Understanding the implication of what “well-defined” means for the operation in quotient group The 2019 Stack Overflow Developer Survey Results Are InFor $Q$ the quaternion group, is $Q/Z(Q)$ a group? For which operation?Is the Axiom of Choice implicitly used when defining a binary operation on a quotient object?Why can quotient groups only be defined for subgroups?Convincing normal subgroup proof?Coset multiplication giving a well defined binary operationShow that the group operation is well definedWhat does well-defined mean? In general or in this context (quotient group)Quotient group is well-definedClosed under an operation which is not well-defined?Showing quotient group operations are well defined
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Understanding the implication of what “well-defined” means for the operation in quotient group
The 2019 Stack Overflow Developer Survey Results Are InFor $Q$ the quaternion group, is $Q/Z(Q)$ a group? For which operation?Is the Axiom of Choice implicitly used when defining a binary operation on a quotient object?Why can quotient groups only be defined for subgroups?Convincing normal subgroup proof?Coset multiplication giving a well defined binary operationShow that the group operation is well definedWhat does well-defined mean? In general or in this context (quotient group)Quotient group is well-definedClosed under an operation which is not well-defined?Showing quotient group operations are well defined
$begingroup$
I want to get an intuitive idea of the operation being "well-defined" for quotient groups. So, let's say I have a group $G$, with subgroup $H$, and let's say my set of left cosets is $G/K$. My lecture note says this: If $H$ is normal, then $G/H$ is a group under binary operation $aH circ bH = (ab)H$. So let's say I am asked whether $G/H$ forms a group where $H$ is not normal, and I have already determined that. My professor referred to the following which I am not sure if I totally understand:
Show that there are $a, b, a', b' in G$ such that $ aH = a'H$ and $bH = b'H$ but $aH circ bH = (ab)H neq a'H circ b'H = (a'b')H$, and you are done, and I did what she suggested, but I am not sure what is going on. So,
$1.$ What did I exactly show by proving what my professor suggested?
$2.$ In general, is it a strategy that every time you have to prove $G/H$ does not form a group for a non-normal $H$, you show that the operation is not well-defined?
Also, a quick google search also showed me that the theorem that talks about $G/K$ forming a group is an "if and only if" statement and not difficult to prove as well. Still, any help on my questions above would be great.
abstract-algebra group-theory
asked 56 mins ago
UfomammutUfomammut
376314
$endgroup$
add a comment |
$begingroup$
I want to get an intuitive idea of the operation being "well-defined" for quotient groups. So, let's say I have a group $G$, with subgroup $H$, and let's say my set of left cosets is $G/K$. My lecture note says this: If $H$ is normal, then $G/H$ is a group under binary operation $aH circ bH = (ab)H$. So let's say I am asked whether $G/H$ forms a group where $H$ is not normal, and I have already determined that. My professor referred to the following which I am not sure if I totally understand:
Show that there are $a, b, a', b' in G$ such that $ aH = a'H$ and $bH = b'H$ but $aH circ bH = (ab)H neq a'H circ b'H = (a'b')H$, and you are done, and I did what she suggested, but I am not sure what is going on. So,
$1.$ What did I exactly show by proving what my professor suggested?
$2.$ In general, is it a strategy that every time you have to prove $G/H$ does not form a group for a non-normal $H$, you show that the operation is not well-defined?
Also, a quick google search also showed me that the theorem that talks about $G/K$ forming a group is an "if and only if" statement and not difficult to prove as well. Still, any help on my questions above would be great.
abstract-algebra group-theory
asked 56 mins ago
UfomammutUfomammut
376314
$endgroup$
4
$begingroup$
For multiplication of cosets to be well-defined you have to get the same answer no matter which representative of the cosets you choose
$endgroup$
– J. W. Tanner
50 mins ago
1
$begingroup$
If $C,D$ are cosets of $H$, we'd like to define $C*D$ by taking a random element of $a in C$ and a random element of $b in D$ and define $C*D = E$ where $E$ is the coset containing $ab$. But we've made a random choice here, so we would expect $E$ to also be "random". But as it turns out, we get the same result $E$ no matter what $a, b$ we pick. That's what we mean when we say the operation is well-defined. In general, to prove something is well-defined means to prove that the "random" choices we made during the construction don't change the result.
