Normal subgroups and functions from cosets to cosets












1












$begingroup$


Let $K ⊂ G$ be a normal subgroup and let $G/K$ be the set of left cosets of $K$.



Consider
the function
$G/K × G/K → G/K$ given by $(g1 · K, g2 · K) → (g1 · g2) · K$.



(a) Prove this function is well-defined.



(b) Prove this function makes $G/K$ a group.



What does it mean for the function to be well defined in this case?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $K ⊂ G$ be a normal subgroup and let $G/K$ be the set of left cosets of $K$.



    Consider
    the function
    $G/K × G/K → G/K$ given by $(g1 · K, g2 · K) → (g1 · g2) · K$.



    (a) Prove this function is well-defined.



    (b) Prove this function makes $G/K$ a group.



    What does it mean for the function to be well defined in this case?






    abstract-algebra group-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $K ⊂ G$ be a normal subgroup and let $G/K$ be the set of left cosets of $K$.



      Consider
      the function
      $G/K × G/K → G/K$ given by $(g1 · K, g2 · K) → (g1 · g2) · K$.



      (a) Prove this function is well-defined.



      (b) Prove this function makes $G/K$ a group.



      What does it mean for the function to be well defined in this case?






      abstract-algebra group-theory







      share|cite|improve this question











      $endgroup$




      Let $K ⊂ G$ be a normal subgroup and let $G/K$ be the set of left cosets of $K$.



      Consider
      the function
      $G/K × G/K → G/K$ given by $(g1 · K, g2 · K) → (g1 · g2) · K$.



      (a) Prove this function is well-defined.



      (b) Prove this function makes $G/K$ a group.



      What does it mean for the function to be well defined in this case?






      abstract-algebra group-theory




      abstract-algebra group-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 9:22









      Math Girl

      633318




      633318










      asked Dec 6 '18 at 8:18









      childishsadbinochildishsadbino

      1148




      1148






















          1 Answer
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          active

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          1












          $begingroup$

          The function has cosets as domain, but the definition is written using representatives of the cosets ($g_1$ and $g_2$ in this case). When writing it that way, there is always the danger of the function taking different values on the same cosets if you choose a different representative. Such a function would not be well-defined.



          Showing that the function is well-defined means showing that choosing different representatives for the same cosets as input to the function gives the same coset as output.



          To show what I mean, here is a function which is not well-defined. Take $f:Bbb Qto Bbb Z$ defined by $f(a/b) = a+b$. Then $f(1/1) = 2$, but $f(2/2) = 4$. The fractions $frac11$ and $frac22$ represent the same rational number, but give different function values. Thus $f$ is not well-defined.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you explain how to go about proving this for this example?
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:47










          • $begingroup$
            @childishsadbino A literal translation of my second paragraph into a concrete algebraic exercise would be this: Take $g_1, g_2, h_1, h_2in G$ such that $g_1K = h_1K$ and $g_2K = h_2K$ (i.e. choose different representatives for the same cosets). Then show that $(g_1g_2)K = (h_1h_2)K$ (i.e. that you get the same coset).
            $endgroup$
            – Arthur
            Dec 6 '18 at 8:50












          • $begingroup$
            Thank you, I was able to prove this from the definition of the normal subgroup K!
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:58










          • $begingroup$
            I am still unsure on how to prove the function makes G/K a group because I am unable to prove the inverse exists in G/K as well.
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:59






          • 2




            $begingroup$
            @childishsadbino That's as simple as it can be. $(gK)^{-1} = (g^{-1})K$.
            $endgroup$
            – Arthur
            Dec 6 '18 at 9:03













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          $begingroup$

          The function has cosets as domain, but the definition is written using representatives of the cosets ($g_1$ and $g_2$ in this case). When writing it that way, there is always the danger of the function taking different values on the same cosets if you choose a different representative. Such a function would not be well-defined.



          Showing that the function is well-defined means showing that choosing different representatives for the same cosets as input to the function gives the same coset as output.



