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If $G$ has inverse of all elements, then $G$ is a group. $($ true / false$) ?$ [duplicate]
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This question already has an answer here:
A set which satisfies all conditions for a Group except associativity
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If $G$ has inverse of all elements, then $G$ is a group. $($ true / false$) ?$
I know a group satisfies closure property, associativity. It has a unique identity element and inverses for all elements.
I can't find any counter example to prove the statement false. If the statement is true, then how existence of inverses of all elements imply that $G$ is a group?
abstract-algebra
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
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marked as duplicate by Dietrich Burde abstract-algebra
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Jan 12 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
A set which satisfies all conditions for a Group except associativity
2 answers
If $G$ has inverse of all elements, then $G$ is a group. $($ true / false$) ?$
I know a group satisfies closure property, associativity. It has a unique identity element and inverses for all elements.
I can't find any counter example to prove the statement false. If the statement is true, then how existence of inverses of all elements imply that $G$ is a group?
abstract-algebra
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
$endgroup$
marked as duplicate by Dietrich Burde abstract-algebra
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Jan 12 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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You haven't given enough information to define $G$. Is $G$ a set equipped with a binary operation? Is $G$ closed under the binary operation? Anything else?
$endgroup$
– parsiad
Jan 12 at 19:38
1
$begingroup$
The unit sphere in $mathbb R^8$ has a product operation with inverses that is not associative.
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– Matt Samuel
Jan 12 at 19:38
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@MattSamuel Is there a (smallish) finite substructure of $mathbb{R}^8$ that also serves as a counterexample?
$endgroup$
– Mike Pierce
Jan 12 at 19:39
$begingroup$
@Mike Probably. Rather than trying to find that though it would probably be easier to directly construct a finite example.
$endgroup$
– Matt Samuel
Jan 12 at 19:40
1
$begingroup$
The question is a bit vague in terms of restriction on set $G$. For example, we can have a trivial example as $G=Bbb{Z}-{0}$, then we can say $G$ has additive inverses but is not closed, hence not a group.
$endgroup$
– Anurag A
Jan 12 at 19:42
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show 1 more comment
$begingroup$
This question already has an answer here:
A set which satisfies all conditions for a Group except associativity
2 answers
If $G$ has inverse of all elements, then $G$ is a group. $($ true / false$) ?$
I know a group satisfies closure property, associativity. It has a unique identity element and inverses for all elements.
I can't find any counter example to prove the statement false. If the statement is true, then how existence of inverses of all elements imply that $G$ is a group?
abstract-algebra
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
$endgroup$
This question already has an answer here:
A set which satisfies all conditions for a Group except associativity
2 answers
If $G$ has inverse of all elements, then $G$ is a group. $($ true / false$) ?$
I know a group satisfies closure property, associativity. It has a unique identity element and inverses for all elements.
I can't find any counter example to prove the statement false. If the statement is true, then how existence of inverses of all elements imply that $G$ is a group?
This question already has an answer here:
A set which satisfies all conditions for a Group except associativity
2 answers
abstract-algebra
abstract-algebra
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
asked Jan 12 at 19:35
MathsaddictMathsaddict
3619
3619
marked as duplicate by Dietrich Burde abstract-algebra
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Jan 12 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde abstract-algebra
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Jan 12 at 19:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You haven't given enough information to define $G$. Is $G$ a set equipped with a binary operation? Is $G$ closed under the binary operation? Anything else?
$endgroup$
– parsiad
Jan 12 at 19:38
1
$begingroup$
The unit sphere in $mathbb R^8$ has a product operation with inverses that is not associative.
$endgroup$
– Matt Samuel
Jan 12 at 19:38
$begingroup$
@MattSamuel Is there a (smallish) finite substructure of $mathbb{R}^8$ that also serves as a counterexample?
$endgroup$
– Mike Pierce
Jan 12 at 19:39
$begingroup$
@Mike Probably. Rather than trying to find that though it would probably be easier to directly construct a finite example.
$endgroup$
– Matt Samuel
Jan 12 at 19:40
1
$begingroup$
The question is a bit vague in terms of restriction on set $G$. For example, we can have a trivial example as $G=Bbb{Z}-{0}$, then we can say $G$ has additive inverses but is not closed, hence not a group.
$endgroup$
– Anurag A
Jan 12 at 19:42
|
show 1 more comment
1
$begingroup$
You haven't given enough information to define $G$. Is $G$ a set equipped with a binary operation? Is $G$ closed under the binary operation? Anything else?
$endgroup$
– parsiad
Jan 12 at 19:38
1
$begingroup$
The unit sphere in $mathbb R^8$ has a product operation with inverses that is not associative.
$endgroup$
– Matt Samuel
Jan 12 at 19:38
$begingroup$
@MattSamuel Is there a (smallish) finite substructure of $mathbb{R}^8$ that also serves as a counterexample?
