The identity element and the inverse [on hold]
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1
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A group is a ordered pair $(G,*)$ which satisfies the following,
$G$ is a set
$*$ is a binary operation on $G$
$*$ is associative
$exists einmathbb{G} (forall ainmathbb{G} a*e=e*a=a)$
$forall ainmathbb{G}exists binmathbb{G}(a*b=b*a=e)$
Note : $e$ is called THE identity element of this group $(G,*)$.
Note : $b$ is called THE inverse of a and denoted by $a^{-1}$.
Prove "THE", in other words, show identity is unique and inverse is unique.
I need a little bit help maybe and idea on how to start such problem.
Thank you
group-theory
put on hold as off-topic by T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh Nov 27 at 5:43
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
1
down vote
favorite
A group is a ordered pair $(G,*)$ which satisfies the following,
$G$ is a set
$*$ is a binary operation on $G$
$*$ is associative
$exists einmathbb{G} (forall ainmathbb{G} a*e=e*a=a)$
$forall ainmathbb{G}exists binmathbb{G}(a*b=b*a=e)$
Note : $e$ is called THE identity element of this group $(G,*)$.
Note : $b$ is called THE inverse of a and denoted by $a^{-1}$.
Prove "THE", in other words, show identity is unique and inverse is unique.
I need a little bit help maybe and idea on how to start such problem.
Thank you
group-theory
put on hold as off-topic by T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh Nov 27 at 5:43
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
It's a bit disappointing that this got onto the hot network questions list... after all, this is a "I have no idea post" that's answered by many elementary books in the field (e.g. Pinter, Dummit & Foote, and innumerable other textbooks...). In general, this is not a do-my-homework site and it's expected to show your own thoughts and the context where you encountered the problem. I'm voting to close as off-topic. But that really doesn't matter much, considering that five enthusiastic users have given you the exact same answers.
– T. Bongers
Nov 26 at 22:30
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A group is a ordered pair $(G,*)$ which satisfies the following,
$G$ is a set
$*$ is a binary operation on $G$
$*$ is associative
$exists einmathbb{G} (forall ainmathbb{G} a*e=e*a=a)$
$forall ainmathbb{G}exists binmathbb{G}(a*b=b*a=e)$
Note : $e$ is called THE identity element of this group $(G,*)$.
Note : $b$ is called THE inverse of a and denoted by $a^{-1}$.
Prove "THE", in other words, show identity is unique and inverse is unique.
I need a little bit help maybe and idea on how to start such problem.
Thank you
group-theory
A group is a ordered pair $(G,*)$ which satisfies the following,
$G$ is a set
$*$ is a binary operation on $G$
$*$ is associative
$exists einmathbb{G} (forall ainmathbb{G} a*e=e*a=a)$
$forall ainmathbb{G}exists binmathbb{G}(a*b=b*a=e)$
Note : $e$ is called THE identity element of this group $(G,*)$.
Note : $b$ is called THE inverse of a and denoted by $a^{-1}$.
Prove "THE", in other words, show identity is unique and inverse is unique.
I need a little bit help maybe and idea on how to start such problem.
Thank you
group-theory
group-theory
edited Nov 26 at 17:12
asked Nov 26 at 16:56
Scott
346
346
put on hold as off-topic by T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh Nov 27 at 5:43
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh Nov 27 at 5:43
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
It's a bit disappointing that this got onto the hot network questions list... after all, this is a "I have no idea post" that's answered by many elementary books in the field (e.g. Pinter, Dummit & Foote, and innumerable other textbooks...). In general, this is not a do-my-homework site and it's expected to show your own thoughts and the context where you encountered the problem. I'm voting to close as off-topic. But that really doesn't matter much, considering that five enthusiastic users have given you the exact same answers.
– T. Bongers
Nov 26 at 22:30
add a comment |
It's a bit disappointing that this got onto the hot network questions list... after all, this is a "I have no idea post" that's answered by many elementary books in the field (e.g. Pinter, Dummit & Foote, and innumerable other textbooks...). In general, this is not a do-my-homework site and it's expected to show your own thoughts and the context where you encountered the problem. I'm voting to close as off-topic. But that really doesn't matter much, considering that five enthusiastic users have given you the exact same answers.
