How do we know $b$ is in $G$? The Next CEO of Stack Overflowa question about group (decomposition of conjugacy classes in normal subgroups)Normal subgroups of $langle(123),(456),(23)(56)rangle$Explanation/How to use the Lattice isomorphism theoremShowing that the intersection of all subgroups of order $n$ is normalLeft Cosets of Cyclic SubgroupOrder of elements in a cyclic group ($mathbb Z_26$)What exactly is the purpose of a coset?Understanding normal subgroup using invariance under conjugationHow do I prove this group is closed under multiplication?If $N/M$ is a normal subgroup of $G/M$ then $N$ is a normal in $G$

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How do we know $b$ is in $G$?



The Next CEO of Stack Overflowa question about group (decomposition of conjugacy classes in normal subgroups)Normal subgroups of $langle(123),(456),(23)(56)rangle$Explanation/How to use the Lattice isomorphism theoremShowing that the intersection of all subgroups of order $n$ is normalLeft Cosets of Cyclic SubgroupOrder of elements in a cyclic group ($mathbb Z_26$)What exactly is the purpose of a coset?Understanding normal subgroup using invariance under conjugationHow do I prove this group is closed under multiplication?If $N/M$ is a normal subgroup of $G/M$ then $N$ is a normal in $G$










2












$begingroup$


I'm having trouble understanding when we can assume that an element is in a group.



Given problem



Let $H$ be a normal subgroup of $G$. Prove that if $ab in H$, then $ba in H$.



Given solution



Let $H$ be a normal subgroup of $G$ and suppose $ab in H$; say $ab = h in H$. Since $H$ is normal, $H$ is closed under conjugation by elements of $G$. In particular, $bhb^-1 in H$. Substituting $h=ab$, we see that $babb^-1 = ba in H$.



Question



Why can we assume $b in G$?



My understanding is that even if the product of two elements is in a group, we can't necessarily assume that the individual elements are also in that group. I'd assume that the same goes for large groups also, i.e., $ab in G nRightarrow a,b in G$.






group-theory







share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    Because all your elements need to be taken from $G$; otherwise, the entire problem doesn't make sense from the start. So $a$ and $b$ are assumed to be elements of $G$, as understood from context.
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:30







  • 4




    $begingroup$
    A more pedantic wording of the original problem would have said something like "Let $H$ be a normal subgroup of $G$ and let $a, b in G$. Prove ...". However, even as written, context makes clear that $a,b in G$ (where else could they come from?)
    $endgroup$
    – Morgan Sherman
    Mar 19 at 20:34






  • 1




    $begingroup$
    What you label "your understanding" is true for subgroups; however, there needs to be a contextual set-with-an-operation from which everything is taken. In some cases this context need not be clear (for example, if we are talking about addition of integers, is our group the integers, the rationals, the reals, the complex numbers, some other set?), in which case it needs to be specified explicitly where the elements are being taken from. But here the only possible context is $G$. So there is an implicity "if $a,bin G$ are such that" in between "Prove that" and "$abin H$".
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:35










  • $begingroup$
    All very helpful. Thank you.
    $endgroup$
    – Alex Johnson
    Mar 19 at 20:38















2












$begingroup$


I'm having trouble understanding when we can assume that an element is in a group.



Given problem



Let $H$ be a normal subgroup of $G$. Prove that if $ab in H$, then $ba in H$.



Given solution



Let $H$ be a normal subgroup of $G$ and suppose $ab in H$; say $ab = h in H$. Since $H$ is normal, $H$ is closed under conjugation by elements of $G$. In particular, $bhb^-1 in H$. Substituting $h=ab$, we see that $babb^-1 = ba in H$.



Question



Why can we assume $b in G$?



My understanding is that even if the product of two elements is in a group, we can't necessarily assume that the individual elements are also in that group. I'd assume that the same goes for large groups also, i.e., $ab in G nRightarrow a,b in G$.






group-theory







share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    Because all your elements need to be taken from $G$; otherwise, the entire problem doesn't make sense from the start. So $a$ and $b$ are assumed to be elements of $G$, as understood from context.
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:30







  • 4




    $begingroup$
    A more pedantic wording of the original problem would have said something like "Let $H$ be a normal subgroup of $G$ and let $a, b in G$. Prove ...". However, even as written, context makes clear that $a,b in G$ (where else could they come from?)
    $endgroup$
    – Morgan Sherman
    Mar 19 at 20:34






  • 1




    $begingroup$
    What you label "your understanding" is true for subgroups; however, there needs to be a contextual set-with-an-operation from which everything is taken. In some cases this context need not be clear (for example, if we are talking about addition of integers, is our group the integers, the rationals, the reals, the complex numbers, some other set?), in which case it needs to be specified explicitly where the elements are being taken from. But here the only possible context is $G$. So there is an implicity "if $a,bin G$ are such that" in between "Prove that" and "$abin H$".
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:35










  • $begingroup$
    All very helpful. Thank you.
    $endgroup$
    – Alex Johnson
    Mar 19 at 20:38













2












2








2





$begingroup$


I'm having trouble understanding when we can assume that an element is in a group.



