Understanding the implication of what “well-defined” means for the operation in quotient group












3












$begingroup$


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.










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$endgroup$








  • 6




    $begingroup$
    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 days 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 days ago
















3












$begingroup$


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









$endgroup$








  • 6




    $begingroup$
    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 days 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 days ago














3












3








3





$begingroup$


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






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share|cite|improve this question











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share|cite|improve this question










asked 2 days ago









UfomammutUfomammut

386314




386314








  • 6




    $begingroup$
    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 days 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 days ago














  • 6




    $begingroup$
    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 days 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 days ago








6




6




$begingroup$
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 days 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 days 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 days 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 days ago










3 Answers
3






active

oldest

votes


















9












$begingroup$

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
$(2mathbb{Z} + 1) circ (2mathbb{Z} + 1)$ is $2mathbb{Z}$.



There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
$$
A circ B = { ab | a in A text{ and } b in B }.
$$

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 days 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 days ago



















7












$begingroup$

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 cosets. Using $H$ itself might be too trivial, so it seems most promising 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
$$(132)Hcirc(123)H =^? eH = H$$



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 it should be two different things! So we're in trouble.






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$endgroup$













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



















1












$begingroup$

Here's a different description that gives a different lens on all of this.



An equivalence relation on a set is a binary relation (call it $sim$) that is reflexive ($xsim x$), symmetric ($xsim yimplies ysim x$), and transitive ($xsim y land ysim z implies xsim z$). Given a group $G$ with a subgroup $H$, we can define an equivalence relation on the underlying set of $G$ via $asim b iff ab^{-1}in H$. Prove that this is indeed an equivalence relation.



An equivalence class of an element $x$ with respect to some equivalence relation $sim$ is the set ${ymid xsim y}$. I'll use $[x]$ as notation for the equivalence class of $x$ (where the relevant equivalence relation is left implicit and resolved from context). For the equivalence relation above on $G$, what does $[g]$ look like for an arbitrary element of $G$? One particularly notable case is $[1]={gmid 1sim g}={gmid g^{-1}in H}=H$.



Since our example is on a group, it would be nice to know that the equivalence relation respects the group operation. The idea is that we want to think of $xsim y$ as $x$ "equals" $y$, and so if $xsim x'$ and $ysim y'$ we would want $xysim x'y'$ as would obviously be the case for equality. (Show that this is the same as $[x][y]=[xy]$ where the multiplication of two sets of group elements is defined as $XY={xymid xin Xland yin Y}$.) An equivalence that respects the group operation is called a congruence. (More generally, a congruence refers to an equivalence relation that respects whatever operations are relevant. For example, a congruence on rings would respect multiplication and addition.) Alternatively, if we're given an equivalence relation, we can say that an operation is well-defined if it takes "equals" to "equals". In other words, "the group operation is well-defined with respect to $sim$" is the same as "$sim$ respects the group operation". The only difference is one of connotation, namely which of the group operation or the equivalence relation we would "blame" if the statement failed to hold.



So the first question is: is the equivalence defined above a congruence?



Given any equivalence relation on a group, we can extend it to a congruence. That is, we can consider the smallest congruence that contains the original equivalence relation. This is called the congruence generated by the equivalence relation. It's quite possible that the generated congruence is the same as the original equivalence relation. Now let's assume we've been given a congruence on $G$ written $approx$. We have $aapprox biff ab^{-1}approx 1$. (Why?) We can thus show $aapprox biff ab^{-1}in[1]$, mimicking our first equivalence relation.



Second and third questions: What does $[1]$ look like in this case? What does $[1]$ look like for the congruence generated by our first equivalence relation, $sim$?






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    3 Answers
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    3 Answers
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    active

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    9












    $begingroup$

    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
    $(2mathbb{Z} + 1) circ (2mathbb{Z} + 1)$ is $2mathbb{Z}$.



