Proving an Intersection of Subgroups is Normal












2












$begingroup$


Let $N$ be a subgroup of the group $G$.



Show that $$X =bigcap_{g in G} g^{-1}Ng $$
is normal in $G$.



It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.



But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$



and also directly from the definition $$Xg=gX quad forall g in G$$
Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.






group-theory normal-subgroups







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $N$ be a subgroup of the group $G$.



    Show that $$X =bigcap_{g in G} g^{-1}Ng $$
    is normal in $G$.



    It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.



    But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$



    and also directly from the definition $$Xg=gX quad forall g in G$$
    Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.






    group-theory normal-subgroups







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $N$ be a subgroup of the group $G$.



      Show that $$X =bigcap_{g in G} g^{-1}Ng $$
      is normal in $G$.



      It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.



      But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$



      and also directly from the definition $$Xg=gX quad forall g in G$$
      Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.






      group-theory normal-subgroups







      share|cite|improve this question











      $endgroup$




      Let $N$ be a subgroup of the group $G$.



      Show that $$X =bigcap_{g in G} g^{-1}Ng $$
      is normal in $G$.



      It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.



      But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg in X quad forall g in G, x in X $$



      and also directly from the definition $$Xg=gX quad forall g in G$$
      Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.






      group-theory normal-subgroups




      group-theory normal-subgroups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 17:19









      ervx

      10.3k31338




      10.3k31338










      asked Dec 2 '18 at 17:07









      B. ActonB. Acton

      376




      376






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
            $endgroup$
            – B. Acton
            Dec 2 '18 at 17:30










          • $begingroup$
            You are very welcome.
            $endgroup$
            – ervx
            Dec 3 '18 at 0:51



















          0












          $begingroup$

          Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
              $endgroup$
              – B. Acton
              Dec 2 '18 at 17:30










            • $begingroup$
              You are very welcome.
              $endgroup$
              – ervx
              Dec 3 '18 at 0:51
















            2












            $begingroup$

            Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
              $endgroup$
              – B. Acton
              Dec 2 '18 at 17:30










            • $begingroup$
              You are very welcome.
              $endgroup$
              – ervx
              Dec 3 '18 at 0:51














            2












            2








            2





            $begingroup$

            Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.






            share|cite|improve this answer









            $endgroup$



            Fix $xin X$ and $gin G$. We'll show that $g^{-1}xgin X$. Since $xin X$, it follows that $xin h^{-1}Nh$ for all $hin G$. Now, fix a particular $tin G$. Our goal is to show that $g^{-1}xgin t^{-1}Nt$. (This will prove that $g^{-1}xgin X$, since $t$ is arbitrary.) Note that $tg^{-1}in G$. Hence, by assumption, $xin (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xgin t^{-1}Nt$, as desired.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 2 '18 at 17:16









            ervxervx

            10.3k31338




            10.3k31338












            • $begingroup$
              Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
              $endgroup$
              – B. Acton
              Dec 2 '18 at 17:30










            • $begingroup$
              You are very welcome.
              $endgroup$
              – ervx
              Dec 3 '18 at 0:51


















            • $begingroup$
              Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
              $endgroup$
              – B. Acton
              Dec 2 '18 at 17:30










            • $begingroup$
              You are very welcome.
              $endgroup$
              – ervx
              Dec 3 '18 at 0:51
















            $begingroup$
            Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
            $endgroup$
            – B. Acton
            Dec 2 '18 at 17:30




            $begingroup$
            Thank you very much, I think I was sniffing around the right result, but messed up what I was keeping arbitrary/ fixed.
            $endgroup$
            – B. Acton
            Dec 2 '18 at 17:30












            $begingroup$
            You are very welcome.
            $endgroup$
            – ervx
            Dec 3 '18 at 0:51




            $begingroup$
            You are very welcome.
            $endgroup$
            – ervx
            Dec 3 '18 at 0:51











            0












            $begingroup$

            Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.






                share|cite|improve this answer









                $endgroup$



                Taking your former approach: for any $x in X$ and any $gin G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h in H$. But, in particular, $x in gh^{-1}Nhg^{-1}$ (since $hg^{-1}in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xgin g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 17:19









                user3482749user3482749

                4,057918




                4,057918






























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