Find a possible neutral such that $1g = g$ and $g1 neq g$, or $g1 = g$ and $1g neq g$
$begingroup$
I'm thinking about some simple counterexamples on Group Theory.
"Choice a set $G$ and find a possible $1 in G$ such that $1g neq g1$ for some $g in G$"
Basically I want to find a possible neutral such that $1g = g$ and $g1 neq g$, or $g1 = g$ and $1g neq g$ for some $g$.
Is it possible?
Edit. I think I did not write correctly. I'll try to be clearer.
In the definition of a group we have the axiom:
"There is a $1 in G$ such that $1g = g = g1$ for all $g in G$."
Why is necessary to verify $1g = g$ and $g1 = g$? I think that the reason is: in a non-group, we can find a element such that $1g = g$ but $g1 neq g$ or $g1 = g$ and $1g neq g$.
abstract-algebra group-theory
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
$endgroup$
add a comment |
$begingroup$
I'm thinking about some simple counterexamples on Group Theory.
"Choice a set $G$ and find a possible $1 in G$ such that $1g neq g1$ for some $g in G$"
Basically I want to find a possible neutral such that $1g = g$ and $g1 neq g$, or $g1 = g$ and $1g neq g$ for some $g$.
Is it possible?
Edit. I think I did not write correctly. I'll try to be clearer.
In the definition of a group we have the axiom:
"There is a $1 in G$ such that $1g = g = g1$ for all $g in G$."
Why is necessary to verify $1g = g$ and $g1 = g$? I think that the reason is: in a non-group, we can find a element such that $1g = g$ but $g1 neq g$ or $g1 = g$ and $1g neq g$.
abstract-algebra group-theory
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
$endgroup$
$begingroup$
If you did find one then that would violate the Axiom of Identity.
$endgroup$
– TheSimpliFire
Dec 1 '18 at 19:31
1
$begingroup$
If you have $1g=gneq g1$, then what you have is not a group. It's very possible in other algebraic structures, though. For instance, the integers where $ab$ is defined as $b-a$, we have $0b=bneq b0$.
$endgroup$
– Arthur
Dec 1 '18 at 19:38
$begingroup$
@Arthur exactly! Is an example of this that I'm trying to find.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:41
add a comment |
$begingroup$
I'm thinking about some simple counterexamples on Group Theory.
"Choice a set $G$ and find a possible $1 in G$ such that $1g neq g1$ for some $g in G$"
Basically I want to find a possible neutral such that $1g = g$ and $g1 neq g$, or $g1 = g$ and $1g neq g$ for some $g$.
Is it possible?
Edit. I think I did not write correctly. I'll try to be clearer.
In the definition of a group we have the axiom:
"There is a $1 in G$ such that $1g = g = g1$ for all $g in G$."
Why is necessary to verify $1g = g$ and $g1 = g$? I think that the reason is: in a non-group, we can find a element such that $1g = g$ but $g1 neq g$ or $g1 = g$ and $1g neq g$.
abstract-algebra group-theory
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
$endgroup$
I'm thinking about some simple counterexamples on Group Theory.
"Choice a set $G$ and find a possible $1 in G$ such that $1g neq g1$ for some $g in G$"
Basically I want to find a possible neutral such that $1g = g$ and $g1 neq g$, or $g1 = g$ and $1g neq g$ for some $g$.
Is it possible?
Edit. I think I did not write correctly. I'll try to be clearer.
In the definition of a group we have the axiom:
"There is a $1 in G$ such that $1g = g = g1$ for all $g in G$."
Why is necessary to verify $1g = g$ and $g1 = g$? I think that the reason is: in a non-group, we can find a element such that $1g = g$ but $g1 neq g$ or $g1 = g$ and $1g neq g$.
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
edited Dec 1 '18 at 19:39
Lucas Corrêa
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
asked Dec 1 '18 at 19:24
Lucas CorrêaLucas Corrêa
1,5581321
1,5581321
$begingroup$
If you did find one then that would violate the Axiom of Identity.
$endgroup$
– TheSimpliFire
Dec 1 '18 at 19:31
1
$begingroup$
If you have $1g=gneq g1$, then what you have is not a group. It's very possible in other algebraic structures, though. For instance, the integers where $ab$ is defined as $b-a$, we have $0b=bneq b0$.
$endgroup$
– Arthur
Dec 1 '18 at 19:38
$begingroup$
@Arthur exactly! Is an example of this that I'm trying to find.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:41
add a comment |
$begingroup$
If you did find one then that would violate the Axiom of Identity.
$endgroup$
– TheSimpliFire
Dec 1 '18 at 19:31
1
$begingroup$
If you have $1g=gneq g1$, then what you have is not a group. It's very possible in other algebraic structures, though. For instance, the integers where $ab$ is defined as $b-a$, we have $0b=bneq b0$.
