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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

2

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.

abstract-algebra group-theory

share|cite|improve this question

edited 1 hour ago

Matt Samuel

39.7k63870

asked 2 hours ago

mkmlpmkmlp

166213

$endgroup$

add a comment | 

2

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.

abstract-algebra group-theory

share|cite|improve this question

edited 1 hour ago

Matt Samuel

39.7k63870

asked 2 hours ago

mkmlpmkmlp

166213

$endgroup$

add a comment | 

$begingroup$

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.

abstract-algebra group-theory

share|cite|improve this question

edited 1 hour ago

Matt Samuel

39.7k63870

asked 2 hours ago

mkmlpmkmlp

166213

$endgroup$

share|cite|improve this question

edited 1 hour ago

Matt Samuel

39.7k63870

asked 2 hours ago

mkmlpmkmlp

166213

share|cite|improve this question

edited 1 hour ago

Matt Samuel

39.7k63870

asked 2 hours ago

mkmlpmkmlp

166213

edited 1 hour ago

Matt Samuel

39.7k63870

edited 1 hour ago

Matt Samuel

39.7k63870

asked 2 hours ago

mkmlpmkmlp

166213

asked 2 hours ago

mkmlpmkmlp

166213

1 Answer
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active

oldest

votes

4

$begingroup$

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

answered 1 hour ago

Matt SamuelMatt Samuel

39.7k63870

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice argument! Thanks a lot, Matt!
  $endgroup$
  – mkmlp
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mkmlp No problem.
  $endgroup$
  – Matt Samuel
  1 hour ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

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

answered 1 hour ago

Matt SamuelMatt Samuel

39.7k63870

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice argument! Thanks a lot, Matt!
  $endgroup$
  – mkmlp
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mkmlp No problem.
  $endgroup$
  – Matt Samuel
  1 hour ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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

answered 1 hour ago

Matt SamuelMatt Samuel

39.7k63870

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice argument! Thanks a lot, Matt!
  $endgroup$
  – mkmlp
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mkmlp No problem.
  $endgroup$
  – Matt Samuel
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 1 hour ago

Matt SamuelMatt Samuel

39.7k63870

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Matt SamuelMatt Samuel

39.7k63870

answered 1 hour ago

Matt SamuelMatt Samuel

39.7k63870

answered 1 hour ago

Matt SamuelMatt Samuel

39.7k63870