Thread:

1. 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?

Thanks!

2. 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?

Thanks!
For the first question, prove that every non-abelian group of order $pq$ has trivial center. (You may wish to first prove that if $G/Z(G)$ is cyclic then $G$ 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?

3. 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| = $p^2$ 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

4. 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| = $p^2$ 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 $p$ or $p^2$. What would happen if it's conjugacy class was $p^2$? What would this "mean"?

Originally Posted by leelooana
Yes
It's the best place to be!
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