Which of the following statements of group theory are true?

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Let $G$ be a group. Which of the following statements are true?



$1.$ The normalizer of a subgroup of $G $ is a normal subgroup of $G.$



$2.$ The centre of $G$ is a normal subgroup of $G.$



$3.$ If $H$ is a normal subgroup of G and is of order $2$, then H is contained in the centre of G.



My attempt:



For option $2)$ is true.



Let $xin Z(G)$ (center of $G$).



Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.



This proves $Z(G)$ is a normal subgroup.



I’m confused about option $1)$ and option $3)$. Are there any counterexamples?






abstract-algebra group-theory







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  • 4




    Did you try some small example to get an idea of whether they might be true?
    – Tobias Kildetoft
    5 hours ago






  • 1




    Your third question have answer here: math.stackexchange.com/questions/267443/…
    – Alan Wang
    4 hours ago














up vote
1
down vote

favorite












Let $G$ be a group. Which of the following statements are true?



$1.$ The normalizer of a subgroup of $G $ is a normal subgroup of $G.$



$2.$ The centre of $G$ is a normal subgroup of $G.$



$3.$ If $H$ is a normal subgroup of G and is of order $2$, then H is contained in the centre of G.



My attempt:



For option $2)$ is true.



Let $xin Z(G)$ (center of $G$).



Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.



This proves $Z(G)$ is a normal subgroup.



I’m confused about option $1)$ and option $3)$. Are there any counterexamples?






abstract-algebra group-theory







share|cite|improve this question



















  • 4




    Did you try some small example to get an idea of whether they might be true?
    – Tobias Kildetoft
    5 hours ago






  • 1




    Your third question have answer here: math.stackexchange.com/questions/267443/…
    – Alan Wang
    4 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $G$ be a group. Which of the following statements are true?



$1.$ The normalizer of a subgroup of $G $ is a normal subgroup of $G.$



$2.$ The centre of $G$ is a normal subgroup of $G.$



$3.$ If $H$ is a normal subgroup of G and is of order $2$, then H is contained in the centre of G.



My attempt:



For option $2)$ is true.



Let $xin Z(G)$ (center of $G$).



Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.



This proves $Z(G)$ is a normal subgroup.



I’m confused about option $1)$ and option $3)$. Are there any counterexamples?






abstract-algebra group-theory







share|cite|improve this question















Let $G$ be a group. Which of the following statements are true?



$1.$ The normalizer of a subgroup of $G $ is a normal subgroup of $G.$



$2.$ The centre of $G$ is a normal subgroup of $G.$



$3.$ If $H$ is a normal subgroup of G and is of order $2$, then H is contained in the centre of G.



My attempt:



For option $2)$ is true.



Let $xin Z(G)$ (center of $G$).



Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.



This proves $Z(G)$ is a normal subgroup.



I’m confused about option $1)$ and option $3)$. Are there any counterexamples?






abstract-algebra group-theory




abstract-algebra group-theory






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edited 3 hours ago









amWhy

190k27221433




190k27221433










asked 5 hours ago









stupid

767112




767112







  • 4




    Did you try some small example to get an idea of whether they might be true?
    – Tobias Kildetoft
    5 hours ago






  • 1




    Your third question have answer here: math.stackexchange.com/questions/267443/…
    – Alan Wang
    4 hours ago












  • 4




    Did you try some small example to get an idea of whether they might be true?
    – Tobias Kildetoft
    5 hours ago






  • 1




    Your third question have answer here: math.stackexchange.com/questions/267443/…
    – Alan Wang
    4 hours ago







4




4




Did you try some small example to get an idea of whether they might be true?
– Tobias Kildetoft
5 hours ago




Did you try some small example to get an idea of whether they might be true?
– Tobias Kildetoft
5 hours ago




1




1




Your third question have answer here: math.stackexchange.com/questions/267443/…
– Alan Wang
4 hours ago




Your third question have answer here: math.stackexchange.com/questions/267443/…
– Alan Wang
4 hours ago










4 Answers
4






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up vote
2
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accepted










For the first argument and as a bit difficult example than @Alan's, consider $G=D_10$. It is as $$e, a, b, ab, b^2, ab^2, b^3, ab^3, b^4, ab^4$$ of order $10$. For the subgroup $H=e, ab^3$ of order $2$ we can check $H=N_G(H)neq G$. So the normalizer could't be a normal subgroup in $G$. Otherwise, for this example $G=N_G(N_G(H))=N_G(H)$ which violets the latter fact.






share|cite|improve this answer



























    up vote
    3
    down vote













    For your first statement:



    Consider $G=S_3$ and $H=langle (12) rangle$.

