most recent 30 from http://mathoverflow.net 2013-05-25T08:26:52Z http://mathoverflow.net/feeds/question/118646 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118646/the-number-of-elements-of-order-k-in-pgl2-q Mart 2013-01-11T16:58:01Z 2013-01-12T03:27:40Z

<p>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?</p>

http://mathoverflow.net/questions/118646/the-number-of-elements-of-order-k-in-pgl2-q/118653#118653 Tom 2013-01-11T17:47:04Z 2013-01-11T17:47:04Z

<p>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 $\phi (k)q(q+1)/2$.</p>

http://mathoverflow.net/questions/118646/the-number-of-elements-of-order-k-in-pgl2-q/118698#118698 Wei Zhou 2013-01-12T03:27:40Z 2013-01-12T03:27:40Z

<p>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 $k$.</p>