– T. Bongers
Nov 26 at 22:30
It's a bit disappointing that this got onto the hot network questions list... after all, this is a "I have no idea post" that's answered by many elementary books in the field (e.g. Pinter, Dummit & Foote, and innumerable other textbooks...). In general, this is not a do-my-homework site and it's expected to show your own thoughts and the context where you encountered the problem. I'm voting to close as off-topic. But that really doesn't matter much, considering that five enthusiastic users have given you the exact same answers.
– T. Bongers
Nov 26 at 22:30
It's a bit disappointing that this got onto the hot network questions list... after all, this is a "I have no idea post" that's answered by many elementary books in the field (e.g. Pinter, Dummit & Foote, and innumerable other textbooks...). In general, this is not a do-my-homework site and it's expected to show your own thoughts and the context where you encountered the problem. I'm voting to close as off-topic. But that really doesn't matter much, considering that five enthusiastic users have given you the exact same answers.
– T. Bongers
Nov 26 at 22:30
add a comment |
5 Answers
5
active
oldest
votes
up vote
5
down vote
Let $e$ and $f$ be two elements where $ea = ae =a$ and $fa = af = a$ for all $a in G$.
Then what is $ef$ equal to? It must be equal to something in $G$ as that is that is the definition of binary operation. And because $ea =ae =a$ for all $ain G$ it's true for $f$ and $ef = fe = f$. But $fa = af =a$ for all $a in G$ so $ef = fe = e$. Is it possible for $e$ and $f$ to be two different elements?
As for inverses: For $a$ let $a'$ be such that $a'a = aa' = e$ and let $a^{-1}$ be such that $a^{-1}a = a a^{-1} = e$.
What is $a^{-1}aa'$ equal to? Hint: use associativity. $(a^{-1}a)a' = ??$ and $a^{-1}(aa') = ???$.
1
Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$?
– David Z
Nov 26 at 19:57
Yes, it was. (Same thing though :) )
– fleablood
Nov 26 at 21:35
add a comment |
up vote
3
down vote
Here's a little help; in both cases consider two such elements (two identity elements, or two inverses of the same element). Use the given relations to show that they are the same.
add a comment |
up vote
2
down vote
Hint. Suppose $a$ and $b$ are both identities. What can you say about $ab$?
If $g$ seems to have two inverses can you find a similar kind of product that will let you conclude they are equal?
Since a and b are both identities then their multiplication $ a ast b$ should be equal to either a or b or ab ?
– Scott
Nov 26 at 17:16
2
@Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal.
– Ethan Bolker
Nov 26 at 17:20
1
@Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$?
– Ethan Bolker
Nov 26 at 17:41
2
@scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now.
– Ethan Bolker
Nov 26 at 18:52
1
1) $e, g^{-1},$ and $frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $alpha$ and $beta$ both act like an identity that $alpha = beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $alpha$ and $beta$ deserve it. You need to prove that only one can have the label i.e $alpha = beta$.
– fleablood
Nov 26 at 21:48
|
show 4 more comments
up vote
1
down vote
Let $e,e'$ be identity elements of $G$. Then $e=e'e$, since $e'$ is an identity element, and $e'e=e'$, since $e$ is an identity element.
Let $u,v$ be inverses of $a$. Then $u= u e = u(av) = (ua)v = ev =v$, where $e$ is the identity element.
add a comment |
up vote
0
down vote
Suppose there exist two identity elements e_1 and e_2
then
e_1*e_2=e_2*e_1=e_2 (since e_1 is an identity).
e_1*e_2=e_2*e_1=e_1 (since e_2 is an identity).
So e_1=e_2.
If y_1 is an inverse of x and y_2 is another one then :
y_1=y_1*e=y_1*(xy_2)=(y_1 x) * y_2=e* y_2=y_2
New contributor
4
Please use MathJax to format mathematical expressions.
