• Support PF! Buy your school textbooks, materials and every day products Here!

Proving Non-Simplicity

  • Thread starter Obraz35
  • Start date
  • #1
31
0

Homework Statement


Let G be a finite group. Prove that if some conjugacy class has exactly two elements then G can't be simple


Homework Equations





The Attempt at a Solution


I originally proved this accidentally assuming that G is abelian, which it isn't. So say x and y are conjugate then we have conjugacy classes {e} and {x,y}. I know that the size of a conjugacy class has to divide the order of a group so |G| is even and must be greater than 2. I'm having trouble showing the existence of a normal subgroup though.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
If G has a conjugacy class with two elements then it has a subgroup of order |G|/2. Look at the centralizer of each element of the conjugacy class. If G has a subgroup H of order |G|/2, H must be normal. Look at left and right cosets.
 

Related Threads for: Proving Non-Simplicity

  • Last Post
Replies
1
Views
1K
Replies
7
Views
2K
Replies
0
Views
594
  • Last Post
Replies
2
Views
1K
Top