$endgroup$
– Jair Taylor
29 mins ago
add a comment |
$begingroup$
I want to get an intuitive idea of the operation being "well-defined" for quotient groups. So, let's say I have a group $G$, with subgroup $H$, and let's say my set of left cosets is $G/K$. My lecture note says this: If $H$ is normal, then $G/H$ is a group under binary operation $aH circ bH = (ab)H$. So let's say I am asked whether $G/H$ forms a group where $H$ is not normal, and I have already determined that. My professor referred to the following which I am not sure if I totally understand:
Show that there are $a, b, a', b' in G$ such that $ aH = a'H$ and $bH = b'H$ but $aH circ bH = (ab)H neq a'H circ b'H = (a'b')H$, and you are done, and I did what she suggested, but I am not sure what is going on. So,
$1.$ What did I exactly show by proving what my professor suggested?
$2.$ In general, is it a strategy that every time you have to prove $G/H$ does not form a group for a non-normal $H$, you show that the operation is not well-defined?
Also, a quick google search also showed me that the theorem that talks about $G/K$ forming a group is an "if and only if" statement and not difficult to prove as well. Still, any help on my questions above would be great.
abstract-algebra group-theory
asked 56 mins ago
UfomammutUfomammut
376314
$endgroup$
I want to get an intuitive idea of the operation being "well-defined" for quotient groups. So, let's say I have a group $G$, with subgroup $H$, and let's say my set of left cosets is $G/K$. My lecture note says this: If $H$ is normal, then $G/H$ is a group under binary operation $aH circ bH = (ab)H$. So let's say I am asked whether $G/H$ forms a group where $H$ is not normal, and I have already determined that. My professor referred to the following which I am not sure if I totally understand:
Show that there are $a, b, a', b' in G$ such that $ aH = a'H$ and $bH = b'H$ but $aH circ bH = (ab)H neq a'H circ b'H = (a'b')H$, and you are done, and I did what she suggested, but I am not sure what is going on. So,
$1.$ What did I exactly show by proving what my professor suggested?
$2.$ In general, is it a strategy that every time you have to prove $G/H$ does not form a group for a non-normal $H$, you show that the operation is not well-defined?
Also, a quick google search also showed me that the theorem that talks about $G/K$ forming a group is an "if and only if" statement and not difficult to prove as well. Still, any help on my questions above would be great.
abstract-algebra group-theory
abstract-algebra group-theory
asked 56 mins ago
UfomammutUfomammut
376314
asked 56 mins ago
UfomammutUfomammut
376314
asked 56 mins ago
UfomammutUfomammut
376314
asked 56 mins ago
UfomammutUfomammut
376314
asked 56 mins ago
UfomammutUfomammut
376314
376314
4
$begingroup$
For multiplication of cosets to be well-defined you have to get the same answer no matter which representative of the cosets you choose
$endgroup$
– J. W. Tanner
50 mins ago
1
$begingroup$
If $C,D$ are cosets of $H$, we'd like to define $C*D$ by taking a random element of $a in C$ and a random element of $b in D$ and define $C*D = E$ where $E$ is the coset containing $ab$. But we've made a random choice here, so we would expect $E$ to also be "random". But as it turns out, we get the same result $E$ no matter what $a, b$ we pick. That's what we mean when we say the operation is well-defined. In general, to prove something is well-defined means to prove that the "random" choices we made during the construction don't change the result.
$endgroup$
– Jair Taylor
29 mins ago
add a comment |
4
$begingroup$
For multiplication of cosets to be well-defined you have to get the same answer no matter which representative of the cosets you choose
$endgroup$
– J. W. Tanner
50 mins ago
1
$begingroup$
If $C,D$ are cosets of $H$, we'd like to define $C*D$ by taking a random element of $a in C$ and a random element of $b in D$ and define $C*D = E$ where $E$ is the coset containing $ab$. But we've made a random choice here, so we would expect $E$ to also be "random". But as it turns out, we get the same result $E$ no matter what $a, b$ we pick. That's what we mean when we say the operation is well-defined. In general, to prove something is well-defined means to prove that the "random" choices we made during the construction don't change the result.
$endgroup$
– Jair Taylor
29 mins ago
4
4
$begingroup$
For multiplication of cosets to be well-defined you have to get the same answer no matter which representative of the cosets you choose
$endgroup$
– J. W. Tanner
50 mins ago
$begingroup$
For multiplication of cosets to be well-defined you have to get the same answer no matter which representative of the cosets you choose
$endgroup$
– J. W. Tanner
50 mins ago
1
1
$begingroup$
If $C,D$ are cosets of $H$, we'd like to define $C*D$ by taking a random element of $a in C$ and a random element of $b in D$ and define $C*D = E$ where $E$ is the coset containing $ab$. But we've made a random choice here, so we would expect $E$ to also be "random". But as it turns out, we get the same result $E$ no matter what $a, b$ we pick. That's what we mean when we say the operation is well-defined. In general, to prove something is well-defined means to prove that the "random" choices we made during the construction don't change the result.
