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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

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    $\begingroup$ Always: rational repn of the norm-one subgroup of the unique quadratic extension, together with Galois automorphism. This can be made as explicit as one's knowledge of a non-square. $\endgroup$ Dec 11, 2014 at 22:40
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    $\begingroup$ @paulgarrett: The galois automorphism is not always in $PSL_2(q)$. If $q$ is odd, then the field extension is of the form $\mathbb{F}_q(\sqrt{a})$ for some non-square $a$ so that the Galois automorphism acts on the basis $\{1,a\}$ as $diag(1,-1)$. If $-1$ is not a square, then this will not lie in $Z\cdot SL_2(q)$. $\endgroup$ Dec 11, 2014 at 22:58
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    $\begingroup$ @JohannesHahn, ah, you're right. So that's the relevant dichotomy, apparently! Good. Thanks! $\endgroup$ Dec 11, 2014 at 23:00
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    $\begingroup$ IMHO this is already in L.Dickson's book "Linear Groups", which was reprinted by Dover in 1958. $\endgroup$ Dec 12, 2014 at 0:21
  • $\begingroup$ @M.B.: Minor edits. You may want to separate the cases $p=2$ and $p$ odd here. For odd $p$, your group has order $q(q−1)(q+1)/2$, where $q−1$ is the order of the group of rational points of a split torus in $\mathrm{SL}(2,q)$ and $q+1$ is the order of the group of rational points of a torus which splits only over a quadratic extension, while $q$ is the order of a Sylow $p$-subgroup. The subgroup structure is not too mysterious, but your specific question may not be answered directly in the literature $\endgroup$ Dec 12, 2014 at 0:41

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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.

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  • $\begingroup$ Dear Geoff: That is perfect. Thanks. $\endgroup$
    – M.B
    Dec 12, 2014 at 3:15
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    $\begingroup$ @Geoff: This is a helpful concise summary. Is there any published source for this particular question? I was reluctant to go back as far as Dickson, but these are pretty well-studied groups. $\endgroup$ Dec 12, 2014 at 21:59
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    $\begingroup$ @Jim: As you said in an earlier comment, such results can easily be reconstructed by various means (you quote Lie type methods and I quote fairly general group theory- transfer, etc). I do not know where the precise question asked is answered in the literature. My arguments are based on what I remember from Gorenstein ( Finite Groups, Chelsea, 1968). I am also reasonably sure that Dickson and or Blichfeldt knew the structure of all subgroups of ${\rm GL}(2,q)$ and hence (implicitly) all subgroups of ${\rm PSL}(2,q).$ $\endgroup$ Dec 12, 2014 at 22:11
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    $\begingroup$ @JimHumphreys: I found a paper of D. Flannery and E.A. O'Brien which contains some useful background history : Linear groups of small degree over finite fields. Internat. J. Algebra Comput. 15 (2005), no. 3, 467–502. $\endgroup$ Dec 12, 2014 at 22:24
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    $\begingroup$ @Geoff: This looks helpful. $\endgroup$ Dec 13, 2014 at 13:11

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