Normal subgroup of even order whose nontrivial elements form a single conjugacy class is abelianIndex of center $Z(G)$ is finite implies the number of elements of conjugacy class is finiteHow is number of conjugacy class related to the order of a group?Center of $G$ is trivial and $p$ divides the order of $G$, show that $G$ has a non-trivial conjugacy class whose order is prime to $p$Prove in two different ways-“If the centre of $G$ is of index $n$,prove that every conjugacy class has atmost $n$ elements .”Conjugacy class in subgroup of index 2Normal Group and Conjugacy ClassIf $n$ is odd, then there are exactly two conjugacy classes of n-cycles in $A_n$ each of which contains $(n - 1)!/2$ elements.Proof of map $varphi(gcdot a) = gG_a$ is a bijectionSuppose $bin O_a$, and $ain A$. Prove that $G_a$ and $G_b$ are isomorphic. Under what conditions are they actually equal?Partition of a conjugacy class to conjugacy classes of a normal subgroup

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Normal subgroup of even order whose nontrivial elements form a single conjugacy class is abelian


Index of center $Z(G)$ is finite implies the number of elements of conjugacy class is finiteHow is number of conjugacy class related to the order of a group?Center of $G$ is trivial and $p$ divides the order of $G$, show that $G$ has a non-trivial conjugacy class whose order is prime to $p$Prove in two different ways-“If the centre of $G$ is of index $n$,prove that every conjugacy class has atmost $n$ elements .”Conjugacy class in subgroup of index 2Normal Group and Conjugacy ClassIf $n$ is odd, then there are exactly two conjugacy classes of n-cycles in $A_n$ each of which contains $(n - 1)!/2$ elements.Proof of map $varphi(gcdot a) = gG_a$ is a bijectionSuppose $bin O_a$, and $ain A$. Prove that $G_a$ and $G_b$ are isomorphic. Under what conditions are they actually equal?Partition of a conjugacy class to conjugacy classes of a normal subgroup













5












$begingroup$


Let $G$ be a finite group and let $Ntrianglelefteq G$, $2mid |N|$. If the non-trivial elements of $N$ form a single conjugacy class of $G$, prove that $N$ is abelian.



My Attempt



I tried to approach by using the Orbit-Stabilizer theorem as follows: let $ain N$ be given, by the Orbit-Stabilizer theorem (the action being $G$ acting on itself by conjugation), we have $$|G| = |O_a||G_a|,$$ where $O_a, G_a$ are the orbit and stabilizer of $a$, respectively.



By assumption, we then have $|G| = (|N|-1)|G_a|$ and we know that $|N|-1$ is odd. Then I am trying to argue that $Nsubseteq G_a$, which would prove that $N$ is abelian since the choice of $a$ is arbitrary. But I couldn't make the connection there.



Also, this approach may be totally wrong. But at the moment I couldn't see any other possible way of proving this.



Any help would be greatly appreciated. Thanks.










share|cite|improve this question











$endgroup$
















    5












    $begingroup$


    Let $G$ be a finite group and let $Ntrianglelefteq G$, $2mid |N|$. If the non-trivial elements of $N$ form a single conjugacy class of $G$, prove that $N$ is abelian.



    My Attempt



    I tried to approach by using the Orbit-Stabilizer theorem as follows: let $ain N$ be given, by the Orbit-Stabilizer theorem (the action being $G$ acting on itself by conjugation), we have $$|G| = |O_a||G_a|,$$ where $O_a, G_a$ are the orbit and stabilizer of $a$, respectively.



    By assumption, we then have $|G| = (|N|-1)|G_a|$ and we know that $|N|-1$ is odd. Then I am trying to argue that $Nsubseteq G_a$, which would prove that $N$ is abelian since the choice of $a$ is arbitrary. But I couldn't make the connection there.



    Also, this approach may be totally wrong. But at the moment I couldn't see any other possible way of proving this.



    Any help would be greatly appreciated. Thanks.










    share|cite|improve this question











    $endgroup$














      5












      5








      5


      0



      $begingroup$


      Let $G$ be a finite group and let $Ntrianglelefteq G$, $2mid |N|$. If the non-trivial elements of $N$ form a single conjugacy class of $G$, prove that $N$ is abelian.



      My Attempt



      I tried to approach by using the Orbit-Stabilizer theorem as follows: let $ain N$ be given, by the Orbit-Stabilizer theorem (the action being $G$ acting on itself by conjugation), we have $$|G| = |O_a||G_a|,$$ where $O_a, G_a$ are the orbit and stabilizer of $a$, respectively.



      By assumption, we then have $|G| = (|N|-1)|G_a|$ and we know that $|N|-1$ is odd. Then I am trying to argue that $Nsubseteq G_a$, which would prove that $N$ is abelian since the choice of $a$ is arbitrary. But I couldn't make the connection there.



      Also, this approach may be totally wrong. But at the moment I couldn't see any other possible way of proving this.