– Servaes
Nov 26 at 17:03
4
Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more.
– Ethan Bolker
Nov 26 at 17:14
Thank you for the answer, i will consider this after i learn how can i conclude to this answer.
– Scott
Nov 26 at 17:19
I get the first part and almost understand the second part. Thanks for the help.
– Scott
Nov 26 at 17:46
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Let $e$ and $f$ be two elements where $ea = ae =a$ and $fa = af = a$ for all $a in G$.
Then what is $ef$ equal to? It must be equal to something in $G$ as that is that is the definition of binary operation. And because $ea =ae =a$ for all $ain G$ it's true for $f$ and $ef = fe = f$. But $fa = af =a$ for all $a in G$ so $ef = fe = e$. Is it possible for $e$ and $f$ to be two different elements?
As for inverses: For $a$ let $a'$ be such that $a'a = aa' = e$ and let $a^{-1}$ be such that $a^{-1}a = a a^{-1} = e$.
What is $a^{-1}aa'$ equal to? Hint: use associativity. $(a^{-1}a)a' = ??$ and $a^{-1}(aa') = ???$.
1
Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$?
– David Z
Nov 26 at 19:57
Yes, it was. (Same thing though :) )
– fleablood
Nov 26 at 21:35
add a comment |
up vote
5
down vote
Let $e$ and $f$ be two elements where $ea = ae =a$ and $fa = af = a$ for all $a in G$.
Then what is $ef$ equal to? It must be equal to something in $G$ as that is that is the definition of binary operation. And because $ea =ae =a$ for all $ain G$ it's true for $f$ and $ef = fe = f$. But $fa = af =a$ for all $a in G$ so $ef = fe = e$. Is it possible for $e$ and $f$ to be two different elements?
As for inverses: For $a$ let $a'$ be such that $a'a = aa' = e$ and let $a^{-1}$ be such that $a^{-1}a = a a^{-1} = e$.
What is $a^{-1}aa'$ equal to? Hint: use associativity. $(a^{-1}a)a' = ??$ and $a^{-1}(aa') = ???$.
1
Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$?
– David Z
Nov 26 at 19:57
Yes, it was. (Same thing though :) )
– fleablood
Nov 26 at 21:35
add a comment |
up vote
5
down vote
up vote
5
down vote
Let $e$ and $f$ be two elements where $ea = ae =a$ and $fa = af = a$ for all $a in G$.
Then what is $ef$ equal to? It must be equal to something in $G$ as that is that is the definition of binary operation. And because $ea =ae =a$ for all $ain G$ it's true for $f$ and $ef = fe = f$. But $fa = af =a$ for all $a in G$ so $ef = fe = e$. Is it possible for $e$ and $f$ to be two different elements?
As for inverses: For $a$ let $a'$ be such that $a'a = aa' = e$ and let $a^{-1}$ be such that $a^{-1}a = a a^{-1} = e$.
What is $a^{-1}aa'$ equal to? Hint: use associativity. $(a^{-1}a)a' = ??$ and $a^{-1}(aa') = ???$.
Let $e$ and $f$ be two elements where $ea = ae =a$ and $fa = af = a$ for all $a in G$.
Then what is $ef$ equal to? It must be equal to something in $G$ as that is that is the definition of binary operation. And because $ea =ae =a$ for all $ain G$ it's true for $f$ and $ef = fe = f$. But $fa = af =a$ for all $a in G$ so $ef = fe = e$. Is it possible for $e$ and $f$ to be two different elements?
As for inverses: For $a$ let $a'$ be such that $a'a = aa' = e$ and let $a^{-1}$ be such that $a^{-1}a = a a^{-1} = e$.
What is $a^{-1}aa'$ equal to? Hint: use associativity. $(a^{-1}a)a' = ??$ and $a^{-1}(aa') = ???$.
edited Nov 26 at 21:35
answered Nov 26 at 17:07
fleablood
66.4k22684
66.4k22684
1
Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$?
– David Z
Nov 26 at 19:57
Yes, it was. (Same thing though :) )
– fleablood
Nov 26 at 21:35
add a comment |
1
Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$?