          To show what I mean, here is a function which is not well-defined. Take $f:Bbb Qto Bbb Z$ defined by $f(a/b) = a+b$. Then $f(1/1) = 2$, but $f(2/2) = 4$. The fractions $frac11$ and $frac22$ represent the same rational number, but give different function values. Thus $f$ is not well-defined.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you explain how to go about proving this for this example?
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:47










          • $begingroup$
            @childishsadbino A literal translation of my second paragraph into a concrete algebraic exercise would be this: Take $g_1, g_2, h_1, h_2in G$ such that $g_1K = h_1K$ and $g_2K = h_2K$ (i.e. choose different representatives for the same cosets). Then show that $(g_1g_2)K = (h_1h_2)K$ (i.e. that you get the same coset).
            $endgroup$
            – Arthur
            Dec 6 '18 at 8:50












          • $begingroup$
            Thank you, I was able to prove this from the definition of the normal subgroup K!
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:58










          • $begingroup$
            I am still unsure on how to prove the function makes G/K a group because I am unable to prove the inverse exists in G/K as well.
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:59






          • 2




            $begingroup$
            @childishsadbino That's as simple as it can be. $(gK)^{-1} = (g^{-1})K$.
            $endgroup$
            – Arthur
            Dec 6 '18 at 9:03


















          1












          $begingroup$

          The function has cosets as domain, but the definition is written using representatives of the cosets ($g_1$ and $g_2$ in this case). When writing it that way, there is always the danger of the function taking different values on the same cosets if you choose a different representative. Such a function would not be well-defined.



          Showing that the function is well-defined means showing that choosing different representatives for the same cosets as input to the function gives the same coset as output.



          To show what I mean, here is a function which is not well-defined. Take $f:Bbb Qto Bbb Z$ defined by $f(a/b) = a+b$. Then $f(1/1) = 2$, but $f(2/2) = 4$. The fractions $frac11$ and $frac22$ represent the same rational number, but give different function values. Thus $f$ is not well-defined.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you explain how to go about proving this for this example?
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:47










          • $begingroup$
            @childishsadbino A literal translation of my second paragraph into a concrete algebraic exercise would be this: Take $g_1, g_2, h_1, h_2in G$ such that $g_1K = h_1K$ and $g_2K = h_2K$ (i.e. choose different representatives for the same cosets). Then show that $(g_1g_2)K = (h_1h_2)K$ (i.e. that you get the same coset).
            $endgroup$
            – Arthur
            Dec 6 '18 at 8:50












          • $begingroup$
            Thank you, I was able to prove this from the definition of the normal subgroup K!
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:58










          • $begingroup$
            I am still unsure on how to prove the function makes G/K a group because I am unable to prove the inverse exists in G/K as well.
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:59






          • 2




            $begingroup$
            @childishsadbino That's as simple as it can be. $(gK)^{-1} = (g^{-1})K$.
            $endgroup$
            – Arthur
            Dec 6 '18 at 9:03
















          1












          1








          1





          $begingroup$

          The function has cosets as domain, but the definition is written using representatives of the cosets ($g_1$ and $g_2$ in this case). When writing it that way, there is always the danger of the function taking different values on the same cosets if you choose a different representative. Such a function would not be well-defined.



          Showing that the function is well-defined means showing that choosing different representatives for the same cosets as input to the function gives the same coset as output.



          To show what I mean, here is a function which is not well-defined. Take $f:Bbb Qto Bbb Z$ defined by $f(a/b) = a+b$. Then $f(1/1) = 2$, but $f(2/2) = 4$. The fractions $frac11$ and $frac22$ represent the same rational number, but give different function values. Thus $f$ is not well-defined.






          share|cite|improve this answer









          $endgroup$



          The function has cosets as domain, but the definition is written using representatives of the cosets ($g_1$ and $g_2$ in this case). When writing it that way, there is always the danger of the function taking different values on the same cosets if you choose a different representative. Such a function would not be well-defined.



          Showing that the function is well-defined means showing that choosing different representatives for the same cosets as input to the function gives the same coset as output.



