Show that $bigcap_{a in G} a H a^{-1}$ is a normal subgroup of $G$.












1












$begingroup$


I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



In this book, there is the following problem.



I think this problem is easy but a little abstract for me.



Is my answer ok?




Let $G$ be a group.

Let $H$ be a subgroup of $G$.

Let $N = bigcap_{a in G} a H a^{-1}$.



(1)

Show that $N$ is a normal subgroup of $G$.



(2)

Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




(1)

Let $g_1, g_2$ be arbitrary elements of $G$.

Let $n$ be an arbitrary element of $N$.

Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
$g_2$ was an arbitrary element of $G$.

So, $g_1 n g_1^{-1} in N$.

So, $N$ is a normal subgroup of $G$.



(2)

Let $m$ be an arbitrary element of $M$.

Let $g_3$ be an arbitrary element of $G$.

Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
$m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
$g_3$ was an arbitrary element of $G$.

So, $m in N$.






abstract-algebra group-theory proof-verification







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



    In this book, there is the following problem.



    I think this problem is easy but a little abstract for me.



    Is my answer ok?




    Let $G$ be a group.

    Let $H$ be a subgroup of $G$.

    Let $N = bigcap_{a in G} a H a^{-1}$.



    (1)

    Show that $N$ is a normal subgroup of $G$.



    (2)

    Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




    (1)

    Let $g_1, g_2$ be arbitrary elements of $G$.

    Let $n$ be an arbitrary element of $N$.

    Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

    So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

    Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
    $g_2$ was an arbitrary element of $G$.

    So, $g_1 n g_1^{-1} in N$.

    So, $N$ is a normal subgroup of $G$.



    (2)

    Let $m$ be an arbitrary element of $M$.

    Let $g_3$ be an arbitrary element of $G$.

    Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
    $m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
    $g_3$ was an arbitrary element of $G$.

    So, $m in N$.






    abstract-algebra group-theory proof-verification







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



      In this book, there is the following problem.



      I think this problem is easy but a little abstract for me.



      Is my answer ok?




      Let $G$ be a group.

      Let $H$ be a subgroup of $G$.

      Let $N = bigcap_{a in G} a H a^{-1}$.



      (1)

      Show that $N$ is a normal subgroup of $G$.



      (2)

      Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




      (1)

      Let $g_1, g_2$ be arbitrary elements of $G$.

      Let $n$ be an arbitrary element of $N$.

      Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

      So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

      Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
      $g_2$ was an arbitrary element of $G$.

      So, $g_1 n g_1^{-1} in N$.

      So, $N$ is a normal subgroup of $G$.



      (2)

      Let $m$ be an arbitrary element of $M$.

      Let $g_3$ be an arbitrary element of $G$.

      Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
      $m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
      $g_3$ was an arbitrary element of $G$.

      So, $m in N$.






      abstract-algebra group-theory proof-verification







      share|cite|improve this question









      $endgroup$




      I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



      In this book, there is the following problem.



      I think this problem is easy but a little abstract for me.



      Is my answer ok?




      Let $G$ be a group.

      Let $H$ be a subgroup of $G$.

      Let $N = bigcap_{a in G} a H a^{-1}$.



      (1)

      Show that $N$ is a normal subgroup of $G$.



      (2)

      Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




      (1)

      Let $g_1, g_2$ be arbitrary elements of $G$.

      Let $n$ be an arbitrary element of $N$.

      Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

      So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

      Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
      $g_2$ was an arbitrary element of $G$.

      So, $g_1 n g_1^{-1} in N$.

      So, $N$ is a normal subgroup of $G$.



      (2)

      Let $m$ be an arbitrary element of $M$.

      Let $g_3$ be an arbitrary element of $G$.

      Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
      $m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
      $g_3$ was an arbitrary element of $G$.

      So, $m in N$.






      abstract-algebra group-theory proof-verification




      abstract-algebra group-theory proof-verification






      share|cite|improve this question













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      asked Jan 23 at 11:49









      tchappy hatchappy ha

      766412




      766412






















          1 Answer
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          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07
















          1












          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07














          1












          1








          1





          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$



          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 11:58









          Sorin TircSorin Tirc

          1,810213




          1,810213








          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07














          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07








          1




          1




          $begingroup$
          Thank you very much, for clear proof, Sorin Tirc.
          $endgroup$
          – tchappy ha
          Jan 23 at 12:07




          $begingroup$
          Thank you very much, for clear proof, Sorin Tirc.
          $endgroup$
          – tchappy ha
          Jan 23 at 12:07


















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