– David Z
Nov 26 at 19:57
Yes, it was. (Same thing though :) )
– fleablood
Nov 26 at 21:35
1
1
Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$?
– David Z
Nov 26 at 19:57
Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$?
– David Z
Nov 26 at 19:57
Yes, it was. (Same thing though :) )
– fleablood
Nov 26 at 21:35
Yes, it was. (Same thing though :) )
– fleablood
Nov 26 at 21:35
add a comment |
up vote
3
down vote
Here's a little help; in both cases consider two such elements (two identity elements, or two inverses of the same element). Use the given relations to show that they are the same.
add a comment |
up vote
3
down vote
Here's a little help; in both cases consider two such elements (two identity elements, or two inverses of the same element). Use the given relations to show that they are the same.
add a comment |
up vote
3
down vote
up vote
3
down vote
Here's a little help; in both cases consider two such elements (two identity elements, or two inverses of the same element). Use the given relations to show that they are the same.
Here's a little help; in both cases consider two such elements (two identity elements, or two inverses of the same element). Use the given relations to show that they are the same.
answered Nov 26 at 16:58
Servaes
21.1k33792
21.1k33792
add a comment |
add a comment |
up vote
2
down vote
Hint. Suppose $a$ and $b$ are both identities. What can you say about $ab$?
If $g$ seems to have two inverses can you find a similar kind of product that will let you conclude they are equal?
Since a and b are both identities then their multiplication $ a ast b$ should be equal to either a or b or ab ?
– Scott
Nov 26 at 17:16
2
@Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal.
– Ethan Bolker
Nov 26 at 17:20
1
@Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$?
– Ethan Bolker
Nov 26 at 17:41
2
@scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now.
– Ethan Bolker
Nov 26 at 18:52
1
1) $e, g^{-1},$ and $frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $alpha$ and $beta$ both act like an identity that $alpha = beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $alpha$ and $beta$ deserve it. You need to prove that only one can have the label i.e $alpha = beta$.
– fleablood
Nov 26 at 21:48
|
show 4 more comments
up vote
2
down vote
Hint. Suppose $a$ and $b$ are both identities. What can you say about $ab$?
If $g$ seems to have two inverses can you find a similar kind of product that will let you conclude they are equal?
Since a and b are both identities then their multiplication $ a ast b$ should be equal to either a or b or ab ?
– Scott
Nov 26 at 17:16
2
@Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal.
– Ethan Bolker
Nov 26 at 17:20
1
@Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$?
– Ethan Bolker
Nov 26 at 17:41
2
@scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now.
– Ethan Bolker
Nov 26 at 18:52
1
1) $e, g^{-1},$ and $frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $alpha$ and $beta$ both act like an identity that $alpha = beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $alpha$ and $beta$ deserve it. You need to prove that only one can have the label i.e $alpha = beta$.
– fleablood
Nov 26 at 21:48
|
show 4 more comments
up vote
2
down vote
up vote
2
down vote
Hint. Suppose $a$ and $b$ are both identities. What can you say about $ab$?
If $g$ seems to have two inverses can you find a similar kind of product that will let you conclude they are equal?
Hint. Suppose $a$ and $b$ are both identities. What can you say about $ab$?
If $g$ seems to have two inverses can you find a similar kind of product that will let you conclude they are equal?
answered Nov 26 at 16:57
Ethan Bolker
39.7k543103
39.7k543103
Since a and b are both identities then their multiplication $ a ast b$ should be equal to either a or b or ab ?
– Scott
Nov 26 at 17:16
2
@Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal.
– Ethan Bolker
Nov 26 at 17:20
1
@Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$?
– Ethan Bolker
Nov 26 at 17:41
2
@scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now.
– Ethan Bolker
Nov 26 at 18:52
1
1) $e, g^{-1},$ and $frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $alpha$ and $beta$ both act like an identity that $alpha = beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $alpha$ and $beta$ deserve it. You need to prove that only one can have the label i.e $alpha = beta$.
