We know PGL(2, $q$) has elements of order $q+1$ or $q1$. Suppose $k\neq 1$, $2$ divide $q+1$ or $q1$. 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?
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By the sizes of conjugacy calasses of PGL(2, q), if $k$ divides $q+1$, then the number of elements of order $k$ is $\phi (k)q(q1)/2$ and if $k$ divides $q1$, then the number of elements of order $k$ is $\phi (k)q(q+1)/2$. 


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 $k$. 
