Mixing
Herstein's Topics in Algebra Section 2.3 Problems 12 and 13
$begingroup$
So, these are the questions:
Let $G$ be a nonempty set closed under associative product, which in addition satisfies:
a) There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b) Give $ain G$, there exists an element $y(a)in G$ such that $a.y(a) = e$.
Prove that G is a group.
I thought that this question is stupid since that's exactly the definition of a group. However I was stumped after reading the problem 13. The question 13 goes like this:
Prove, by an example, that the conclusion of problem 12 is false if we assume instead:
a') There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b') Give $ain G$, there exists an element $y(a)in G$ such that $y(a).a = e$.
After realising that I'll need to prove that $a.y(a)=y(a).a=e$ in question 12, I thought:
In 12 b, $a.y(a) = e Rightarrow (y(a).a).(y(a).a) = (y(a).e).a = y(a).a$. So using 12 a $(a.e=a)$, we have y(a).a = e.
This part is certainly false since same method proves that 13 is also a group, and also it could be that $e$ may not be unique in $G$ so I cannot assert that $y(a).a=a$ (Am I correct about this?).
So, my question is what do I need to show here to prove that G is a group. Not to ask too much, But I couldn't find any example for ques 13 either. I think that I'll need to find a closed set where $a.e=a$ does not necessarily imply $e.a=a$. But I'm hopeless.
group-theory semigroups
edited Jan 3 at 3:37
Shaun
8,888113681
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
$endgroup$
add a comment |
$begingroup$
So, these are the questions:
Let $G$ be a nonempty set closed under associative product, which in addition satisfies:
a) There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b) Give $ain G$, there exists an element $y(a)in G$ such that $a.y(a) = e$.
Prove that G is a group.
I thought that this question is stupid since that's exactly the definition of a group. However I was stumped after reading the problem 13. The question 13 goes like this:
Prove, by an example, that the conclusion of problem 12 is false if we assume instead:
a') There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b') Give $ain G$, there exists an element $y(a)in G$ such that $y(a).a = e$.
After realising that I'll need to prove that $a.y(a)=y(a).a=e$ in question 12, I thought:
In 12 b, $a.y(a) = e Rightarrow (y(a).a).(y(a).a) = (y(a).e).a = y(a).a$. So using 12 a $(a.e=a)$, we have y(a).a = e.
This part is certainly false since same method proves that 13 is also a group, and also it could be that $e$ may not be unique in $G$ so I cannot assert that $y(a).a=a$ (Am I correct about this?).
So, my question is what do I need to show here to prove that G is a group. Not to ask too much, But I couldn't find any example for ques 13 either. I think that I'll need to find a closed set where $a.e=a$ does not necessarily imply $e.a=a$. But I'm hopeless.
group-theory semigroups
edited Jan 3 at 3:37
Shaun
8,888113681
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
$endgroup$
5
$begingroup$
12. isn't the definition of a group. The definition requires both a "2-sided" identity and "2-sided" inverses. These questions say that one can get away with 1-sided identity and inverses, provided the sides in question match up suitably.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 3:09
$begingroup$
See math.stackexchange.com/q/2532072/444015
$endgroup$
– Lucas Corrêa
Jan 3 at 3:15
$begingroup$
Oh, sorry for the duplicate.
$endgroup$
– Nutan Nepal
Jan 3 at 3:16
$begingroup$
See math.stackexchange.com/q/2070106/444015. In this link you will find a good discussion about this.
$endgroup$
– Lucas Corrêa
Jan 3 at 3:32
add a comment |
$begingroup$
So, these are the questions:
Let $G$ be a nonempty set closed under associative product, which in addition satisfies:
a) There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b) Give $ain G$, there exists an element $y(a)in G$ such that $a.y(a) = e$.
Prove that G is a group.
I thought that this question is stupid since that's exactly the definition of a group. However I was stumped after reading the problem 13. The question 13 goes like this:
Prove, by an example, that the conclusion of problem 12 is false if we assume instead:
a') There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b') Give $ain G$, there exists an element $y(a)in G$ such that $y(a).a = e$.
After realising that I'll need to prove that $a.y(a)=y(a).a=e$ in question 12, I thought:
In 12 b, $a.y(a) = e Rightarrow (y(a).a).(y(a).a) = (y(a).e).a = y(a).a$. So using 12 a $(a.e=a)$, we have y(a).a = e.
This part is certainly false since same method proves that 13 is also a group, and also it could be that $e$ may not be unique in $G$ so I cannot assert that $y(a).a=a$ (Am I correct about this?).
So, my question is what do I need to show here to prove that G is a group. Not to ask too much, But I couldn't find any example for ques 13 either. I think that I'll need to find a closed set where $a.e=a$ does not necessarily imply $e.a=a$. But I'm hopeless.
group-theory semigroups
edited Jan 3 at 3:37
Shaun
8,888113681
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
$endgroup$
So, these are the questions:
Let $G$ be a nonempty set closed under associative product, which in addition satisfies:
a) There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b) Give $ain G$, there exists an element $y(a)in G$ such that $a.y(a) = e$.