– fleablood
Nov 26 at 21:48
|
show 4 more comments
Since a and b are both identities then their multiplication $ a ast b$ should be equal to either a or b or ab ?
– Scott
Nov 26 at 17:16
2
@Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal.
– Ethan Bolker
Nov 26 at 17:20
1
@Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$?
– Ethan Bolker
Nov 26 at 17:41
2
@scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now.
– Ethan Bolker
Nov 26 at 18:52
1
1) $e, g^{-1},$ and $frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $alpha$ and $beta$ both act like an identity that $alpha = beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $alpha$ and $beta$ deserve it. You need to prove that only one can have the label i.e $alpha = beta$.
– fleablood
Nov 26 at 21:48
Since a and b are both identities then their multiplication $ a ast b$ should be equal to either a or b or ab ?
– Scott
Nov 26 at 17:16
Since a and b are both identities then their multiplication $ a ast b$ should be equal to either a or b or ab ?
– Scott
Nov 26 at 17:16
2
2
@Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal.
– Ethan Bolker
Nov 26 at 17:20
@Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal.
– Ethan Bolker
Nov 26 at 17:20
1
1
@Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$?
– Ethan Bolker
Nov 26 at 17:41
@Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$?
– Ethan Bolker
Nov 26 at 17:41
2
2
@scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now.
– Ethan Bolker
Nov 26 at 18:52
@scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now.
– Ethan Bolker
Nov 26 at 18:52
1
1
1) $e, g^{-1},$ and $frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $alpha$ and $beta$ both act like an identity that $alpha = beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $alpha$ and $beta$ deserve it. You need to prove that only one can have the label i.e $alpha = beta$.
– fleablood
Nov 26 at 21:48
1) $e, g^{-1},$ and $frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $alpha$ and $beta$ both act like an identity that $alpha = beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $alpha$ and $beta$ deserve it. You need to prove that only one can have the label i.e $alpha = beta$.
– fleablood
Nov 26 at 21:48
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show 4 more comments
up vote
1
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Let $e,e'$ be identity elements of $G$. Then $e=e'e$, since $e'$ is an identity element, and $e'e=e'$, since $e$ is an identity element.
Let $u,v$ be inverses of $a$. Then $u= u e = u(av) = (ua)v = ev =v$, where $e$ is the identity element.
add a comment |
up vote
1
down vote
Let $e,e'$ be identity elements of $G$. Then $e=e'e$, since $e'$ is an identity element, and $e'e=e'$, since $e$ is an identity element.
Let $u,v$ be inverses of $a$. Then $u= u e = u(av) = (ua)v = ev =v$, where $e$ is the identity element.
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $e,e'$ be identity elements of $G$. Then $e=e'e$, since $e'$ is an identity element, and $e'e=e'$, since $e$ is an identity element.
Let $u,v$ be inverses of $a$. Then $u= u e = u(av) = (ua)v = ev =v$, where $e$ is the identity element.
Let $e,e'$ be identity elements of $G$. Then $e=e'e$, since $e'$ is an identity element, and $e'e=e'$, since $e$ is an identity element.
Let $u,v$ be inverses of $a$. Then $u= u e = u(av) = (ua)v = ev =v$, where $e$ is the identity element.
answered Nov 26 at 17:59
Wuestenfux
2,6621410
2,6621410
add a comment |
add a comment |
up vote
0
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Suppose there exist two identity elements e_1 and e_2
then
e_1*e_2=e_2*e_1=e_2 (since e_1 is an identity).
e_1*e_2=e_2*e_1=e_1 (since e_2 is an identity).
So e_1=e_2.
If y_1 is an inverse of x and y_2 is another one then :
y_1=y_1*e=y_1*(xy_2)=(y_1 x) * y_2=e* y_2=y_2
New contributor
4
Please use MathJax to format mathematical expressions.
– Servaes
Nov 26 at 17:03
4
Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more.
– Ethan Bolker
Nov 26 at 17:14
Thank you for the answer, i will consider this after i learn how can i conclude to this answer.
