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

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

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

Now, since p divides q-1, $\displaystyle Aut(Q) \cong C_{q-1} $ has a unique subgroup P' such that $\displaystyle 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 $\displaystyle x \mapsto x^{i_0} \text{, whereby } i_0 \neq 1 $ in $\displaystyle \mathbb{Z}/q\mathbb{Z} $. Then

$\displaystyle

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

$

Since a different $\displaystyle 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? - 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 $\displaystyle |G|= |Z(G)|+ \sum\limits_{a \notin Z(G)} |G|/|N(a)|$ where $\displaystyle N(a)={ x \in G | xa=ax\} $.

It should be something like $\displaystyle pq= 1 + \sum\limits_{a \notin Z(G)} |G|/|N(a)|$. Since for this group the center is trivial. - Jan 10th 2010, 05:44 AMabender
Let $\displaystyle G$ be a non abelian group of order $\displaystyle pq$, whereby $\displaystyle p<q$ are prime and $\displaystyle p|q-1$. Describe the class equation of $\displaystyle G$.

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

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

We have the conjugacy class $\displaystyle \{1\}$.

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

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

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

The total number of conjugacy classes is

$\displaystyle 1+\frac{q-1}{p}+(p-1) $. - Jan 10th 2010, 07:25 AMChandru1excellent
Hi--

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