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Math Help - Class Equation for a non abelian group of order pq

  1. #1
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    Class 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
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    Quote Originally Posted by Chandru1 View Post
    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

    Let  n_p \text{  and  } n_q denote, respectively, the number of sylow-p subgroups and sylow-q subgroups of G.

    From Sylow's Theorems,  n_q = 1+kq for some integer  k \text{,  and  also that  } n_q is a factor of pq. Now, since p and q are (co)prime and p<q,  n_q = 1 , and that one sylow-q subgroup is normal in G.

    Also, since PQ = G and   P \cap Q= \{1\},  G \text{  is a semi-direct-product of  } P \text{ and  } Q .

    Now, since p divides q-1,   Aut(Q) \cong C_{q-1} has a unique subgroup P' such that   P' = \{x \mapsto x^i | i \in \mathbb{Z}/q\mathbb{Z}, i^p = 1\}  \text{  and   } |P'|=p .

    So, we let a and b be generators for P and Q, respectively. Suppose the action of a on Q by conjugation is  x \mapsto x^{i_0} \text{,  whereby  } i_0 \neq 1 in  \mathbb{Z}/q\mathbb{Z}  . Then

     <br />
G = <a,b | a^p = b^q = 1 , aba^-1 = b^{i_0}> .<br />

    Since a different  i_0 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?
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  3. #3
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    Not 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
     |G|= |Z(G)|+ \sum\limits_{a \notin Z(G)} |G|/|N(a)| where  N(a)={ x \in G | xa=ax\} .


    It should be something like pq= 1 + \sum\limits_{a \notin Z(G)} |G|/|N(a)|. Since for this group the center is trivial.
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    Let G be a non abelian group of order pq, whereby p<q are prime and p|q-1. Describe the class equation of G.

    Let  P = C_p = < x > \text{  and  }  Q = C_q = < y > .

    Any nontrivial action  yxy^{-1} = x^a must have order   q  , but since  q >p and thus clearly does not divide  p-1 = |aut(P)| , this such action can't exist. Therefore,  xyx^{-1} = y^{a} ,  a^p = 1(q) .

    We have the conjugacy class \{1\}.

    Now let's consider the conjugacy class of  y , which is  \{y,y^a,...,y^{a^{p-1}}\}    . All the elements in this conjugacy class are distinct. This is because if  y^{a^{r}} = y for  r<p , then  a^r=1=a^p and  (p,r)=1 . Hence,  a^1=1 , which is a contradiction to our assumption that the action of P on Q was nontrivial. If p=q-1, this class will have all the elements  y^i . Otherwise, some of the powers of y will not be in this class. We will wind up with  \frac{q-1}{p} conjugacy classes, which are of the form  y=y^{i_1},y^{i_2},...,y^{i_{\frac{q-1}{p}}} . Its union is all of the powers of y, excluding 1.

    Now, for   i\neq0  , we want to shed light on all of the elements that are conjugate to  y^jx^i . When we conjugate by x, we get  xy^jx^ix^{-1} = (xyx^{-1})^jx^i . Thus, we get by conjugation by powers of x all of the elements of the form   y^{ja^{k}}x^i , with fixed i, and k going from 1 to p. Moreover, conjugation by  y^{-1} gives  y^{-1}y^jx^iy = y^{j-1}y^{ai}x^i = y^{ai+j-1}x^i . Conjugation by a power  y^{-k}  gives us   y^{-k}y^jx^iy^k = y^{j-k}y^{ka^{i}}x^i = y^{j+(a^{i}-1)k}x^i.

    Now, since  a^i \neq 1 for  i<p ,  a^i-1 \neq 0 , and k goes from 1 to q-1,  J+(a^i-1)k also takes on every value from 1 to q-1. So, each  x^i, i=1,...,q-1 is a conjugacy class, whose elements are  y^jx^i,  j=1,...,q-1 .

    The total number of conjugacy classes is

      1+\frac{q-1}{p}+(p-1)  .
    Last edited by abender; July 9th 2010 at 02:50 PM. Reason: correcting latex
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  5. #5
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    excellent

    Hi--

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