– Scott
Nov 26 at 17:19
I get the first part and almost understand the second part. Thanks for the help.
– Scott
Nov 26 at 17:46
add a comment |
up vote
0
down vote
Suppose there exist two identity elements e_1 and e_2
then
e_1*e_2=e_2*e_1=e_2 (since e_1 is an identity).
e_1*e_2=e_2*e_1=e_1 (since e_2 is an identity).
So e_1=e_2.
If y_1 is an inverse of x and y_2 is another one then :
y_1=y_1*e=y_1*(xy_2)=(y_1 x) * y_2=e* y_2=y_2
New contributor
4
Please use MathJax to format mathematical expressions.
– Servaes
Nov 26 at 17:03
4
Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more.
– Ethan Bolker
Nov 26 at 17:14
Thank you for the answer, i will consider this after i learn how can i conclude to this answer.
– Scott
Nov 26 at 17:19
I get the first part and almost understand the second part. Thanks for the help.
– Scott
Nov 26 at 17:46
add a comment |
up vote
0
down vote
up vote
0
down vote
Suppose there exist two identity elements e_1 and e_2
then
e_1*e_2=e_2*e_1=e_2 (since e_1 is an identity).
e_1*e_2=e_2*e_1=e_1 (since e_2 is an identity).
So e_1=e_2.
If y_1 is an inverse of x and y_2 is another one then :
y_1=y_1*e=y_1*(xy_2)=(y_1 x) * y_2=e* y_2=y_2
New contributor
Suppose there exist two identity elements e_1 and e_2
then
e_1*e_2=e_2*e_1=e_2 (since e_1 is an identity).
e_1*e_2=e_2*e_1=e_1 (since e_2 is an identity).
So e_1=e_2.
If y_1 is an inverse of x and y_2 is another one then :
y_1=y_1*e=y_1*(xy_2)=(y_1 x) * y_2=e* y_2=y_2
New contributor
edited Nov 26 at 17:04
New contributor
answered Nov 26 at 17:03
Profahk
92
92
New contributor
New contributor
4
Please use MathJax to format mathematical expressions.
– Servaes
Nov 26 at 17:03
4
Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more.
– Ethan Bolker
Nov 26 at 17:14
Thank you for the answer, i will consider this after i learn how can i conclude to this answer.
– Scott
Nov 26 at 17:19
I get the first part and almost understand the second part. Thanks for the help.
– Scott
Nov 26 at 17:46
add a comment |
4
Please use MathJax to format mathematical expressions.
– Servaes
Nov 26 at 17:03
4
Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more.
– Ethan Bolker
Nov 26 at 17:14
Thank you for the answer, i will consider this after i learn how can i conclude to this answer.
– Scott
Nov 26 at 17:19
I get the first part and almost understand the second part. Thanks for the help.
– Scott
Nov 26 at 17:46
4
4
Please use MathJax to format mathematical expressions.
– Servaes
Nov 26 at 17:03
Please use MathJax to format mathematical expressions.
– Servaes
Nov 26 at 17:03
4
4
Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more.
– Ethan Bolker
Nov 26 at 17:14
Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more.
– Ethan Bolker
Nov 26 at 17:14
Thank you for the answer, i will consider this after i learn how can i conclude to this answer.
– Scott
Nov 26 at 17:19
Thank you for the answer, i will consider this after i learn how can i conclude to this answer.
– Scott
Nov 26 at 17:19
I get the first part and almost understand the second part. Thanks for the help.
– Scott
Nov 26 at 17:46
I get the first part and almost understand the second part. Thanks for the help.
– Scott
Nov 26 at 17:46
add a comment |
It's a bit disappointing that this got onto the hot network questions list... after all, this is a "I have no idea post" that's answered by many elementary books in the field (e.g. Pinter, Dummit & Foote, and innumerable other textbooks...). In general, this is not a do-my-homework site and it's expected to show your own thoughts and the context where you encountered the problem. I'm voting to close as off-topic. But that really doesn't matter much, considering that five enthusiastic users have given you the exact same answers.
– T. Bongers
Nov 26 at 22:30