    There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
    $$
    A circ B = { ab | a in A text{ and } b in B }.
    $$

    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 days 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 days ago
















    9












    $begingroup$

    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
    $(2mathbb{Z} + 1) circ (2mathbb{Z} + 1)$ is $2mathbb{Z}$.



    There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
    $$
    A circ B = { ab | a in A text{ and } b in B }.
    $$

    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 days 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 days ago














    9












    9








    9





    $begingroup$

    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
    $(2mathbb{Z} + 1) circ (2mathbb{Z} + 1)$ is $2mathbb{Z}$.



    There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
    $$
    A circ B = { ab | a in A text{ and } b in B }.
    $$

    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
    $(2mathbb{Z} + 1) circ (2mathbb{Z} + 1)$ is $2mathbb{Z}$.



    There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
    $$
    A circ B = { ab | a in A text{ and } b in B }.
    $$

    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 days ago

























    answered 2 days ago









    Ethan BolkerEthan Bolker

    46k553120




    46k553120












    • $begingroup$
      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 days 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 days 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 days 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 days 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 days 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 days 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 days 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 days ago











    7












    $begingroup$

    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 cosets. Using $H$ itself might be too trivial, so it seems most promising 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
    $$(132)Hcirc(123)H =^? eH = H$$



    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 it 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 days ago
















    7












    $begingroup$

    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 cosets. Using $H$ itself might be too trivial, so it seems most promising 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
    $$(132)Hcirc(123)H =^? eH = H$$



    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 it 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 days ago














    7












    7








    7





    $begingroup$

    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 cosets. Using $H$ itself might be too trivial, so it seems most promising 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
    $$(132)Hcirc(123)H =^? eH = H$$



    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 it 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 cosets. Using $H$ itself might be too trivial, so it seems most promising 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
    $$(132)Hcirc(123)H =^? eH = H$$



    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 it 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








    edited yesterday

























    answered 2 days ago









    Henning MakholmHenning Makholm

    243k17311555




    243k17311555












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


















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
















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




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











    1












    $begingroup$

    Here's a different description that gives a different lens on all of this.



    An equivalence relation on a set is a binary relation (call it $sim$) that is reflexive ($xsim x$), symmetric ($xsim yimplies ysim x$), and transitive ($xsim y land ysim z implies xsim z$). Given a group $G$ with a subgroup $H$, we can define an equivalence relation on the underlying set of $G$ via $asim b iff ab^{-1}in H$. Prove that this is indeed an equivalence relation.



    An equivalence class of an element $x$ with respect to some equivalence relation $sim$ is the set ${ymid xsim y}$. I'll use $[x]$ as notation for the equivalence class of $x$ (where the relevant equivalence relation is left implicit and resolved from context). For the equivalence relation above on $G$, what does $[g]$ look like for an arbitrary element of $G$? One particularly notable case is $[1]={gmid 1sim g}={gmid g^{-1}in H}=H$.



    Since our example is on a group, it would be nice to know that the equivalence relation respects the group operation. The idea is that we want to think of $xsim y$ as $x$ "equals" $y$, and so if $xsim x'$ and $ysim y'$ we would want $xysim x'y'$ as would obviously be the case for equality. (Show that this is the same as $[x][y]=[xy]$ where the multiplication of two sets of group elements is defined as $XY={xymid xin Xland yin Y}$.) An equivalence that respects the group operation is called a congruence. (More generally, a congruence refers to an equivalence relation that respects whatever operations are relevant. For example, a congruence on rings would respect multiplication and addition.) Alternatively, if we're given an equivalence relation, we can say that an operation is well-defined if it takes "equals" to "equals". In other words, "the group operation is well-defined with respect to $sim$" is the same as "$sim$ respects the group operation". The only difference is one of connotation, namely which of the group operation or the equivalence relation we would "blame" if the statement failed to hold.



    So the first question is: is the equivalence defined above a congruence?



