Cyclic subgroup of PSL

Is it 'expected' that PSL(2,p) contains a cyclic subgroup of order $\tfrac{p+1}{2}$? I read such an element exists and generates a "nonsplit torus", but I'm not sure what that means.

Is there an easy explanation why such an element should exist and have a cycle structure $\tfrac{p+1}{2},\tfrac{p+1}{2}$ in its action on the projective line?

(Edit: On a sidenote, it is on the other hand obvious that PSL(2,p) has cyclic subgroups of order (p-1)/2 and p, generated by $(\begin{smallmatrix}r&0\\0&\tfrac{1}{r}\end{smallmatrix})$ and $(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix})$ where r generates $GF(p)^*$ )

-

Elements of $\text{SL}_2(\mathbb{F}_p)$ have eigenvalues in $\mathbb{F}_{p^2}$. The multiplicative group of $\mathbb{F}_{p^2}$ is cyclic of order $p^2 - 1$ generated by some element $g$, so it follows that $\text{GL}_2(\mathbb{F}_p)$ has a cyclic subgroup of order $p^2 - 1$ (just write out the matrix of multiplication by $g$). The determinant of this matrix is some element of $\mathbb{F}_p^{\ast}$, so $g^{p-1}$ has determinant $1$ and order $p+1$. Finally, the image of $g^{p-1}$ in $\text{PSL}_2(\mathbb{F}_p)$ has order $\frac{p+1}{2}$ since $g^{ \frac{p^2 - 1}{2} } = -1$. You can work out the cycle structure from here.