$endgroup$
– Arthur
Dec 1 '18 at 19:38
$begingroup$
@Arthur exactly! Is an example of this that I'm trying to find.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:41
$begingroup$
If you did find one then that would violate the Axiom of Identity.
$endgroup$
– TheSimpliFire
Dec 1 '18 at 19:31
$begingroup$
If you did find one then that would violate the Axiom of Identity.
$endgroup$
– TheSimpliFire
Dec 1 '18 at 19:31
1
1
$begingroup$
If you have $1g=gneq g1$, then what you have is not a group. It's very possible in other algebraic structures, though. For instance, the integers where $ab$ is defined as $b-a$, we have $0b=bneq b0$.
$endgroup$
– Arthur
Dec 1 '18 at 19:38
$begingroup$
If you have $1g=gneq g1$, then what you have is not a group. It's very possible in other algebraic structures, though. For instance, the integers where $ab$ is defined as $b-a$, we have $0b=bneq b0$.
$endgroup$
– Arthur
Dec 1 '18 at 19:38
$begingroup$
@Arthur exactly! Is an example of this that I'm trying to find.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:41
$begingroup$
@Arthur exactly! Is an example of this that I'm trying to find.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:41
add a comment |
1 Answer
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$begingroup$
No. In any group $G$ with identity $1$, $g1 = 1g = g$ for any $g in G$.
Suppose $1g = g$. Then $gg = g(1g) = (g1)g$. Right-multiplying both sides by $g^{-1}$ leaves $g = g1$.
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
$endgroup$
$begingroup$
Yes, I dont know if I wrote correctly, but I mean a possible neutral $1$ (sure, that $1$ is not really a neutral). Its just for justify the definition $g1 = g = 1g$ and not just $1g = g$ or $g1 = g$
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:31
$begingroup$
I'm not quite sure what you are asking for. There is only one identity (I think this is what you mean by neutral, so correct me if this is wrong) element in any group.
$endgroup$
– platty
Dec 1 '18 at 19:33
$begingroup$
I've edited. I tried to be clearer.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:40
1
$begingroup$
Okay. As noted in my answer, if such a set is closed under the associative operation of multiplication and inverses, we always have $1g = g implies g = 1g$. So really, if the other group axioms hold, then we only need that there exists a $1 in G$ such that $1g = g$ for all $g in G$.
$endgroup$
– platty
Dec 1 '18 at 19:43
add a comment |
Your Answer
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1 Answer
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$begingroup$
No. In any group $G$ with identity $1$, $g1 = 1g = g$ for any $g in G$.
Suppose $1g = g$. Then $gg = g(1g) = (g1)g$. Right-multiplying both sides by $g^{-1}$ leaves $g = g1$.
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
$endgroup$
$begingroup$
Yes, I dont know if I wrote correctly, but I mean a possible neutral $1$ (sure, that $1$ is not really a neutral). Its just for justify the definition $g1 = g = 1g$ and not just $1g = g$ or $g1 = g$
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:31
$begingroup$
I'm not quite sure what you are asking for. There is only one identity (I think this is what you mean by neutral, so correct me if this is wrong) element in any group.
$endgroup$
– platty
Dec 1 '18 at 19:33
$begingroup$
I've edited. I tried to be clearer.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:40
1
$begingroup$
Okay. As noted in my answer, if such a set is closed under the associative operation of multiplication and inverses, we always have $1g = g implies g = 1g$. So really, if the other group axioms hold, then we only need that there exists a $1 in G$ such that $1g = g$ for all $g in G$.
$endgroup$
– platty
Dec 1 '18 at 19:43
add a comment |
$begingroup$
No. In any group $G$ with identity $1$, $g1 = 1g = g$ for any $g in G$.
Suppose $1g = g$. Then $gg = g(1g) = (g1)g$. Right-multiplying both sides by $g^{-1}$ leaves $g = g1$.
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
$endgroup$
$begingroup$
Yes, I dont know if I wrote correctly, but I mean a possible neutral $1$ (sure, that $1$ is not really a neutral). Its just for justify the definition $g1 = g = 1g$ and not just $1g = g$ or $g1 = g$
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:31
$begingroup$
I'm not quite sure what you are asking for. There is only one identity (I think this is what you mean by neutral, so correct me if this is wrong) element in any group.
$endgroup$
– platty
Dec 1 '18 at 19:33
$begingroup$
I've edited. I tried to be clearer.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:40
1
$begingroup$
Okay. As noted in my answer, if such a set is closed under the associative operation of multiplication and inverses, we always have $1g = g implies g = 1g$. So really, if the other group axioms hold, then we only need that there exists a $1 in G$ such that $1g = g$ for all $g in G$.
