Let $\mathbb{F}\_q$ be a finite field with $q=p^f$ elements. I need to know when $\mathrm{PSL}\_2(\mathbb{F}\_q)$ contains the group $D\_{(q+1)/2}$, where by $D\_n$ I mean the dihedral group of order $2n$. Also if such a classification of $q$ exists, can we write explicit generators of $D\_{(q+1)/2}$? 


Thanks



Dihedral subgroups of $\mathrm{PSL}_2(\mathbb{F}_q)$
The answer is indeed "always" when $q$ is odd, though the analysis when $q \equiv 1$ (mod $4$) and $q \equiv -1$ (mod $4$) differs. If $q \equiv 1$ (mod $4$), then $G = {\rm PSL}(2,q)$ has a cyclic Hall subgroup $H$ of odd order $\frac{q+1}{2}$ which is the centralizer of all its non-identity subgroups. All odd order subgroups of $G$ are Abelian when $q$ is odd.
It follows by a transfer argument that $N\_{G}(H) \neq C\_{G}(H)$ and then that $N\_{G}(H)$ is dihedral with $q+1$ elements.
 When $q \equiv -1$ ( mod $4$), $G$ has one conjugacy of involutions, and $C\_{G}(t)$ has a cyclic normal $2$-complement and a dihedral (or Klein $4$) Sylow $2$-subgroup. If $\frac{q+1}{2}$ is a power of $2,$ then we have the required dihedral subgroup. Otherwise, there is non-identity element $h$ of odd order whose centralizer has order $\frac{q+1}{2}$ and whose extended centralizer is the required dihedral group.


Later edit: Let me add some detail, which gives some insight into the second part of the question. When $q$ is odd, ${\rm PSL}(2,q)$ has a dihedral Sylow $2$-subgroup (counting a Klein 4-group as dihedral), since $X = {\rm SL}(2,q)$ has a (generalized) quaternion Sylow $2$-subgroup. This deals with the case that $q+1$ is a power of $2$, so suppose otherwise. Since there is an irreducible polynomial of degree $2$ in $\mathbb{F}\_{q}[x],$ Schur's Lemma tells us that the centralizer of its companion matrix in ${\rm GL}(2,q)$ is isomorphic to the multiplicative group of $\mathbb{F}\_{q^{2}},$ and that its centralizer in $X = {\rm SL}(2,q)$ is cyclic of order $q+1.$ It follows that if $u \neq 1$ is an element of odd order in that centralizer (and such exists under current assumptions), then $C\_{X}(u)$ is cyclic of order $q+1.$ It follows that the image ($v$ say) of $u$ in $G = {\rm PSL}(2,q)$ has a centralizer which is is cyclic of order $\frac{q+1}{2}.$ 


Now $N\_{G}(\langle v \rangle)$ has cyclic or dihedral Sylow $2$-subgroups since $G$ has dihedral Sylow $2$-subgroups. As explained earlier, $v$ is conjugate to its inverse in $G.$ If $C\_{G}(v)$ contains no involution, then we are done. If $C\_{G}(v)$ contains an involution and $N\_{G}(\langle v \rangle)$ has dihedral Sylow $2$-subgroups, then $v$ is inverted by an involution, and we are done. If $C\_{G}(v)$ contains an element $x$ of order $4,$ then $C\_{G}(v) = C\_{G}(x)$, and $N\_{G}(\langle v \rangle)$ contains a Sylow $2$-subgroup of $G,$ so we are done. If $C\_{G}(v)$ has a Sylow $2$-subgroup $\langle t \rangle$ of order $2$, then $C\_{G}(t)$ has order $q+1$, contains a Sylow $2$-subgroup of $G,$ and normalizes $\langle v \rangle,$ so we are done.


To summarize: if $\frac{q+1}{2}$ is even, then for any involution $t \in G,$ we see that $C\_{G}(t)$ is dihedral with $q+1$ elements. If $\frac{q+1}{2}$ is odd, then $G$ has a subgroup $H$ of order $\frac{q+1}{2}$ (unique up to conjugacy) such that $N\_{G}(H)$ is dihedral with $q+1$ elements.