          To show what I mean, here is a function which is not well-defined. Take $f:Bbb Qto Bbb Z$ defined by $f(a/b) = a+b$. Then $f(1/1) = 2$, but $f(2/2) = 4$. The fractions $frac11$ and $frac22$ represent the same rational number, but give different function values. Thus $f$ is not well-defined.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 8:34









          ArthurArthur

          112k7107190




          112k7107190












          • $begingroup$
            Could you explain how to go about proving this for this example?
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:47










          • $begingroup$
            @childishsadbino A literal translation of my second paragraph into a concrete algebraic exercise would be this: Take $g_1, g_2, h_1, h_2in G$ such that $g_1K = h_1K$ and $g_2K = h_2K$ (i.e. choose different representatives for the same cosets). Then show that $(g_1g_2)K = (h_1h_2)K$ (i.e. that you get the same coset).
            $endgroup$
            – Arthur
            Dec 6 '18 at 8:50












          • $begingroup$
            Thank you, I was able to prove this from the definition of the normal subgroup K!
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:58










          • $begingroup$
            I am still unsure on how to prove the function makes G/K a group because I am unable to prove the inverse exists in G/K as well.
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:59






          • 2




            $begingroup$
            @childishsadbino That's as simple as it can be. $(gK)^{-1} = (g^{-1})K$.
            $endgroup$
            – Arthur
            Dec 6 '18 at 9:03




















          • $begingroup$
            Could you explain how to go about proving this for this example?
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:47










          • $begingroup$
            @childishsadbino A literal translation of my second paragraph into a concrete algebraic exercise would be this: Take $g_1, g_2, h_1, h_2in G$ such that $g_1K = h_1K$ and $g_2K = h_2K$ (i.e. choose different representatives for the same cosets). Then show that $(g_1g_2)K = (h_1h_2)K$ (i.e. that you get the same coset).
            $endgroup$
            – Arthur
            Dec 6 '18 at 8:50












          • $begingroup$
            Thank you, I was able to prove this from the definition of the normal subgroup K!
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:58










          • $begingroup$
            I am still unsure on how to prove the function makes G/K a group because I am unable to prove the inverse exists in G/K as well.
            $endgroup$
            – childishsadbino
            Dec 6 '18 at 8:59






          • 2




            $begingroup$
            @childishsadbino That's as simple as it can be. $(gK)^{-1} = (g^{-1})K$.
            $endgroup$
            – Arthur
            Dec 6 '18 at 9:03


















          $begingroup$
          Could you explain how to go about proving this for this example?
          $endgroup$
          – childishsadbino
          Dec 6 '18 at 8:47




          $begingroup$
          Could you explain how to go about proving this for this example?
          $endgroup$
          – childishsadbino
          Dec 6 '18 at 8:47












          $begingroup$
          @childishsadbino A literal translation of my second paragraph into a concrete algebraic exercise would be this: Take $g_1, g_2, h_1, h_2in G$ such that $g_1K = h_1K$ and $g_2K = h_2K$ (i.e. choose different representatives for the same cosets). Then show that $(g_1g_2)K = (h_1h_2)K$ (i.e. that you get the same coset).
          $endgroup$
          – Arthur
          Dec 6 '18 at 8:50






          $begingroup$
          @childishsadbino A literal translation of my second paragraph into a concrete algebraic exercise would be this: Take $g_1, g_2, h_1, h_2in G$ such that $g_1K = h_1K$ and $g_2K = h_2K$ (i.e. choose different representatives for the same cosets). Then show that $(g_1g_2)K = (h_1h_2)K$ (i.e. that you get the same coset).
          $endgroup$
          – Arthur
          Dec 6 '18 at 8:50














          $begingroup$
          Thank you, I was able to prove this from the definition of the normal subgroup K!
          $endgroup$
          – childishsadbino
          Dec 6 '18 at 8:58




          $begingroup$
          Thank you, I was able to prove this from the definition of the normal subgroup K!
          $endgroup$
          – childishsadbino
          Dec 6 '18 at 8:58












          $begingroup$
          I am still unsure on how to prove the function makes G/K a group because I am unable to prove the inverse exists in G/K as well.
          $endgroup$
          – childishsadbino
          Dec 6 '18 at 8:59




          $begingroup$
          I am still unsure on how to prove the function makes G/K a group because I am unable to prove the inverse exists in G/K as well.
          $endgroup$
          – childishsadbino
          Dec 6 '18 at 8:59




          2




          2




          $begingroup$
          @childishsadbino That's as simple as it can be. $(gK)^{-1} = (g^{-1})K$.
          $endgroup$
          – Arthur
          Dec 6 '18 at 9:03






          $begingroup$
          @childishsadbino That's as simple as it can be. $(gK)^{-1} = (g^{-1})K$.
          $endgroup$
          – Arthur
          Dec 6 '18 at 9:03




















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