$endgroup$
– Jair Taylor
29 mins ago
$begingroup$
If $C,D$ are cosets of $H$, we'd like to define $C*D$ by taking a random element of $a in C$ and a random element of $b in D$ and define $C*D = E$ where $E$ is the coset containing $ab$. But we've made a random choice here, so we would expect $E$ to also be "random". But as it turns out, we get the same result $E$ no matter what $a, b$ we pick. That's what we mean when we say the operation is well-defined. In general, to prove something is well-defined means to prove that the "random" choices we made during the construction don't change the result.
$endgroup$
– Jair Taylor
29 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In general, mathematicians use the phrase "well defined" when a definition is written in a form that depends (or, rather, seems to depend) on some more or less arbitrary choice. If you make such a definition, you are obligated to show that another choice that satisfied appropriate conditions would lead to the same result.
In a group the product $abc$ is well defined to be $(ab)c$ because its value does not depend on your choice of where to put the parentheses: associativity guarantees $(ab)c = a(bc)$. This fact is so intuitively clear that it's often not made explicit in a beginning algebra course.
When considering quotient groups, you want to define the multiplication of two cosets by choosing an element from each, multiplying them together, and taking the coset of the product. This coset product will be well defined only when the coset of the product of the two group elements does not depend on which ones you happened to choose. The sum of any two odd numbers will be even, so the product of cosets
$(2mathbbZ + 1) circ (2mathbbZ + 1)$ is $2mathbbZ$.
There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
$$
A circ B = ab .
$$
This definition does not make any arbitrary choices, but you don't know that the set so defined is really a coset until you prove it.
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
$begingroup$
I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation?
$endgroup$
– Ufomammut
35 mins ago
1
$begingroup$
Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition.
$endgroup$
– Ethan Bolker
30 mins ago
add a comment |
$begingroup$
Perhaps a concrete example will make it clearer?
We need a non-abelian group. Let's take the simplest one there is, namely $G=S_3$.
We need a non-normal subgroup. Let's take $H=e,(12)$.
We need two different cosets, and it won't work with H itself, so we have to take
$$ a=(23) qquad aH = (23),(132) = a'H qquad a'=(132) $$
$$ b=(13) qquad bH = (13),(123) = b'H qquad b'=(123) $$
Now, if we had a quotient group what should the product $(23),(132)circ(13),(123)$ be?
From one perspective we have
$$(23)Hcirc(13)H =^? (123)H$$
But we could also say
$$(123)Hcirc(132)H =^? eH = H$$
But the product of the coset $(12),(132)$ with the coset $(23),(123)$ cannot be allowed to depend on what we choose to call those cosets. And here we have two calculations that say they should be two different things! So we're in trouble.
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
$endgroup$
$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
32 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
In general, mathematicians use the phrase "well defined" when a definition is written in a form that depends (or, rather, seems to depend) on some more or less arbitrary choice. If you make such a definition, you are obligated to show that another choice that satisfied appropriate conditions would lead to the same result.
In a group the product $abc$ is well defined to be $(ab)c$ because its value does not depend on your choice of where to put the parentheses: associativity guarantees $(ab)c = a(bc)$. This fact is so intuitively clear that it's often not made explicit in a beginning algebra course.
When considering quotient groups, you want to define the multiplication of two cosets by choosing an element from each, multiplying them together, and taking the coset of the product. This coset product will be well defined only when the coset of the product of the two group elements does not depend on which ones you happened to choose. The sum of any two odd numbers will be even, so the product of cosets
$(2mathbbZ + 1) circ (2mathbbZ + 1)$ is $2mathbbZ$.
There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
$$
A circ B = ab .
$$
This definition does not make any arbitrary choices, but you don't know that the set so defined is really a coset until you prove it.
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
$begingroup$
I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation?
$endgroup$
– Ufomammut
35 mins ago
1
$begingroup$
Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition.
$endgroup$
– Ethan Bolker
30 mins ago
add a comment |
$begingroup$
In general, mathematicians use the phrase "well defined" when a definition is written in a form that depends (or, rather, seems to depend) on some more or less arbitrary choice. If you make such a definition, you are obligated to show that another choice that satisfied appropriate conditions would lead to the same result.