Given problem



Let $H$ be a normal subgroup of $G$. Prove that if $ab in H$, then $ba in H$.



Given solution



Let $H$ be a normal subgroup of $G$ and suppose $ab in H$; say $ab = h in H$. Since $H$ is normal, $H$ is closed under conjugation by elements of $G$. In particular, $bhb^-1 in H$. Substituting $h=ab$, we see that $babb^-1 = ba in H$.



Question



Why can we assume $b in G$?



My understanding is that even if the product of two elements is in a group, we can't necessarily assume that the individual elements are also in that group. I'd assume that the same goes for large groups also, i.e., $ab in G nRightarrow a,b in G$.






group-theory







share|cite|improve this question











$endgroup$




I'm having trouble understanding when we can assume that an element is in a group.



Given problem



Let $H$ be a normal subgroup of $G$. Prove that if $ab in H$, then $ba in H$.



Given solution



Let $H$ be a normal subgroup of $G$ and suppose $ab in H$; say $ab = h in H$. Since $H$ is normal, $H$ is closed under conjugation by elements of $G$. In particular, $bhb^-1 in H$. Substituting $h=ab$, we see that $babb^-1 = ba in H$.



Question



Why can we assume $b in G$?



My understanding is that even if the product of two elements is in a group, we can't necessarily assume that the individual elements are also in that group. I'd assume that the same goes for large groups also, i.e., $ab in G nRightarrow a,b in G$.






group-theory




group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 19 at 20:30









Arturo Magidin

266k34590920




266k34590920










asked Mar 19 at 20:26









Alex JohnsonAlex Johnson

1468




1468







  • 4




    $begingroup$
    Because all your elements need to be taken from $G$; otherwise, the entire problem doesn't make sense from the start. So $a$ and $b$ are assumed to be elements of $G$, as understood from context.
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:30







  • 4




    $begingroup$
    A more pedantic wording of the original problem would have said something like "Let $H$ be a normal subgroup of $G$ and let $a, b in G$. Prove ...". However, even as written, context makes clear that $a,b in G$ (where else could they come from?)
    $endgroup$
    – Morgan Sherman
    Mar 19 at 20:34






  • 1




    $begingroup$
    What you label "your understanding" is true for subgroups; however, there needs to be a contextual set-with-an-operation from which everything is taken. In some cases this context need not be clear (for example, if we are talking about addition of integers, is our group the integers, the rationals, the reals, the complex numbers, some other set?), in which case it needs to be specified explicitly where the elements are being taken from. But here the only possible context is $G$. So there is an implicity "if $a,bin G$ are such that" in between "Prove that" and "$abin H$".
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:35










  • $begingroup$
    All very helpful. Thank you.
    $endgroup$
    – Alex Johnson
    Mar 19 at 20:38












  • 4




    $begingroup$
    Because all your elements need to be taken from $G$; otherwise, the entire problem doesn't make sense from the start. So $a$ and $b$ are assumed to be elements of $G$, as understood from context.
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:30







  • 4




    $begingroup$
    A more pedantic wording of the original problem would have said something like "Let $H$ be a normal subgroup of $G$ and let $a, b in G$. Prove ...". However, even as written, context makes clear that $a,b in G$ (where else could they come from?)
    $endgroup$
    – Morgan Sherman
    Mar 19 at 20:34






  • 1




    $begingroup$
    What you label "your understanding" is true for subgroups; however, there needs to be a contextual set-with-an-operation from which everything is taken. In some cases this context need not be clear (for example, if we are talking about addition of integers, is our group the integers, the rationals, the reals, the complex numbers, some other set?), in which case it needs to be specified explicitly where the elements are being taken from. But here the only possible context is $G$. So there is an implicity "if $a,bin G$ are such that" in between "Prove that" and "$abin H$".
    $endgroup$
    – Arturo Magidin
    Mar 19 at 20:35










  • $begingroup$
    All very helpful. Thank you.
    $endgroup$
    – Alex Johnson
    Mar 19 at 20:38







4




4




$begingroup$
Because all your elements need to be taken from $G$; otherwise, the entire problem doesn't make sense from the start. So $a$ and $b$ are assumed to be elements of $G$, as understood from context.
$endgroup$
– Arturo Magidin
Mar 19 at 20:30





$begingroup$
Because all your elements need to be taken from $G$; otherwise, the entire problem doesn't make sense from the start. So $a$ and $b$ are assumed to be elements of $G$, as understood from context.
$endgroup$
– Arturo Magidin
Mar 19 at 20:30





4




4




$begingroup$
A more pedantic wording of the original problem would have said something like "Let $H$ be a normal subgroup of $G$ and let $a, b in G$. Prove ...". However, even as written, context makes clear that $a,b in G$ (where else could they come from?)
$endgroup$
– Morgan Sherman
Mar 19 at 20:34




