mar 20 at 10:15

nathannathan

334

edited Mar 20 at 16:13

Javi

3,1242932

edited Mar 20 at 16:13

Javi

3,1242932

asked Mar 20 at 10:15

nathannathan

334

asked Mar 20 at 10:15

nathannathan

334

4 Answers
4

active

oldest

votes

1

$begingroup$

Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$

share|cite|improve this answer

answered Mar 20 at 10:24

FredFred

48.7k11849

$endgroup$

add a comment | 

1

$begingroup$

The identity element (if it exists) is a value $e$ such that

$x*e = x space forall x$

Note that the symmetry of the definition of $x*y$ tells us that

$x*e = x space forall x Rightarrow e*y = y space forall y$

so if we can find $e$ then it will be a two-sided identity.

$x*e = x space forall x \
Rightarrow xe -3x - 3e + 12 = x space forall x \
Rightarrow x(e-4) -3e +12 = 0 space forall x \
Rightarrow (x-3)(e-4)= 0 space forall x$

From this can you see a value for $e$ which will act as an identity ?

share|cite|improve this answer

answered Mar 20 at 10:29

gandalf61gandalf61

9,174825

$endgroup$

add a comment | 

0

$begingroup$

Note that in any group $G$, the identity is the unique solution to the equation $g^2=g$.

Hence, the identity in your group is the unique solution to the equation $x^2-6x+12=x$ in $mathbf Rsetminus 3$.

share|cite|improve this answer

answered Mar 20 at 16:58

tomasztomasz

24k23482

$endgroup$

add a comment | 

0

$begingroup$

Hint: Show that the function $f(x)=x-3$ defines an isomorphism between your group and the group of nonzero real numbers with multiplication. Note that this way, you can simultaneously check that this formula really defines a group operation.

share|cite|improve this answer

answered Mar 20 at 17:01

tomasztomasz

24k23482

$endgroup$

add a comment | 

1

$begingroup$

Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$

share|cite|improve this answer

answered Mar 20 at 10:24

FredFred

48.7k11849

$endgroup$

add a comment | 

1

$begingroup$

Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$

share|cite|improve this answer

answered Mar 20 at 10:24

FredFred

48.7k11849

$endgroup$

add a comment | 

$begingroup$

Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$

share|cite|improve this answer

answered Mar 20 at 10:24

FredFred

48.7k11849

$endgroup$

share|cite|improve this answer

answered Mar 20 at 10:24

FredFred

48.7k11849

answered Mar 20 at 10:24

FredFred

48.7k11849

answered Mar 20 at 10:24

FredFred

48.7k11849

1

$begingroup$

The identity element (if it exists) is a value $e$ such that

$x*e = x space forall x$

Note that the symmetry of the definition of $x*y$ tells us that

$x*e = x space forall x Rightarrow e*y = y space forall y$

so if we can find $e$ then it will be a two-sided identity.

$x*e = x space forall x \
Rightarrow xe -3x - 3e + 12 = x space forall x \
Rightarrow x(e-4) -3e +12 = 0 space forall x \
Rightarrow (x-3)(e-4)= 0 space forall x$

From this can you see a value for $e$ which will act as an identity ?

share|cite|improve this answer

answered Mar 20 at 10:29

gandalf61gandalf61

9,174825

$endgroup$

add a comment | 

1

$begingroup$

The identity element (if it exists) is a value $e$ such that

$x*e = x space forall x$

Note that the symmetry of the definition of $x*y$ tells us that

$x*e = x space forall x Rightarrow e*y = y space forall y$

so if we can find $e$ then it will be a two-sided identity.

$x*e = x space forall x \
Rightarrow xe -3x - 3e + 12 = x space forall x \
Rightarrow x(e-4) -3e +12 = 0 space forall x \
Rightarrow (x-3)(e-4)= 0 space forall x$

From this can you see a value for $e$ which will act as an identity ?

share|cite|improve this answer

answered Mar 20 at 10:29

gandalf61gandalf61

9,174825

$endgroup$

add a comment | 

$begingroup$

The identity element (if it exists) is a value $e$ such that

$x*e = x space forall x$

Note that the symmetry of the definition of $x*y$ tells us that

$x*e = x space forall x Rightarrow e*y = y space forall y$

so if we can find $e$ then it will be a two-sided identity.

$x*e = x space forall x \
Rightarrow xe -3x - 3e + 12 = x space forall x \
Rightarrow x(e-4) -3e +12 = 0 space forall x \
Rightarrow (x-3)(e-4)= 0 space forall x$

From this can you see a value for $e$ which will act as an identity ?

share|cite|improve this answer

answered Mar 20 at 10:29

gandalf61gandalf61

9,174825

$endgroup$

share|cite|improve this answer

answered Mar 20 at 10:29

gandalf61gandalf61

9,174825

answered Mar 20 at 10:29

gandalf61gandalf61

9,174825

answered Mar 20 at 10:29

gandalf61gandalf61

9,174825

0

$begingroup$

Note that in any group $G$, the identity is the unique solution to the equation $g^2=g$.

Hence, the identity in your group is the unique solution to the equation $x^2-6x+12=x$ in $mathbf Rsetminus 3$.

share|cite|improve this answer

answered Mar 20 at 16:58

tomasztomasz

24k23482

$endgroup$

add a comment | 

0

$begingroup$

Note that in any group $G$, the identity is the unique solution to the equation $g^2=g$.

Hence, the identity in your group is the unique solution to the equation $x^2-6x+12=x$ in $mathbf Rsetminus 3$.

share|cite|improve this answer

answered Mar 20 at 16:58

tomasztomasz

24k23482

$endgroup$

add a comment | 

$begingroup$

Note that in any group $G$, the identity is the unique solution to the equation $g^2=g$.

Hence, the identity in your group is the unique solution to the equation $x^2-6x+12=x$ in $mathbf Rsetminus 3$.

share|cite|improve this answer

answered Mar 20 at 16:58

tomasztomasz

24k23482

$endgroup$

share|cite|improve this answer

answered Mar 20 at 16:58

tomasztomasz

24k23482

answered Mar 20 at 16:58

tomasztomasz

24k23482

answered Mar 20 at 16:58

tomasztomasz

24k23482

0

$begingroup$

Hint: Show that the function $f(x)=x-3$ defines an isomorphism between your group and the group of nonzero real numbers with multiplication. Note that this way, you can simultaneously check that this formula really defines a group operation.

share|cite|improve this answer

answered Mar 20 at 17:01

tomasztomasz

24k23482

$endgroup$

add a comment | 

0

$begingroup$

Hint: Show that the function $f(x)=x-3$ defines an isomorphism between your group and the group of nonzero real numbers with multiplication. Note that this way, you can simultaneously check that this formula really defines a group operation.

share|cite|improve this answer

answered Mar 20 at 17:01

tomasztomasz

24k23482

$endgroup$

add a comment | 

$begingroup$

Hint: Show that the function $f(x)=x-3$ defines an isomorphism between your group and the group of nonzero real numbers with multiplication. Note that this way, you can simultaneously check that this formula really defines a group operation.

share|cite|improve this answer

answered Mar 20 at 17:01

tomasztomasz

24k23482

$endgroup$

share|cite|improve this answer

answered Mar 20 at 17:01

tomasztomasz

24k23482

answered Mar 20 at 17:01

tomasztomasz

24k23482

answered Mar 20 at 17:01

tomasztomasz

24k23482