Few group theory questions












2












$begingroup$


I am trying to solve the following;



First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$



I can show that it is reflexive as the identity is always in the subgroup.



if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.



Now I must determine if $G$ being abelian is required for this to be transitive.



My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?



And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.



My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?



Thank you all






group-theory relations group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
    $endgroup$
    – fleablood
    Dec 8 '15 at 21:32
















2












$begingroup$


I am trying to solve the following;



First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$



I can show that it is reflexive as the identity is always in the subgroup.



if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.



Now I must determine if $G$ being abelian is required for this to be transitive.



My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?



And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.



My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?



Thank you all






group-theory relations group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
    $endgroup$
    – fleablood
    Dec 8 '15 at 21:32














2












2








2





$begingroup$


I am trying to solve the following;



First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$



I can show that it is reflexive as the identity is always in the subgroup.



if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.



Now I must determine if $G$ being abelian is required for this to be transitive.



My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?



And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.



My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?



Thank you all






group-theory relations group-homomorphism







share|cite|improve this question











$endgroup$




I am trying to solve the following;



First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$



I can show that it is reflexive as the identity is always in the subgroup.



if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.



Now I must determine if $G$ being abelian is required for this to be transitive.



My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?



And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.



My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?



Thank you all






group-theory relations group-homomorphism




group-theory relations group-homomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 22:43









the_fox

2,57711432




2,57711432










asked Dec 8 '15 at 21:17









PersonaAPersonaA

329617




329617












  • $begingroup$
    I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
    $endgroup$
    – fleablood
    Dec 8 '15 at 21:32


















  • $begingroup$
    I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
    $endgroup$
    – fleablood
    Dec 8 '15 at 21:32
















$begingroup$
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
$endgroup$
– fleablood
Dec 8 '15 at 21:32




$begingroup$
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
$endgroup$
– fleablood
Dec 8 '15 at 21:32










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.



So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.



Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$



And $an = 0$ for all $a$ in $Bbb Z/n$.



So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.



So $phi$ is the zero homomorphism. It's the only possible homomorphism.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$


    1. Correct, commutativity is not needed and not used in your proof.

    2. The group operation is rather addition.
      Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
      $endgroup$
      – PersonaA
      Dec 8 '15 at 21:35










    • $begingroup$
      The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
      $endgroup$
      – fleablood
      Dec 8 '15 at 21:43










    • $begingroup$
      Wouldn't it be all of them relatively ptime?
      $endgroup$
      – PersonaA
      Dec 8 '15 at 21:47












    • $begingroup$
      But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
      $endgroup$
      – fleablood
      Dec 8 '15 at 21:48










    • $begingroup$
      Only a=0 then ?
      $endgroup$
      – PersonaA
      Dec 8 '15 at 21:48











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    2 Answers
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    2 Answers
    2






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    active

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    active

    oldest

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    1












    $begingroup$

    Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.



    So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.



    Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$



    And $an = 0$ for all $a$ in $Bbb Z/n$.



    So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.



    So $phi$ is the zero homomorphism. It's the only possible homomorphism.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.



      So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.



      Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$



      And $an = 0$ for all $a$ in $Bbb Z/n$.



      So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.



      So $phi$ is the zero homomorphism. It's the only possible homomorphism.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.



        So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.



        Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$



        And $an = 0$ for all $a$ in $Bbb Z/n$.



        So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.



        So $phi$ is the zero homomorphism. It's the only possible homomorphism.






        share|cite|improve this answer











        $endgroup$



        Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.



        So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.



        Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$



        And $an = 0$ for all $a$ in $Bbb Z/n$.



        So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.



        So $phi$ is the zero homomorphism. It's the only possible homomorphism.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 22:21









        Shaun

        8,832113681




        8,832113681










        answered Dec 8 '15 at 21:53









        fleabloodfleablood

        68.7k22685




        68.7k22685























            0












            $begingroup$


            1. Correct, commutativity is not needed and not used in your proof.

            2. The group operation is rather addition.
              Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:35










            • $begingroup$
              The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:43










            • $begingroup$
              Wouldn't it be all of them relatively ptime?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:47












            • $begingroup$
              But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:48










            • $begingroup$
              Only a=0 then ?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:48
















            0












            $begingroup$


            1. Correct, commutativity is not needed and not used in your proof.

            2. The group operation is rather addition.
              Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:35










            • $begingroup$
              The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:43










            • $begingroup$
              Wouldn't it be all of them relatively ptime?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:47












            • $begingroup$
              But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:48










            • $begingroup$
              Only a=0 then ?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:48














            0












            0








            0





            $begingroup$


            1. Correct, commutativity is not needed and not used in your proof.

            2. The group operation is rather addition.
              Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?






            share|cite|improve this answer









            $endgroup$




            1. Correct, commutativity is not needed and not used in your proof.

            2. The group operation is rather addition.
              Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '15 at 21:23









            BerciBerci

            60k23672




            60k23672












            • $begingroup$
              So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:35










            • $begingroup$
              The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:43










            • $begingroup$
              Wouldn't it be all of them relatively ptime?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:47












            • $begingroup$
              But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:48










            • $begingroup$
              Only a=0 then ?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:48


















            • $begingroup$
              So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:35










            • $begingroup$
              The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:43










            • $begingroup$
              Wouldn't it be all of them relatively ptime?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:47












            • $begingroup$
              But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
              $endgroup$
              – fleablood
              Dec 8 '15 at 21:48










            • $begingroup$
              Only a=0 then ?
              $endgroup$
              – PersonaA
              Dec 8 '15 at 21:48
















            $begingroup$
            So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
            $endgroup$
            – PersonaA
            Dec 8 '15 at 21:35




            $begingroup$
            So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
            $endgroup$
            – PersonaA
            Dec 8 '15 at 21:35












            $begingroup$
            The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
            $endgroup$
            – fleablood
            Dec 8 '15 at 21:43




            $begingroup$
            The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
            $endgroup$
            – fleablood
            Dec 8 '15 at 21:43












            $begingroup$
            Wouldn't it be all of them relatively ptime?
            $endgroup$
            – PersonaA
            Dec 8 '15 at 21:47






            $begingroup$
            Wouldn't it be all of them relatively ptime?
            $endgroup$
            – PersonaA
            Dec 8 '15 at 21:47














            $begingroup$
            But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
            $endgroup$
            – fleablood
            Dec 8 '15 at 21:48




            $begingroup$
            But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
            $endgroup$
            – fleablood
            Dec 8 '15 at 21:48












            $begingroup$
            Only a=0 then ?
            $endgroup$
            – PersonaA
            Dec 8 '15 at 21:48




            $begingroup$
            Only a=0 then ?
            $endgroup$
            – PersonaA
            Dec 8 '15 at 21:48


















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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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