In a group the product $abc$ is well defined to be $(ab)c$ because its value does not depend on your choice of where to put the parentheses: associativity guarantees $(ab)c = a(bc)$. This fact is so intuitively clear that it's often not made explicit in a beginning algebra course.
When considering quotient groups, you want to define the multiplication of two cosets by choosing an element from each, multiplying them together, and taking the coset of the product. This coset product will be well defined only when the coset of the product of the two group elements does not depend on which ones you happened to choose. The sum of any two odd numbers will be even, so the product of cosets
$(2mathbbZ + 1) circ (2mathbbZ + 1)$ is $2mathbbZ$.
There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
$$
A circ B = ab .
$$
This definition does not make any arbitrary choices, but you don't know that the set so defined is really a coset until you prove it.
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
$begingroup$
I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation?
$endgroup$
– Ufomammut
35 mins ago
1
$begingroup$
Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition.
$endgroup$
– Ethan Bolker
30 mins ago
add a comment |
$begingroup$
In general, mathematicians use the phrase "well defined" when a definition is written in a form that depends (or, rather, seems to depend) on some more or less arbitrary choice. If you make such a definition, you are obligated to show that another choice that satisfied appropriate conditions would lead to the same result.
In a group the product $abc$ is well defined to be $(ab)c$ because its value does not depend on your choice of where to put the parentheses: associativity guarantees $(ab)c = a(bc)$. This fact is so intuitively clear that it's often not made explicit in a beginning algebra course.
When considering quotient groups, you want to define the multiplication of two cosets by choosing an element from each, multiplying them together, and taking the coset of the product. This coset product will be well defined only when the coset of the product of the two group elements does not depend on which ones you happened to choose. The sum of any two odd numbers will be even, so the product of cosets
$(2mathbbZ + 1) circ (2mathbbZ + 1)$ is $2mathbbZ$.
There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
$$
A circ B = ab .
$$
This definition does not make any arbitrary choices, but you don't know that the set so defined is really a coset until you prove it.
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
In general, mathematicians use the phrase "well defined" when a definition is written in a form that depends (or, rather, seems to depend) on some more or less arbitrary choice. If you make such a definition, you are obligated to show that another choice that satisfied appropriate conditions would lead to the same result.
In a group the product $abc$ is well defined to be $(ab)c$ because its value does not depend on your choice of where to put the parentheses: associativity guarantees $(ab)c = a(bc)$. This fact is so intuitively clear that it's often not made explicit in a beginning algebra course.
When considering quotient groups, you want to define the multiplication of two cosets by choosing an element from each, multiplying them together, and taking the coset of the product. This coset product will be well defined only when the coset of the product of the two group elements does not depend on which ones you happened to choose. The sum of any two odd numbers will be even, so the product of cosets
$(2mathbbZ + 1) circ (2mathbbZ + 1)$ is $2mathbbZ$.
There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
$$
A circ B = ab .
$$
This definition does not make any arbitrary choices, but you don't know that the set so defined is really a coset until you prove it.
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
edited 27 mins ago
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
answered 43 mins ago
Ethan BolkerEthan Bolker
45.9k553120
45.9k553120
$begingroup$
I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation?
$endgroup$
– Ufomammut
35 mins ago
1
$begingroup$
Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition.
$endgroup$
– Ethan Bolker
30 mins ago
add a comment |
$begingroup$
I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation?
$endgroup$
– Ufomammut
35 mins ago
1
$begingroup$
Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition.
$endgroup$
– Ethan Bolker
30 mins ago
$begingroup$
I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation?
$endgroup$
– Ufomammut
35 mins ago
$begingroup$
I do not know if what I am asking makes sense, but is the way the binary operation between two sets of the set of cosets is defined always the same? Also, getting back to my question, is this equivalent to proving closure under the binary operation?
$endgroup$
– Ufomammut
35 mins ago
1
1
$begingroup$
Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition.
$endgroup$
– Ethan Bolker
30 mins ago
$begingroup$
Closure is a bit of a red herring. In your definition of the coset product you always get a coset as the result. The issue is proving that the particular coset is independent of the choices. In my alternative definition you have to prove that a particular set is a coset. Closure only comes up when you already have an operation defined and you want to show you don't leave some subset. So the set of odd integers is not closed under addition.
$endgroup$
– Ethan Bolker
30 mins ago
add a comment |
$begingroup$
Perhaps a concrete example will make it clearer?
We need a non-abelian group. Let's take the simplest one there is, namely $G=S_3$.