      Any help would be greatly appreciated. Thanks.










      share|cite|improve this question











      $endgroup$




      Let $G$ be a finite group and let $Ntrianglelefteq G$, $2mid |N|$. If the non-trivial elements of $N$ form a single conjugacy class of $G$, prove that $N$ is abelian.



      My Attempt



      I tried to approach by using the Orbit-Stabilizer theorem as follows: let $ain N$ be given, by the Orbit-Stabilizer theorem (the action being $G$ acting on itself by conjugation), we have $$|G| = |O_a||G_a|,$$ where $O_a, G_a$ are the orbit and stabilizer of $a$, respectively.



      By assumption, we then have $|G| = (|N|-1)|G_a|$ and we know that $|N|-1$ is odd. Then I am trying to argue that $Nsubseteq G_a$, which would prove that $N$ is abelian since the choice of $a$ is arbitrary. But I couldn't make the connection there.



      Also, this approach may be totally wrong. But at the moment I couldn't see any other possible way of proving this.



      Any help would be greatly appreciated. Thanks.







      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 Apr 28 at 16:10









      Matt Samuel

      40k63870




      40k63870










      asked Apr 28 at 15:46









      mkmlpmkmlp

      329313




      329313




















          1 Answer
          1






          active

          oldest

          votes


















          8












          $begingroup$

          Well actually, by Cauchy's theorem, $N$ has an element of order $2$. By conjugacy it follows that every nonidentity element is of order $2$. It's a very common exercise that I'm sure you've done to show that if every nonidentity element of a group is of order $2$, then the group is abelian. This actually doesn't require the hypothesis that the subgroup is normal.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Really nice argument! Thanks a lot, Matt!
            $endgroup$
            – mkmlp
            Apr 28 at 16:19






          • 1




            $begingroup$
            @mkmlp No problem.
            $endgroup$
            – Matt Samuel
            Apr 28 at 16:20











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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8












          $begingroup$

          Well actually, by Cauchy's theorem, $N$ has an element of order $2$. By conjugacy it follows that every nonidentity element is of order $2$. It's a very common exercise that I'm sure you've done to show that if every nonidentity element of a group is of order $2$, then the group is abelian. This actually doesn't require the hypothesis that the subgroup is normal.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Really nice argument! Thanks a lot, Matt!
            $endgroup$
            – mkmlp
            Apr 28 at 16:19






          • 1




            $begingroup$
            @mkmlp No problem.
            $endgroup$
            – Matt Samuel
            Apr 28 at 16:20















          8












          $begingroup$

          Well actually, by Cauchy's theorem, $N$ has an element of order $2$. By conjugacy it follows that every nonidentity element is of order $2$. It's a very common exercise that I'm sure you've done to show that if every nonidentity element of a group is of order $2$, then the group is abelian. This actually doesn't require the hypothesis that the subgroup is normal.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Really nice argument! Thanks a lot, Matt!
            $endgroup$
            – mkmlp
            Apr 28 at 16:19






          • 1




            $begingroup$
            @mkmlp No problem.
            $endgroup$
            – Matt Samuel
            Apr 28 at 16:20













          8












          8








          8





          $begingroup$

          Well actually, by Cauchy's theorem, $N$ has an element of order $2$. By conjugacy it follows that every nonidentity element is of order $2$. It's a very common exercise that I'm sure you've done to show that if every nonidentity element of a group is of order $2$, then the group is abelian. This actually doesn't require the hypothesis that the subgroup is normal.






          share|cite|improve this answer









          $endgroup$



          Well actually, by Cauchy's theorem, $N$ has an element of order $2$. By conjugacy it follows that every nonidentity element is of order $2$. It's a very common exercise that I'm sure you've done to show that if every nonidentity element of a group is of order $2$, then the group is abelian. This actually doesn't require the hypothesis that the subgroup is normal.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 28 at 16:09









          Matt SamuelMatt Samuel

          40k63870




          40k63870











          • $begingroup$
            Really nice argument! Thanks a lot, Matt!
            $endgroup$
            – mkmlp
            Apr 28 at 16:19






          • 1




            $begingroup$
            @mkmlp No problem.
            $endgroup$
            – Matt Samuel
            Apr 28 at 16:20
















          • $begingroup$
            Really nice argument! Thanks a lot, Matt!
            $endgroup$
            – mkmlp
            Apr 28 at 16:19






          • 1




            $begingroup$
            @mkmlp No problem.
            $endgroup$
            – Matt Samuel
            Apr 28 at 16:20















          $begingroup$
          Really nice argument! Thanks a lot, Matt!
          $endgroup$
          – mkmlp
          Apr 28 at 16:19




          $begingroup$
          Really nice argument! Thanks a lot, Matt!
          $endgroup$
          – mkmlp
          Apr 28 at 16:19




          1




          1




          $begingroup$
          @mkmlp No problem.
          $endgroup$
          – Matt Samuel
          Apr 28 at 16:20




          $begingroup$
          @mkmlp No problem.
          $endgroup$
          – Matt Samuel
          Apr 28 at 16:20

















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