    Since $Hleq N_G(H)leq G$, by Lagrange's Theorem, $N_G(H)=H$ or $G$.

    But since $H$ is not normal, $N_G(H)=H$ which is also not normal.






    share|cite|improve this answer



























      up vote
      1
      down vote













      Not as easy an answer as is probably expected, but whenever normal subgroups are in a true / false question non-abelian simple groups are good place to start, so for statement 1:



      Take any maximal subgroup $M$ of a non-abelian simple group $G$. Since $G$ is simple and $Mtrianglelefteq N_G(M)le G$ we must have $M=N_G(M)$.






      share|cite|improve this answer



























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        Ad 3. Let $H unlhd G$ and $|H|=2$. Observe that since $H$ is normal, $N_G(H)=G$. In general $N_G(H)/C_G(H)$ embeds homomorphically in $Aut(H)$ (this is called the "N/C theorem"). But $H cong C_2$ so, $Aut(H)=1$. Hence $G=C_G(H)$, that is $H subseteq Z(G)$.






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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          For the first argument and as a bit difficult example than @Alan's, consider $G=D_10$. It is as $$e, a, b, ab, b^2, ab^2, b^3, ab^3, b^4, ab^4$$ of order $10$. For the subgroup $H=e, ab^3$ of order $2$ we can check $H=N_G(H)neq G$. So the normalizer could't be a normal subgroup in $G$. Otherwise, for this example $G=N_G(N_G(H))=N_G(H)$ which violets the latter fact.






          share|cite|improve this answer
























            up vote
            2
            down vote



            accepted










            For the first argument and as a bit difficult example than @Alan's, consider $G=D_10$. It is as $$e, a, b, ab, b^2, ab^2, b^3, ab^3, b^4, ab^4$$ of order $10$. For the subgroup $H=e, ab^3$ of order $2$ we can check $H=N_G(H)neq G$. So the normalizer could't be a normal subgroup in $G$. Otherwise, for this example $G=N_G(N_G(H))=N_G(H)$ which violets the latter fact.






            share|cite|improve this answer






















              up vote
              2
              down vote



              accepted







              up vote
              2
              down vote



              accepted






              For the first argument and as a bit difficult example than @Alan's, consider $G=D_10$. It is as $$e, a, b, ab, b^2, ab^2, b^3, ab^3, b^4, ab^4$$ of order $10$. For the subgroup $H=e, ab^3$ of order $2$ we can check $H=N_G(H)neq G$. So the normalizer could't be a normal subgroup in $G$. Otherwise, for this example $G=N_G(N_G(H))=N_G(H)$ which violets the latter fact.






              share|cite|improve this answer












              For the first argument and as a bit difficult example than @Alan's, consider $G=D_10$. It is as $$e, a, b, ab, b^2, ab^2, b^3, ab^3, b^4, ab^4$$ of order $10$. For the subgroup $H=e, ab^3$ of order $2$ we can check $H=N_G(H)neq G$. So the normalizer could't be a normal subgroup in $G$. Otherwise, for this example $G=N_G(N_G(H))=N_G(H)$ which violets the latter fact.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 4 hours ago









              mrs

              58.5k750143




              58.5k750143




















                  up vote
                  3
                  down vote













                  For your first statement:



                  Consider $G=S_3$ and $H=langle (12) rangle$.

                  Since $Hleq N_G(H)leq G$, by Lagrange's Theorem, $N_G(H)=H$ or $G$.

                  But since $H$ is not normal, $N_G(H)=H$ which is also not normal.






                  share|cite|improve this answer
























                    up vote
                    3
                    down vote













                    For your first statement:



                    Consider $G=S_3$ and $H=langle (12) rangle$.

                    Since $Hleq N_G(H)leq G$, by Lagrange's Theorem, $N_G(H)=H$ or $G$.