We need a non-normal subgroup. Let's take $H=e,(12)$.
We need two different cosets, and it won't work with H itself, so we have to take
$$ a=(23) qquad aH = (23),(132) = a'H qquad a'=(132) $$
$$ b=(13) qquad bH = (13),(123) = b'H qquad b'=(123) $$
Now, if we had a quotient group what should the product $(23),(132)circ(13),(123)$ be?
From one perspective we have
$$(23)Hcirc(13)H =^? (123)H$$
But we could also say
$$(123)Hcirc(132)H =^? eH = H$$
But the product of the coset $(12),(132)$ with the coset $(23),(123)$ cannot be allowed to depend on what we choose to call those cosets. And here we have two calculations that say they should be two different things! So we're in trouble.
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
$endgroup$
$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
32 mins ago
add a comment |
$begingroup$
Perhaps a concrete example will make it clearer?
We need a non-abelian group. Let's take the simplest one there is, namely $G=S_3$.
We need a non-normal subgroup. Let's take $H=e,(12)$.
We need two different cosets, and it won't work with H itself, so we have to take
$$ a=(23) qquad aH = (23),(132) = a'H qquad a'=(132) $$
$$ b=(13) qquad bH = (13),(123) = b'H qquad b'=(123) $$
Now, if we had a quotient group what should the product $(23),(132)circ(13),(123)$ be?
From one perspective we have
$$(23)Hcirc(13)H =^? (123)H$$
But we could also say
$$(123)Hcirc(132)H =^? eH = H$$
But the product of the coset $(12),(132)$ with the coset $(23),(123)$ cannot be allowed to depend on what we choose to call those cosets. And here we have two calculations that say they should be two different things! So we're in trouble.
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
$endgroup$
$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
32 mins ago
add a comment |
$begingroup$
Perhaps a concrete example will make it clearer?
We need a non-abelian group. Let's take the simplest one there is, namely $G=S_3$.
We need a non-normal subgroup. Let's take $H=e,(12)$.
We need two different cosets, and it won't work with H itself, so we have to take
$$ a=(23) qquad aH = (23),(132) = a'H qquad a'=(132) $$
$$ b=(13) qquad bH = (13),(123) = b'H qquad b'=(123) $$
Now, if we had a quotient group what should the product $(23),(132)circ(13),(123)$ be?
From one perspective we have
$$(23)Hcirc(13)H =^? (123)H$$
But we could also say
$$(123)Hcirc(132)H =^? eH = H$$
But the product of the coset $(12),(132)$ with the coset $(23),(123)$ cannot be allowed to depend on what we choose to call those cosets. And here we have two calculations that say they should be two different things! So we're in trouble.
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
$endgroup$
Perhaps a concrete example will make it clearer?
We need a non-abelian group. Let's take the simplest one there is, namely $G=S_3$.
We need a non-normal subgroup. Let's take $H=e,(12)$.
We need two different cosets, and it won't work with H itself, so we have to take
$$ a=(23) qquad aH = (23),(132) = a'H qquad a'=(132) $$
$$ b=(13) qquad bH = (13),(123) = b'H qquad b'=(123) $$
Now, if we had a quotient group what should the product $(23),(132)circ(13),(123)$ be?
From one perspective we have
$$(23)Hcirc(13)H =^? (123)H$$
But we could also say
$$(123)Hcirc(132)H =^? eH = H$$
But the product of the coset $(12),(132)$ with the coset $(23),(123)$ cannot be allowed to depend on what we choose to call those cosets. And here we have two calculations that say they should be two different things! So we're in trouble.
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
answered 34 mins ago
Henning MakholmHenning Makholm
243k17310554
243k17310554
$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
32 mins ago
add a comment |
$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
32 mins ago
$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
32 mins ago
$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
32 mins ago
add a comment |
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$begingroup$
For multiplication of cosets to be well-defined you have to get the same answer no matter which representative of the cosets you choose
$endgroup$
– J. W. Tanner
50 mins ago
1
$begingroup$
If $C,D$ are cosets of $H$, we'd like to define $C*D$ by taking a random element of $a in C$ and a random element of $b in D$ and define $C*D = E$ where $E$ is the coset containing $ab$. But we've made a random choice here, so we would expect $E$ to also be "random". But as it turns out, we get the same result $E$ no matter what $a, b$ we pick. That's what we mean when we say the operation is well-defined. In general, to prove something is well-defined means to prove that the "random" choices we made during the construction don't change the result.
$endgroup$
– Jair Taylor
29 mins ago