Why does $xin H Rightarrow x^{-1} in H$ if $(H,*)$ is a Group [closed]
I have a question about Groups. Why is it that in Groups, if you take an element in the set from which the Group is built on, it implies the membership of its inverse?
The definition of Group presented to me was:
A Group $(G,*)$ is a monoid in which every element of $G$ is invertible with repect to $*$.
(A monoid $(X,*)$ is an algebraic structure that satisfies associativity and the existence of an identity)
My question is, more specifically:
Let $(X,*)$ be a monoid with the existence of invertible elements and $(G,*)$ be a Group in which all invertible elements of $X$ are contained. Then if we take a Subgroup $(H,*)$ of $(G,*)$, does a membership of an element of H imply the membership of its inverse? If so, why?
Thank you in advance.
abstract-algebra group-theory
asked Dec 26 at 16:23
Daàvid
415
closed as off-topic by Dietrich Burde, Shaun, Shailesh, amWhy, metamorphy Dec 26 at 19:41
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, Shailesh, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 5 more comments
I have a question about Groups. Why is it that in Groups, if you take an element in the set from which the Group is built on, it implies the membership of its inverse?
The definition of Group presented to me was:
A Group $(G,*)$ is a monoid in which every element of $G$ is invertible with repect to $*$.
(A monoid $(X,*)$ is an algebraic structure that satisfies associativity and the existence of an identity)
My question is, more specifically:
Let $(X,*)$ be a monoid with the existence of invertible elements and $(G,*)$ be a Group in which all invertible elements of $X$ are contained. Then if we take a Subgroup $(H,*)$ of $(G,*)$, does a membership of an element of H imply the membership of its inverse? If so, why?
Thank you in advance.
abstract-algebra group-theory
asked Dec 26 at 16:23
Daàvid
415
closed as off-topic by Dietrich Burde, Shaun, Shailesh, amWhy, metamorphy Dec 26 at 19:41
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, Shailesh, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
1
This is part of the definition of what it means to be a group. There are several ways of phrasing the definition. Can you write your definition in the question?
– Alex Kruckman
Dec 26 at 16:24
By definition, for $xin G$ the inverse $x^{-1}$ exists with $x^{-1}in G$, "closed", see this question.
– Dietrich Burde
Dec 26 at 16:36
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Dec 26 at 16:40
This is what is meant by the phrase "every element is invertible" in the first clause of the definition.
– saulspatz
Dec 26 at 16:40
But every element of $H$ being invertible is not the same as every invertible element being contained in $H$, is it?
– Daàvid
Dec 26 at 16:56
|
show 5 more comments
I have a question about Groups. Why is it that in Groups, if you take an element in the set from which the Group is built on, it implies the membership of its inverse?
The definition of Group presented to me was:
A Group $(G,*)$ is a monoid in which every element of $G$ is invertible with repect to $*$.
(A monoid $(X,*)$ is an algebraic structure that satisfies associativity and the existence of an identity)
My question is, more specifically:
Let $(X,*)$ be a monoid with the existence of invertible elements and $(G,*)$ be a Group in which all invertible elements of $X$ are contained. Then if we take a Subgroup $(H,*)$ of $(G,*)$, does a membership of an element of H imply the membership of its inverse? If so, why?
Thank you in advance.
abstract-algebra group-theory
asked Dec 26 at 16:23
Daàvid
415
I have a question about Groups. Why is it that in Groups, if you take an element in the set from which the Group is built on, it implies the membership of its inverse?
The definition of Group presented to me was:
A Group $(G,*)$ is a monoid in which every element of $G$ is invertible with repect to $*$.
(A monoid $(X,*)$ is an algebraic structure that satisfies associativity and the existence of an identity)
My question is, more specifically:
Let $(X,*)$ be a monoid with the existence of invertible elements and $(G,*)$ be a Group in which all invertible elements of $X$ are contained. Then if we take a Subgroup $(H,*)$ of $(G,*)$, does a membership of an element of H imply the membership of its inverse? If so, why?