    Given any equivalence relation on a group, we can extend it to a congruence. That is, we can consider the smallest congruence that contains the original equivalence relation. This is called the congruence generated by the equivalence relation. It's quite possible that the generated congruence is the same as the original equivalence relation. Now let's assume we've been given a congruence on $G$ written $approx$. We have $aapprox biff ab^{-1}approx 1$. (Why?) We can thus show $aapprox biff ab^{-1}in[1]$, mimicking our first equivalence relation.



    Second and third questions: What does $[1]$ look like in this case? What does $[1]$ look like for the congruence generated by our first equivalence relation, $sim$?






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Here's a different description that gives a different lens on all of this.



      An equivalence relation on a set is a binary relation (call it $sim$) that is reflexive ($xsim x$), symmetric ($xsim yimplies ysim x$), and transitive ($xsim y land ysim z implies xsim z$). Given a group $G$ with a subgroup $H$, we can define an equivalence relation on the underlying set of $G$ via $asim b iff ab^{-1}in H$. Prove that this is indeed an equivalence relation.



      An equivalence class of an element $x$ with respect to some equivalence relation $sim$ is the set ${ymid xsim y}$. I'll use $[x]$ as notation for the equivalence class of $x$ (where the relevant equivalence relation is left implicit and resolved from context). For the equivalence relation above on $G$, what does $[g]$ look like for an arbitrary element of $G$? One particularly notable case is $[1]={gmid 1sim g}={gmid g^{-1}in H}=H$.



      Since our example is on a group, it would be nice to know that the equivalence relation respects the group operation. The idea is that we want to think of $xsim y$ as $x$ "equals" $y$, and so if $xsim x'$ and $ysim y'$ we would want $xysim x'y'$ as would obviously be the case for equality. (Show that this is the same as $[x][y]=[xy]$ where the multiplication of two sets of group elements is defined as $XY={xymid xin Xland yin Y}$.) An equivalence that respects the group operation is called a congruence. (More generally, a congruence refers to an equivalence relation that respects whatever operations are relevant. For example, a congruence on rings would respect multiplication and addition.) Alternatively, if we're given an equivalence relation, we can say that an operation is well-defined if it takes "equals" to "equals". In other words, "the group operation is well-defined with respect to $sim$" is the same as "$sim$ respects the group operation". The only difference is one of connotation, namely which of the group operation or the equivalence relation we would "blame" if the statement failed to hold.



      So the first question is: is the equivalence defined above a congruence?



      Given any equivalence relation on a group, we can extend it to a congruence. That is, we can consider the smallest congruence that contains the original equivalence relation. This is called the congruence generated by the equivalence relation. It's quite possible that the generated congruence is the same as the original equivalence relation. Now let's assume we've been given a congruence on $G$ written $approx$. We have $aapprox biff ab^{-1}approx 1$. (Why?) We can thus show $aapprox biff ab^{-1}in[1]$, mimicking our first equivalence relation.



      Second and third questions: What does $[1]$ look like in this case? What does $[1]$ look like for the congruence generated by our first equivalence relation, $sim$?






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Here's a different description that gives a different lens on all of this.



        An equivalence relation on a set is a binary relation (call it $sim$) that is reflexive ($xsim x$), symmetric ($xsim yimplies ysim x$), and transitive ($xsim y land ysim z implies xsim z$). Given a group $G$ with a subgroup $H$, we can define an equivalence relation on the underlying set of $G$ via $asim b iff ab^{-1}in H$. Prove that this is indeed an equivalence relation.



        An equivalence class of an element $x$ with respect to some equivalence relation $sim$ is the set ${ymid xsim y}$. I'll use $[x]$ as notation for the equivalence class of $x$ (where the relevant equivalence relation is left implicit and resolved from context). For the equivalence relation above on $G$, what does $[g]$ look like for an arbitrary element of $G$? One particularly notable case is $[1]={gmid 1sim g}={gmid g^{-1}in H}=H$.



