Let G be a non abelian group of order pq where p<q are primes and p|q-1. describe the class equation of G

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- Jan 9th 2010, 11:30 PMChandru1Class Equation for a non abelian group of order pq
Let G be a non abelian group of order pq where p<q are primes and p|q-1. describe the class equation of G

- Jan 10th 2010, 03:54 AMabender

Let denote, respectively, the number of sylow-p subgroups and sylow-q subgroups of G.

From Sylow's Theorems, for some integer is a factor of pq. Now, since p and q are (co)prime and p<q, , and that one sylow-q subgroup is normal in G.

Also, since PQ = G and .

Now, since p divides q-1, has a unique subgroup P' such that .

So, we let a and b be generators for P and Q, respectively. Suppose the action of a on Q by conjugation is in . Then

Since a different amounts to choosing a different generator a for P, it does not result in a new isomorphism class. THEREFORE, there are exactly two isomorphism classes of groups of order pq.

Does this help? - Jan 10th 2010, 04:00 AMChandru1Not really
Well not really. You have told me something about the isomorphism classes. For explaining about the class equation you should look at the conjugacy classes.

We have the class equation when G acts on G by conjugation as where .

It should be something like . Since for this group the center is trivial. - Jan 10th 2010, 05:44 AMabender
Let be a non abelian group of order , whereby are prime and . Describe the class equation of .

Let .

Any nontrivial action must have order , but since and thus clearly does not divide , this such action can't exist. Therefore, , .

We have the conjugacy class .

Now let's consider the conjugacy class of , which is . All the elements in this conjugacy class are distinct. This is because if for , then and . Hence, , which is a contradiction to our assumption that the action of P on Q was nontrivial. If , this class will have all the elements . Otherwise, some of the powers of will not be in this class. We will wind up with conjugacy classes, which are of the form . Its union is all of the powers of , excluding .

Now, for , we want to shed light on all of the elements that are conjugate to . When we conjugate by , we get . Thus, we get by conjugation by powers of all of the elements of the form , with fixed , and going from to . Moreover, conjugation by gives . Conjugation by a power gives us

Now, since for , , and goes from to , also takes on every value from to . So, each is a conjugacy class, whose elements are .

The total number of conjugacy classes is

. - Jan 10th 2010, 07:25 AMChandru1excellent
Hi--

You, beauty!!! This is what i exactly wanted!!!