Thread: Conjugacy classes - two problems.

Hi, I was wondering if I could get some help with these questions?

1.Let G be a non-abelian group of order 21. Show that if m is the number of conjugacy classes of size 3 and n is the number of conjugacy classes of size 7 then 3m+7n = 20. Deduce that m = n = 2.

2. Let G be a group of order p^2 where p is a prime number. Show that all conjugacy classes in G have size 1 or p.


I guess that I need to use The Class Equation or The Orbit-Stabilizer Theorem but I do not know how?

Can anyone help me, please?
Thanks!

Originally Posted by leelooana

Hi, I was wondering if I could get some help with these questions?

1.Let G be a non-abelian group of order 21. Show that if m is the number of conjugacy classes of size 3 and n is the number of conjugacy classes of size 7 then 3m+7n = 20. Deduce that m = n = 2.

2. Let G be a group of order p^2 where p is a prime number. Show that all conjugacy classes in G have size 1 or p.


I guess that I need to use The Class Equation or The Orbit-Stabilizer Theorem but I do not know how?

Can anyone help me, please?
Thanks!

For the first question, prove that every non-abelian group of order has trivial center. (You may wish to first prove that if is cyclic then is abelian). You can then apply the class equation.

For the second result, use the orbit-stabiliser theorem to investigate the action of conjugation. What is an element's orbit? What is the corresponding stabiliser? Is one of these a subgroup?

Are you at uni in Scotland?

Thanks for Your reply! I just proved in previous problem that if p and q are distinct primes and G is non abelian then Z(G) is trivial but I did not notice the relation between these problems.

so quick proof:
From Lagrange Theorem |Z(G)| could be pq, p, q or 1. It is easy to prove that order of Z(G) can not be pq ( because in this case Z(G) = G therefore G is abelian what contradicts the initial claim ).
Also can not be p because the quotient group G/Z(G) would be cyclic of order q but G/Z(G) cannot be cyclic unless G is abelian ( because G/Z(G) cyclic implies G abelian.)
The same for q.

So Z(G) must be trivial.

And now I see that if |G| = 21 then |Z(G)| = 1 and from class equation I get that 3m+7n = 20 and must be n=m=2.


But I still can not done the 2nd question. I know that :
-the size of each orbit divides the group order
-in each conjugacy classes all elements have the same order
-in all abelian groups every conjugacy class is a set containing one element (if |G| = and p is prime then G is abelian)

and I can infer that order of conjugacy classes of abelian group equals 1.

how to show that conjugacy classes in G have size p?

Originally Posted by Swlabr

Are you at uni in Scotland?

Yes

Originally Posted by leelooana

Thanks for Your reply! I just proved in previous problem that if p and q are distinct primes and G is non abelian then Z(G) is trivial but I did not notice the relation between these problems.

so quick proof:
From Lagrange Theorem |Z(G)| could be pq, p, q or 1. It is easy to prove that order of Z(G) can not be pq ( because in this case Z(G) = G therefore G is abelian what contradicts the initial claim ).
Also can not be p because the quotient group G/Z(G) would be cyclic of order q but G/Z(G) cannot be cyclic unless G is abelian ( because G/Z(G) cyclic implies G abelian.)
The same for q.

So Z(G) must be trivial.

And now I see that if |G| = 21 then |Z(G)| = 1 and from class equation I get that 3m+7n = 20 and must be n=m=2.


But I still can not done the 2nd question. I know that :
-the size of each orbit divides the group order
-in each conjugacy classes all elements have the same order
-in all abelian groups every conjugacy class is a set containing one element (if |G| = and p is prime then G is abelian)

and I can infer that order of conjugacy classes of abelian group equals 1.

how to show that conjugacy classes in G have size p?

Well, the conjugacy class of the identity is 1. Assume your group is non-abelian and take an element which is non-central. What is it's conjugacy class? It must have order or . What would happen if it's conjugacy class was ? What would this "mean"?

Originally Posted by leelooana

Yes

It's the best place to be!