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










1












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 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
    2 hours 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
    2 hours ago















1












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 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
    2 hours 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
    2 hours ago













1












1








1





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours 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
    2 hours 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
    2 hours ago












  • 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
    2 hours 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
    2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    2 hours 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
    2 hours ago


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    The example helped a lot. Great explanation.
    $endgroup$
    – Ufomammut
    2 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    2 hours 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
    2 hours ago















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    2 hours 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
    2 hours ago













5












5








5





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 2 hours 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
    2 hours 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
    2 hours 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
    2 hours 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
    2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago











4












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    The example helped a lot. Great explanation.
    $endgroup$
    – Ufomammut
    2 hours ago















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    The example helped a lot. Great explanation.
    $endgroup$
    – Ufomammut
    2 hours ago













4












4








4





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Henning MakholmHenning Makholm

243k17310554




243k17310554











  • $begingroup$
    The example helped a lot. Great explanation.
    $endgroup$
    – Ufomammut
    2 hours ago
















  • $begingroup$
    The example helped a lot. Great explanation.
    $endgroup$
    – Ufomammut
    2 hours ago















$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
2 hours ago




$begingroup$
The example helped a lot. Great explanation.
$endgroup$
– Ufomammut
2 hours ago

















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