                    But since $H$ is not normal, $N_G(H)=H$ which is also not normal.






                    share|cite|improve this answer






















                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      For your first statement:



                      Consider $G=S_3$ and $H=langle (12) rangle$.

                      Since $Hleq N_G(H)leq G$, by Lagrange's Theorem, $N_G(H)=H$ or $G$.

                      But since $H$ is not normal, $N_G(H)=H$ which is also not normal.






                      share|cite|improve this answer












                      For your first statement:



                      Consider $G=S_3$ and $H=langle (12) rangle$.

                      Since $Hleq N_G(H)leq G$, by Lagrange's Theorem, $N_G(H)=H$ or $G$.

                      But since $H$ is not normal, $N_G(H)=H$ which is also not normal.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 4 hours ago









                      Alan Wang

                      4,696932




                      4,696932




















                          up vote
                          1
                          down vote













                          Not as easy an answer as is probably expected, but whenever normal subgroups are in a true / false question non-abelian simple groups are good place to start, so for statement 1:



                          Take any maximal subgroup $M$ of a non-abelian simple group $G$. Since $G$ is simple and $Mtrianglelefteq N_G(M)le G$ we must have $M=N_G(M)$.






                          share|cite|improve this answer
























                            up vote
                            1
                            down vote













                            Not as easy an answer as is probably expected, but whenever normal subgroups are in a true / false question non-abelian simple groups are good place to start, so for statement 1:



                            Take any maximal subgroup $M$ of a non-abelian simple group $G$. Since $G$ is simple and $Mtrianglelefteq N_G(M)le G$ we must have $M=N_G(M)$.






                            share|cite|improve this answer






















                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              Not as easy an answer as is probably expected, but whenever normal subgroups are in a true / false question non-abelian simple groups are good place to start, so for statement 1:



                              Take any maximal subgroup $M$ of a non-abelian simple group $G$. Since $G$ is simple and $Mtrianglelefteq N_G(M)le G$ we must have $M=N_G(M)$.






                              share|cite|improve this answer












                              Not as easy an answer as is probably expected, but whenever normal subgroups are in a true / false question non-abelian simple groups are good place to start, so for statement 1:



                              Take any maximal subgroup $M$ of a non-abelian simple group $G$. Since $G$ is simple and $Mtrianglelefteq N_G(M)le G$ we must have $M=N_G(M)$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 3 hours ago









                              Robert Chamberlain

                              3,9301421




                              3,9301421




















                                  up vote
                                  0
                                  down vote













                                  Ad 3. Let $H unlhd G$ and $|H|=2$. Observe that since $H$ is normal, $N_G(H)=G$. In general $N_G(H)/C_G(H)$ embeds homomorphically in $Aut(H)$ (this is called the "N/C theorem"). But $H cong C_2$ so, $Aut(H)=1$. Hence $G=C_G(H)$, that is $H subseteq Z(G)$.






                                  share|cite|improve this answer
























                                    up vote
                                    0
                                    down vote













                                    Ad 3. Let $H unlhd G$ and $|H|=2$. Observe that since $H$ is normal, $N_G(H)=G$. In general $N_G(H)/C_G(H)$ embeds homomorphically in $Aut(H)$ (this is called the "N/C theorem"). But $H cong C_2$ so, $Aut(H)=1$. Hence $G=C_G(H)$, that is $H subseteq Z(G)$.






                                    share|cite|improve this answer






















                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Ad 3. Let $H unlhd G$ and $|H|=2$. Observe that since $H$ is normal, $N_G(H)=G$. In general $N_G(H)/C_G(H)$ embeds homomorphically in $Aut(H)$ (this is called the "N/C theorem"). But $H cong C_2$ so, $Aut(H)=1$. Hence $G=C_G(H)$, that is $H subseteq Z(G)$.






                                      share|cite|improve this answer












                                      Ad 3. Let $H unlhd G$ and $|H|=2$. Observe that since $H$ is normal, $N_G(H)=G$. In general $N_G(H)/C_G(H)$ embeds homomorphically in $Aut(H)$ (this is called the "N/C theorem"). But $H cong C_2$ so, $Aut(H)=1$. Hence $G=C_G(H)$, that is $H subseteq Z(G)$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 41 mins ago









                                      Nicky Hekster

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