Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We know PGL(2, $q$) has elements of order $q+1$ or $q-1$. Suppose $k\neq 1$, $2$ divide $q+1$ or $q-1$. It is clear that PGL(2, $q$) has an elements of order $k$. I would like to know what is the number of the elements of order $k$ and how we can get it?

share|improve this question
    
You can extract the formula you seek from various sources, for instance from the webpage groupprops.subwiki.org/wiki/… Here is a standard way to get the answer. First compute the size of the conjugacy class of an order-k cyclic subgroup C of PGL(2,q), by computing the normalizer N of C in PGL(2,q) and noting that this size is [PGL(2,q):N]. Then determine the number of conjugacy classes of such subgroups C. –  Michael Zieve Jan 11 '13 at 17:40
    
@Michael Zieve: Thank you. If possible give me more details of your answer. –  Mart Jan 11 '13 at 18:01
    
@Mart: You can find the details in many standard sources, for instance in Suzuki's group theory book, but of course you would gain the most by trying to work them out on your own. –  Michael Zieve Jan 11 '13 at 18:11

2 Answers 2

up vote 0 down vote accepted

By the sizes of conjugacy calasses of PGL(2, q), if $k$ divides $q+1$, then the number of elements of order $k$ is $\phi (k)q(q-1)/2$ and if $k$ divides $q-1$, then the number of elements of order $k$ is $\phi (k)q(q+1)/2$.

share|improve this answer
    
@Tom: I don't know you how get the number of elements of order k. –  Mart Jan 11 '13 at 18:03
    
Now I know what is the size of the conjugacy class of an order-k cyclic subgroup C, but still I don't know what is the number of conjugacy classes of such subgroups C. By Tom's answer it must be $\phi (k)/2$. But why? Can anybody help me? –  Mart Jan 11 '13 at 20:14

Let $C$ be a cyclic subgroup of order $k$. Then there are $|G:N_G(C)|$ subgroups conjugate to $C$. We say $|G:N_G(C)|$ is the size of conjugacy class of $C$. In a cyclic group of order $k$, there are just $\phi(k)$ elements of order $k$.

share|improve this answer
    
@Mart: Sorry I have no reference on PGL at hand. But I think $1/2$ come from $|G:N_G(C)|$. –  Wei Zhou Jan 12 '13 at 5:43
    
@Wei Zhou: By the sizes of conjugacy classes of PGL the size of the conjugacy class of an order-k cyclic subgroup $C$ must be $q(q-1)$ or $q(q+1)$. –  Mart Jan 12 '13 at 7:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.