Timeline for The number of elements of order k in PGL(2, q)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2013 at 17:19 | vote | accept | Mart | ||
| Jan 12, 2013 at 3:27 | answer | added | Wei Zhou | timeline score: 0 | |
| Jan 11, 2013 at 18:11 | comment | added | Michael Zieve | @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. | |
| Jan 11, 2013 at 18:01 | comment | added | Mart | @Michael Zieve: Thank you. If possible give me more details of your answer. | |
| Jan 11, 2013 at 17:47 | answer | added | Tom | timeline score: 0 | |
| Jan 11, 2013 at 17:40 | comment | added | Michael Zieve | 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. | |
| Jan 11, 2013 at 16:58 | history | asked | Mart | CC BY-SA 3.0 |