$endgroup$
– platty
Dec 1 '18 at 19:43
add a comment |
$begingroup$
No. In any group $G$ with identity $1$, $g1 = 1g = g$ for any $g in G$.
Suppose $1g = g$. Then $gg = g(1g) = (g1)g$. Right-multiplying both sides by $g^{-1}$ leaves $g = g1$.
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
$endgroup$
No. In any group $G$ with identity $1$, $g1 = 1g = g$ for any $g in G$.
Suppose $1g = g$. Then $gg = g(1g) = (g1)g$. Right-multiplying both sides by $g^{-1}$ leaves $g = g1$.
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
answered Dec 1 '18 at 19:27
plattyplatty
3,370320
3,370320
$begingroup$
Yes, I dont know if I wrote correctly, but I mean a possible neutral $1$ (sure, that $1$ is not really a neutral). Its just for justify the definition $g1 = g = 1g$ and not just $1g = g$ or $g1 = g$
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:31
$begingroup$
I'm not quite sure what you are asking for. There is only one identity (I think this is what you mean by neutral, so correct me if this is wrong) element in any group.
$endgroup$
– platty
Dec 1 '18 at 19:33
$begingroup$
I've edited. I tried to be clearer.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:40
1
$begingroup$
Okay. As noted in my answer, if such a set is closed under the associative operation of multiplication and inverses, we always have $1g = g implies g = 1g$. So really, if the other group axioms hold, then we only need that there exists a $1 in G$ such that $1g = g$ for all $g in G$.
$endgroup$
– platty
Dec 1 '18 at 19:43
add a comment |
$begingroup$
Yes, I dont know if I wrote correctly, but I mean a possible neutral $1$ (sure, that $1$ is not really a neutral). Its just for justify the definition $g1 = g = 1g$ and not just $1g = g$ or $g1 = g$
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:31
$begingroup$
I'm not quite sure what you are asking for. There is only one identity (I think this is what you mean by neutral, so correct me if this is wrong) element in any group.
$endgroup$
– platty
Dec 1 '18 at 19:33
$begingroup$
I've edited. I tried to be clearer.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:40
1
$begingroup$
Okay. As noted in my answer, if such a set is closed under the associative operation of multiplication and inverses, we always have $1g = g implies g = 1g$. So really, if the other group axioms hold, then we only need that there exists a $1 in G$ such that $1g = g$ for all $g in G$.
$endgroup$
– platty
Dec 1 '18 at 19:43
$begingroup$
Yes, I dont know if I wrote correctly, but I mean a possible neutral $1$ (sure, that $1$ is not really a neutral). Its just for justify the definition $g1 = g = 1g$ and not just $1g = g$ or $g1 = g$
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:31
$begingroup$
Yes, I dont know if I wrote correctly, but I mean a possible neutral $1$ (sure, that $1$ is not really a neutral). Its just for justify the definition $g1 = g = 1g$ and not just $1g = g$ or $g1 = g$
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:31
$begingroup$
I'm not quite sure what you are asking for. There is only one identity (I think this is what you mean by neutral, so correct me if this is wrong) element in any group.
$endgroup$
– platty
Dec 1 '18 at 19:33
$begingroup$
I'm not quite sure what you are asking for. There is only one identity (I think this is what you mean by neutral, so correct me if this is wrong) element in any group.
$endgroup$
– platty
Dec 1 '18 at 19:33
$begingroup$
I've edited. I tried to be clearer.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:40
$begingroup$
I've edited. I tried to be clearer.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:40
1
1
$begingroup$
Okay. As noted in my answer, if such a set is closed under the associative operation of multiplication and inverses, we always have $1g = g implies g = 1g$. So really, if the other group axioms hold, then we only need that there exists a $1 in G$ such that $1g = g$ for all $g in G$.
$endgroup$
– platty
Dec 1 '18 at 19:43
$begingroup$
Okay. As noted in my answer, if such a set is closed under the associative operation of multiplication and inverses, we always have $1g = g implies g = 1g$. So really, if the other group axioms hold, then we only need that there exists a $1 in G$ such that $1g = g$ for all $g in G$.
$endgroup$
– platty
Dec 1 '18 at 19:43
add a comment |
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$begingroup$
If you did find one then that would violate the Axiom of Identity.
$endgroup$
– TheSimpliFire
Dec 1 '18 at 19:31
1
$begingroup$
If you have $1g=gneq g1$, then what you have is not a group. It's very possible in other algebraic structures, though. For instance, the integers where $ab$ is defined as $b-a$, we have $0b=bneq b0$.
$endgroup$
– Arthur
Dec 1 '18 at 19:38
$begingroup$
@Arthur exactly! Is an example of this that I'm trying to find.
$endgroup$
– Lucas Corrêa
Dec 1 '18 at 19:41