Prove that G is a group.
I thought that this question is stupid since that's exactly the definition of a group. However I was stumped after reading the problem 13. The question 13 goes like this:
Prove, by an example, that the conclusion of problem 12 is false if we assume instead:
a') There exists an $e in G$ such that $a.e=a$ for all $a in G.$
b') Give $ain G$, there exists an element $y(a)in G$ such that $y(a).a = e$.
After realising that I'll need to prove that $a.y(a)=y(a).a=e$ in question 12, I thought:
In 12 b, $a.y(a) = e Rightarrow (y(a).a).(y(a).a) = (y(a).e).a = y(a).a$. So using 12 a $(a.e=a)$, we have y(a).a = e.
This part is certainly false since same method proves that 13 is also a group, and also it could be that $e$ may not be unique in $G$ so I cannot assert that $y(a).a=a$ (Am I correct about this?).
So, my question is what do I need to show here to prove that G is a group. Not to ask too much, But I couldn't find any example for ques 13 either. I think that I'll need to find a closed set where $a.e=a$ does not necessarily imply $e.a=a$. But I'm hopeless.
group-theory semigroups
group-theory semigroups
edited Jan 3 at 3:37
Shaun
8,888113681
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
edited Jan 3 at 3:37
Shaun
8,888113681
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
edited Jan 3 at 3:37
Shaun
8,888113681
edited Jan 3 at 3:37
Shaun
8,888113681
edited Jan 3 at 3:37
Shaun
8,888113681
8,888113681
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
asked Jan 3 at 3:06
Nutan NepalNutan Nepal
455
455
5
$begingroup$
12. isn't the definition of a group. The definition requires both a "2-sided" identity and "2-sided" inverses. These questions say that one can get away with 1-sided identity and inverses, provided the sides in question match up suitably.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 3:09
$begingroup$
See math.stackexchange.com/q/2532072/444015
$endgroup$
– Lucas Corrêa
Jan 3 at 3:15
$begingroup$
Oh, sorry for the duplicate.
$endgroup$
– Nutan Nepal
Jan 3 at 3:16
$begingroup$
See math.stackexchange.com/q/2070106/444015. In this link you will find a good discussion about this.
$endgroup$
– Lucas Corrêa
Jan 3 at 3:32
add a comment |
5
$begingroup$
12. isn't the definition of a group. The definition requires both a "2-sided" identity and "2-sided" inverses. These questions say that one can get away with 1-sided identity and inverses, provided the sides in question match up suitably.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 3:09
$begingroup$
See math.stackexchange.com/q/2532072/444015
$endgroup$
– Lucas Corrêa
Jan 3 at 3:15
$begingroup$
Oh, sorry for the duplicate.
$endgroup$
– Nutan Nepal
Jan 3 at 3:16
$begingroup$
See math.stackexchange.com/q/2070106/444015. In this link you will find a good discussion about this.
$endgroup$
– Lucas Corrêa
Jan 3 at 3:32
5
5
$begingroup$
12. isn't the definition of a group. The definition requires both a "2-sided" identity and "2-sided" inverses. These questions say that one can get away with 1-sided identity and inverses, provided the sides in question match up suitably.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 3:09
$begingroup$
12. isn't the definition of a group. The definition requires both a "2-sided" identity and "2-sided" inverses. These questions say that one can get away with 1-sided identity and inverses, provided the sides in question match up suitably.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 3:09
$begingroup$
See math.stackexchange.com/q/2532072/444015
$endgroup$
– Lucas Corrêa
Jan 3 at 3:15
$begingroup$
See math.stackexchange.com/q/2532072/444015
$endgroup$
– Lucas Corrêa
Jan 3 at 3:15
$begingroup$
Oh, sorry for the duplicate.
$endgroup$
– Nutan Nepal
Jan 3 at 3:16
$begingroup$
Oh, sorry for the duplicate.
$endgroup$
– Nutan Nepal
Jan 3 at 3:16
$begingroup$
See math.stackexchange.com/q/2070106/444015. In this link you will find a good discussion about this.
$endgroup$
– Lucas Corrêa
Jan 3 at 3:32
$begingroup$
See math.stackexchange.com/q/2070106/444015. In this link you will find a good discussion about this.
$endgroup$
– Lucas Corrêa
Jan 3 at 3:32
add a comment |
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$begingroup$
12. isn't the definition of a group. The definition requires both a "2-sided" identity and "2-sided" inverses. These questions say that one can get away with 1-sided identity and inverses, provided the sides in question match up suitably.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 3:09
$begingroup$
See math.stackexchange.com/q/2532072/444015
$endgroup$
– Lucas Corrêa
Jan 3 at 3:15
$begingroup$
Oh, sorry for the duplicate.
$endgroup$
– Nutan Nepal
Jan 3 at 3:16
$begingroup$
See math.stackexchange.com/q/2070106/444015. In this link you will find a good discussion about this.
$endgroup$
– Lucas Corrêa
Jan 3 at 3:32