$begingroup$
A more pedantic wording of the original problem would have said something like "Let $H$ be a normal subgroup of $G$ and let $a, b in G$. Prove ...". However, even as written, context makes clear that $a,b in G$ (where else could they come from?)
$endgroup$
– Morgan Sherman
Mar 19 at 20:34




1




1




$begingroup$
What you label "your understanding" is true for subgroups; however, there needs to be a contextual set-with-an-operation from which everything is taken. In some cases this context need not be clear (for example, if we are talking about addition of integers, is our group the integers, the rationals, the reals, the complex numbers, some other set?), in which case it needs to be specified explicitly where the elements are being taken from. But here the only possible context is $G$. So there is an implicity "if $a,bin G$ are such that" in between "Prove that" and "$abin H$".
$endgroup$
– Arturo Magidin
Mar 19 at 20:35




$begingroup$
What you label "your understanding" is true for subgroups; however, there needs to be a contextual set-with-an-operation from which everything is taken. In some cases this context need not be clear (for example, if we are talking about addition of integers, is our group the integers, the rationals, the reals, the complex numbers, some other set?), in which case it needs to be specified explicitly where the elements are being taken from. But here the only possible context is $G$. So there is an implicity "if $a,bin G$ are such that" in between "Prove that" and "$abin H$".
$endgroup$
– Arturo Magidin
Mar 19 at 20:35












$begingroup$
All very helpful. Thank you.
$endgroup$
– Alex Johnson
Mar 19 at 20:38




$begingroup$
All very helpful. Thank you.
$endgroup$
– Alex Johnson
Mar 19 at 20:38










1 Answer
1






active

oldest

votes


















1












$begingroup$

Well, your binary group operation $cdot$ is only defined on $G$. That is, its domain is $G times G$. When you write $ab$, this a shorthand for $a cdot b$. But then for the expression to make sense, $a$ and $b$ must be in $G$. Therefore it's a literature convention to imply that by writing $acdot b in H$, we actually mean "$a in G, b in G, a cdot b in H$".



Note that is very different than saying that "$acdot b in H$" implies "$ain H, b in H, ab in H$", because $cdot$ is well-defined on a superset of $H times H$. So, usually literature just implies the minimal requirement for the expression to make sense, that is, $a$ and $b$ are in $G$.






share|cite|improve this answer











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    active

    oldest

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    1












    $begingroup$

    Well, your binary group operation $cdot$ is only defined on $G$. That is, its domain is $G times G$. When you write $ab$, this a shorthand for $a cdot b$. But then for the expression to make sense, $a$ and $b$ must be in $G$. Therefore it's a literature convention to imply that by writing $acdot b in H$, we actually mean "$a in G, b in G, a cdot b in H$".



    Note that is very different than saying that "$acdot b in H$" implies "$ain H, b in H, ab in H$", because $cdot$ is well-defined on a superset of $H times H$. So, usually literature just implies the minimal requirement for the expression to make sense, that is, $a$ and $b$ are in $G$.






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      Well, your binary group operation $cdot$ is only defined on $G$. That is, its domain is $G times G$. When you write $ab$, this a shorthand for $a cdot b$. But then for the expression to make sense, $a$ and $b$ must be in $G$. Therefore it's a literature convention to imply that by writing $acdot b in H$, we actually mean "$a in G, b in G, a cdot b in H$".



      Note that is very different than saying that "$acdot b in H$" implies "$ain H, b in H, ab in H$", because $cdot$ is well-defined on a superset of $H times H$. So, usually literature just implies the minimal requirement for the expression to make sense, that is, $a$ and $b$ are in $G$.






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        Well, your binary group operation $cdot$ is only defined on $G$. That is, its domain is $G times G$. When you write $ab$, this a shorthand for $a cdot b$. But then for the expression to make sense, $a$ and $b$ must be in $G$. Therefore it's a literature convention to imply that by writing $acdot b in H$, we actually mean "$a in G, b in G, a cdot b in H$".



        Note that is very different than saying that "$acdot b in H$" implies "$ain H, b in H, ab in H$", because $cdot$ is well-defined on a superset of $H times H$. So, usually literature just implies the minimal requirement for the expression to make sense, that is, $a$ and $b$ are in $G$.






        share|cite|improve this answer











        $endgroup$



        Well, your binary group operation $cdot$ is only defined on $G$. That is, its domain is $G times G$. When you write $ab$, this a shorthand for $a cdot b$. But then for the expression to make sense, $a$ and $b$ must be in $G$. Therefore it's a literature convention to imply that by writing $acdot b in H$, we actually mean "$a in G, b in G, a cdot b in H$".



        Note that is very different than saying that "$acdot b in H$" implies "$ain H, b in H, ab in H$", because $cdot$ is well-defined on a superset of $H times H$. So, usually literature just implies the minimal requirement for the expression to make sense, that is, $a$ and $b$ are in $G$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 19 at 20:47









        J. W. Tanner

        4,1961320




        4,1961320










        answered Mar 19 at 20:38









        LærneLærne

        2,189518




        2,189518



























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