$endgroup$
– Mike Pierce
Jan 12 at 19:39
$begingroup$
@Mike Probably. Rather than trying to find that though it would probably be easier to directly construct a finite example.
$endgroup$
– Matt Samuel
Jan 12 at 19:40
1
$begingroup$
The question is a bit vague in terms of restriction on set $G$. For example, we can have a trivial example as $G=Bbb{Z}-{0}$, then we can say $G$ has additive inverses but is not closed, hence not a group.
$endgroup$
– Anurag A
Jan 12 at 19:42
1
1
$begingroup$
You haven't given enough information to define $G$. Is $G$ a set equipped with a binary operation? Is $G$ closed under the binary operation? Anything else?
$endgroup$
– parsiad
Jan 12 at 19:38
$begingroup$
You haven't given enough information to define $G$. Is $G$ a set equipped with a binary operation? Is $G$ closed under the binary operation? Anything else?
$endgroup$
– parsiad
Jan 12 at 19:38
1
1
$begingroup$
The unit sphere in $mathbb R^8$ has a product operation with inverses that is not associative.
$endgroup$
– Matt Samuel
Jan 12 at 19:38
$begingroup$
The unit sphere in $mathbb R^8$ has a product operation with inverses that is not associative.
$endgroup$
– Matt Samuel
Jan 12 at 19:38
$begingroup$
@MattSamuel Is there a (smallish) finite substructure of $mathbb{R}^8$ that also serves as a counterexample?
$endgroup$
– Mike Pierce
Jan 12 at 19:39
$begingroup$
@MattSamuel Is there a (smallish) finite substructure of $mathbb{R}^8$ that also serves as a counterexample?
$endgroup$
– Mike Pierce
Jan 12 at 19:39
$begingroup$
@Mike Probably. Rather than trying to find that though it would probably be easier to directly construct a finite example.
$endgroup$
– Matt Samuel
Jan 12 at 19:40
$begingroup$
@Mike Probably. Rather than trying to find that though it would probably be easier to directly construct a finite example.
$endgroup$
– Matt Samuel
Jan 12 at 19:40
1
1
$begingroup$
The question is a bit vague in terms of restriction on set $G$. For example, we can have a trivial example as $G=Bbb{Z}-{0}$, then we can say $G$ has additive inverses but is not closed, hence not a group.
$endgroup$
– Anurag A
Jan 12 at 19:42
$begingroup$
The question is a bit vague in terms of restriction on set $G$. For example, we can have a trivial example as $G=Bbb{Z}-{0}$, then we can say $G$ has additive inverses but is not closed, hence not a group.
$endgroup$
– Anurag A
Jan 12 at 19:42
|
show 1 more comment
1 Answer
1
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$begingroup$
No, groupoids (https://en.wikipedia.org/wiki/Groupoid?wprov=sfla1) are a counterexample.
answered Jan 12 at 19:43
ecrinecrin
3497
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, groupoids (https://en.wikipedia.org/wiki/Groupoid?wprov=sfla1) are a counterexample.
answered Jan 12 at 19:43
ecrinecrin
3497
$endgroup$
add a comment |
$begingroup$
No, groupoids (https://en.wikipedia.org/wiki/Groupoid?wprov=sfla1) are a counterexample.
answered Jan 12 at 19:43
ecrinecrin
3497
$endgroup$
add a comment |
$begingroup$
No, groupoids (https://en.wikipedia.org/wiki/Groupoid?wprov=sfla1) are a counterexample.
answered Jan 12 at 19:43
ecrinecrin
3497
$endgroup$
No, groupoids (https://en.wikipedia.org/wiki/Groupoid?wprov=sfla1) are a counterexample.
answered Jan 12 at 19:43
ecrinecrin
3497
answered Jan 12 at 19:43
ecrinecrin
3497
answered Jan 12 at 19:43
ecrinecrin
3497
answered Jan 12 at 19:43
ecrinecrin
3497
3497
add a comment |
add a comment |
1
$begingroup$
You haven't given enough information to define $G$. Is $G$ a set equipped with a binary operation? Is $G$ closed under the binary operation? Anything else?
$endgroup$
– parsiad
Jan 12 at 19:38
1
$begingroup$
The unit sphere in $mathbb R^8$ has a product operation with inverses that is not associative.
$endgroup$
– Matt Samuel
Jan 12 at 19:38
$begingroup$
@MattSamuel Is there a (smallish) finite substructure of $mathbb{R}^8$ that also serves as a counterexample?
$endgroup$
– Mike Pierce
Jan 12 at 19:39
$begingroup$
@Mike Probably. Rather than trying to find that though it would probably be easier to directly construct a finite example.
$endgroup$
– Matt Samuel
Jan 12 at 19:40
1
$begingroup$
The question is a bit vague in terms of restriction on set $G$. For example, we can have a trivial example as $G=Bbb{Z}-{0}$, then we can say $G$ has additive inverses but is not closed, hence not a group.
$endgroup$
– Anurag A
Jan 12 at 19:42