Thank you in advance.
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 26 at 16:23
Daàvid
415
asked Dec 26 at 16:23
Daàvid
415
edited Dec 26 at 16:53
asked Dec 26 at 16:23
Daàvid
415
asked Dec 26 at 16:23
Daàvid
415
asked Dec 26 at 16:23
Daàvid
415
415
closed as off-topic by Dietrich Burde, Shaun, Shailesh, amWhy, metamorphy Dec 26 at 19:41
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, Shailesh, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Dietrich Burde, Shaun, Shailesh, amWhy, metamorphy Dec 26 at 19:41
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, Shailesh, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
1
This is part of the definition of what it means to be a group. There are several ways of phrasing the definition. Can you write your definition in the question?
– Alex Kruckman
Dec 26 at 16:24
By definition, for $xin G$ the inverse $x^{-1}$ exists with $x^{-1}in G$, "closed", see this question.
– Dietrich Burde
Dec 26 at 16:36
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Dec 26 at 16:40
This is what is meant by the phrase "every element is invertible" in the first clause of the definition.
– saulspatz
Dec 26 at 16:40
But every element of $H$ being invertible is not the same as every invertible element being contained in $H$, is it?
– Daàvid
Dec 26 at 16:56
|
show 5 more comments
1
This is part of the definition of what it means to be a group. There are several ways of phrasing the definition. Can you write your definition in the question?
– Alex Kruckman
Dec 26 at 16:24
By definition, for $xin G$ the inverse $x^{-1}$ exists with $x^{-1}in G$, "closed", see this question.
– Dietrich Burde
Dec 26 at 16:36
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Dec 26 at 16:40
This is what is meant by the phrase "every element is invertible" in the first clause of the definition.
– saulspatz
Dec 26 at 16:40
But every element of $H$ being invertible is not the same as every invertible element being contained in $H$, is it?
– Daàvid
Dec 26 at 16:56
1
1
This is part of the definition of what it means to be a group. There are several ways of phrasing the definition. Can you write your definition in the question?
– Alex Kruckman
Dec 26 at 16:24
This is part of the definition of what it means to be a group. There are several ways of phrasing the definition. Can you write your definition in the question?
– Alex Kruckman
Dec 26 at 16:24
By definition, for $xin G$ the inverse $x^{-1}$ exists with $x^{-1}in G$, "closed", see this question.
– Dietrich Burde
Dec 26 at 16:36
By definition, for $xin G$ the inverse $x^{-1}$ exists with $x^{-1}in G$, "closed", see this question.
– Dietrich Burde
Dec 26 at 16:36
1
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Dec 26 at 16:40
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Dec 26 at 16:40
This is what is meant by the phrase "every element is invertible" in the first clause of the definition.
– saulspatz
Dec 26 at 16:40
This is what is meant by the phrase "every element is invertible" in the first clause of the definition.
– saulspatz
Dec 26 at 16:40
But every element of $H$ being invertible is not the same as every invertible element being contained in $H$, is it?
– Daàvid
Dec 26 at 16:56
But every element of $H$ being invertible is not the same as every invertible element being contained in $H$, is it?
– Daàvid
Dec 26 at 16:56
|
show 5 more comments
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1
This is part of the definition of what it means to be a group. There are several ways of phrasing the definition. Can you write your definition in the question?
– Alex Kruckman
Dec 26 at 16:24
By definition, for $xin G$ the inverse $x^{-1}$ exists with $x^{-1}in G$, "closed", see this question.
– Dietrich Burde
Dec 26 at 16:36
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Dec 26 at 16:40
This is what is meant by the phrase "every element is invertible" in the first clause of the definition.
– saulspatz
Dec 26 at 16:40
But every element of $H$ being invertible is not the same as every invertible element being contained in $H$, is it?
– Daàvid
Dec 26 at 16:56