# Class Equation for a non abelian group of order pq

• January 9th 2010, 11:30 PM
Chandru1
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
• January 10th 2010, 03:54 AM
abender
Quote:

Originally Posted by Chandru1
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

$
G = .
$

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?
• January 10th 2010, 04:00 AM
Chandru1
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.
• January 10th 2010, 05:44 AM
abender
Let $G$ be a non abelian group of order $pq$, whereby $p 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, 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, $a^i-1 \neq 0$, 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

$1+\frac{q-1}{p}+(p-1)$.
• January 10th 2010, 07:25 AM
Chandru1
excellent
Hi--

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