Identity element of a group, help [closed] The Next CEO of Stack OverflowProve that $a * b = a + b - ab$ defines a group operation on $Bbb R setminus 1$Group of order $63$ has an element of order $3$, without using Cauchy's or Sylow's theoremsIdentity element of a groupIdentity Element and Identity PropertiesFinding the identity element of a groupIdentity element of matrix groupThe identity element in a sub groupFinding identity element and inverse element of a groupIdentity elementThe identity element of the group of all grouphomomorphismsDetermine the idempotent element of a group
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Identity element of a group, help [closed]
The Next CEO of Stack OverflowProve that $a * b = a + b - ab$ defines a group operation on $Bbb R setminus 1$Group of order $63$ has an element of order $3$, without using Cauchy's or Sylow's theoremsIdentity element of a groupIdentity Element and Identity PropertiesFinding the identity element of a groupIdentity element of matrix groupThe identity element in a sub groupFinding identity element and inverse element of a groupIdentity elementThe identity element of the group of all grouphomomorphismsDetermine the idempotent element of a group
$begingroup$
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
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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
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
$endgroup$
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
$begingroup$
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
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$e=4$ 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
$endgroup$
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
abstract-algebra group-theory
edited Mar 20 at 16:13
Javi
3,1242932
3,1242932
asked Mar 20 at 10:15
nathannathan
334
334
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
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
$begingroup$
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 |
$begingroup$
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
$begingroup$
Apply definition of identity element. Put $y=0$ in $x*y=y$.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:18
$begingroup$
Apply definition of identity element. Put $y=0$ in $x*y=y$.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:18
1
1
$begingroup$
Again one of the various variations of this question.
$endgroup$
– Dietrich Burde
Mar 20 at 10:21
$begingroup$
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
$begingroup$
$e=4$ 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
$begingroup$
You factored wrong. It factors as $(x-3)(y-3)+3$.
$endgroup$
– jgon
Mar 20 at 15:49
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$
$endgroup$
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 ?
$endgroup$
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 $mathbf Rsetminus 3$.
$endgroup$
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.
$endgroup$
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$
$endgroup$
add a comment |
$begingroup$
Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$
$endgroup$
add a comment |
$begingroup$
Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$
$endgroup$
Let $e$ denote the identity element. Then: $0=0*e=-3e+12$. This gives $e=4.$
answered Mar 20 at 10:24
FredFred
48.7k11849
48.7k11849
add a comment |
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 ?
$endgroup$
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 ?
$endgroup$
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 ?
$endgroup$
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 ?
answered Mar 20 at 10:29
gandalf61gandalf61
9,174825
9,174825
add a comment |
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 $mathbf Rsetminus 3$.
$endgroup$
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 $mathbf Rsetminus 3$.
$endgroup$
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 $mathbf Rsetminus 3$.
$endgroup$
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 $mathbf Rsetminus 3$.
answered Mar 20 at 16:58
tomasztomasz
24k23482
24k23482
add a comment |
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
answered Mar 20 at 17:01
tomasztomasz
24k23482
24k23482
add a comment |
add a comment |
$begingroup$
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