        Since our example is on a group, it would be nice to know that the equivalence relation respects the group operation. The idea is that we want to think of $xsim y$ as $x$ "equals" $y$, and so if $xsim x'$ and $ysim y'$ we would want $xysim x'y'$ as would obviously be the case for equality. (Show that this is the same as $[x][y]=[xy]$ where the multiplication of two sets of group elements is defined as $XY={xymid xin Xland yin Y}$.) An equivalence that respects the group operation is called a congruence. (More generally, a congruence refers to an equivalence relation that respects whatever operations are relevant. For example, a congruence on rings would respect multiplication and addition.) Alternatively, if we're given an equivalence relation, we can say that an operation is well-defined if it takes "equals" to "equals". In other words, "the group operation is well-defined with respect to $sim$" is the same as "$sim$ respects the group operation". The only difference is one of connotation, namely which of the group operation or the equivalence relation we would "blame" if the statement failed to hold.



        So the first question is: is the equivalence defined above a congruence?



        Given any equivalence relation on a group, we can extend it to a congruence. That is, we can consider the smallest congruence that contains the original equivalence relation. This is called the congruence generated by the equivalence relation. It's quite possible that the generated congruence is the same as the original equivalence relation. Now let's assume we've been given a congruence on $G$ written $approx$. We have $aapprox biff ab^{-1}approx 1$. (Why?) We can thus show $aapprox biff ab^{-1}in[1]$, mimicking our first equivalence relation.



        Second and third questions: What does $[1]$ look like in this case? What does $[1]$ look like for the congruence generated by our first equivalence relation, $sim$?






        share|cite|improve this answer









        $endgroup$



        Here's a different description that gives a different lens on all of this.



        An equivalence relation on a set is a binary relation (call it $sim$) that is reflexive ($xsim x$), symmetric ($xsim yimplies ysim x$), and transitive ($xsim y land ysim z implies xsim z$). Given a group $G$ with a subgroup $H$, we can define an equivalence relation on the underlying set of $G$ via $asim b iff ab^{-1}in H$. Prove that this is indeed an equivalence relation.



        An equivalence class of an element $x$ with respect to some equivalence relation $sim$ is the set ${ymid xsim y}$. I'll use $[x]$ as notation for the equivalence class of $x$ (where the relevant equivalence relation is left implicit and resolved from context). For the equivalence relation above on $G$, what does $[g]$ look like for an arbitrary element of $G$? One particularly notable case is $[1]={gmid 1sim g}={gmid g^{-1}in H}=H$.



        Since our example is on a group, it would be nice to know that the equivalence relation respects the group operation. The idea is that we want to think of $xsim y$ as $x$ "equals" $y$, and so if $xsim x'$ and $ysim y'$ we would want $xysim x'y'$ as would obviously be the case for equality. (Show that this is the same as $[x][y]=[xy]$ where the multiplication of two sets of group elements is defined as $XY={xymid xin Xland yin Y}$.) An equivalence that respects the group operation is called a congruence. (More generally, a congruence refers to an equivalence relation that respects whatever operations are relevant. For example, a congruence on rings would respect multiplication and addition.) Alternatively, if we're given an equivalence relation, we can say that an operation is well-defined if it takes "equals" to "equals". In other words, "the group operation is well-defined with respect to $sim$" is the same as "$sim$ respects the group operation". The only difference is one of connotation, namely which of the group operation or the equivalence relation we would "blame" if the statement failed to hold.



        So the first question is: is the equivalence defined above a congruence?



        Given any equivalence relation on a group, we can extend it to a congruence. That is, we can consider the smallest congruence that contains the original equivalence relation. This is called the congruence generated by the equivalence relation. It's quite possible that the generated congruence is the same as the original equivalence relation. Now let's assume we've been given a congruence on $G$ written $approx$. We have $aapprox biff ab^{-1}approx 1$. (Why?) We can thus show $aapprox biff ab^{-1}in[1]$, mimicking our first equivalence relation.



        Second and third questions: What does $[1]$ look like in this case? What does $[1]$ look like for the congruence generated by our first equivalence relation, $sim$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Derek ElkinsDerek Elkins

        17.6k11437




        17.6k11437






























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