# Let $G = PSL(2,q)$ with $q$ odd and $H$ be a subgroup of even order. Then the centralizer of an involution in $H$ is a dihedral group.

Let $G = PSL(2, q)$ with $q = p^n$ and $p \ne 2$. Suppose $H$ is a subgroup of even order. Then $H$ contains an involution $u$. Assume that $N_G(H)$ contains the centralizer of $u$.

The centralizer of $u$ is a dihedral group $D$ of order $q + 1$ or $q - 1$ according as $\frac{1}{2}(q+1)$ or $\frac{1}{2}(q-1)$ is even and $D$ is a maximal subgroup of $G$ except when $q = 9$ or $q = 7$.

I do not understand that $D$ (the centralizer of $u$) is a dihedral group of order $q + 1$ or $q - 1$, and that it is maximal for $q \notin \{ 7, 9 \}$?

I collected some facts about $PSL(2, q)$. I know it could be thought of as the set of all mappings on $\mathbb F_q \cup \{\infty\}$ (where $\mathbb F_q$ denotes the finite field of order $q$) $$x \mapsto \frac{ax + b}{cx + d}$$ with $a,b,c,d \in \mathbb F_q$ such that $ad - bc$ is a square in $\mathbb F_q$. Then I know that the Sylow $2$-subgroups are dihedral or isomorphic to $C_2 \times C_2$ for $p \ne 2$. Also what I guess might be relevant here (for $p \ne 2$), that $G$ is partitioned by the subgroups $$\{ P^x, U^x, S^x : x \in G \}$$ where $P$ is a Sylow $p$-subgroup, $U$ is cyclic of order $(q - 1)/2$ and $S$ is cyclc of order $(q + 1)/2$. In its natural action on $\mathbb F_q \cup \{\infty\}$ the subgroups $P^x$ are the subgroups fixing exactly one point, the subgroups $U^x$ are the ones fixing exactly two points, and the $S^x$ are the subgroups fixing no point.

Now for the $U^x$'s I can choose an $U = \{ x \mapsto a^2 x : a \in \mathbb F_q^{\ast} \}$, then it has the stated properties, i.e. $|U| = (q-1)/2$ and it fixes $\infty$. Suppose that the $u$ of the question is contained in $U$. Then it corresponds to a map $x^u = a^2 x$ with $a^4 = 1$ as it is an involution. Let $s$ be the map $x^s = -x^{-1}$. Then $x^{us} = -(a^2 x)^{-1} = -x^{-1}a^2 = a^2(-x^{-1}) = x^{su}$, hence $us = su$. Now let $v \in U$ be a generator of $U$ with $x^v = b^2 x$, then a similar calculation shows $v^s = v^{-1}$ and $s^2 = 1$, hence $V := \langle v, s \rangle$ is a dihedral group, and as $u = v^k$ for some $k$ and it commutes with $s$, it lies in the center of $V$, and as the center is just nontrivial if for dihedral groups its order is divisble by $4$ we have $4 \mid |V|$ and $|V| = q-1$. Further by the above computations $V \le C_G(u)$. As $N_G(\langle u \rangle) = N_G(U)$ and $|N_G(U)| = q-1$ (this fact I and the above mentioned I have from B. Huppert, Endliche Gruppen I), we have $C_G(u) = N_G(U)$. Which shows that the centralizer is a dihedral group in this case.

But if $u \in S^x$ I have no idea how to proceed, and maybe my arguments above are to complicated. So I am asking for help!?

• I think Huppert's book includes a description (due to Dickson originally) of all subgroups of ${\rm PSL}(2,q)$. Given that, you can see immediately that, except in some small cases, $U$ and $S$ are maximal and since (in the two cases) they have an involution in their centres, the fact that they are the complete centralizer follows from their maximality. – Derek Holt Dec 23 '15 at 14:55
• You mean $N_G(U)$ and $N_G(S)$ are maximal? – StefanH Dec 23 '15 at 15:38
• That's right, yes - the dihedral groups of order $p-1$ and $p+1$ are maximal. – Derek Holt Dec 23 '15 at 15:53
• Ok, I guess I understand it now. I posted an answer myself. I would be glad if you, or anyone else points me to it if I have made anything wrong or my arguments are to complicated. – StefanH Dec 23 '15 at 17:07

## 1 Answer

I guess I have the answer now, thanks to Derek's comment. Notation like in the question. By the mentioned partitioning property we have $u \in U$ or $u \in S$ (or one of its conjugates, but we can assume that its $U$ or $S$). Suppose $u \in N_G(U)$, then as $N_G(U)$ is a maximal dihedral group with cyclic group $U$, its involution as part of the cyclic group is in the center of $N_G(U)$, hence $N_G(U) \le C_G(u)$. But as $C_G(u) \ne G$ (I have no proof yet, but see my question) and maximality, we have $N_G(U) = C_G(u)$, hence $D = C_G(u)$ has order $q-1$, is maximal, and has order divisible by four.

If $u \in S$, then the reasoning is similar, as $N_G(S)$ is also a dihedral group with cyclic $S$. As just one of the numbers $q - 1$ or $q + 1$ is divisible by four, and $u \in U$ implies $|N_G(U)| = q - 1$ is divisible by four respectively for $u \in S$, the number which is divisible by four determines if $u \in U$ or $u \in S$.

To give a more detailed analysis with respect to Dickson's list of the subgroups of $PSL(2, q)$. I have posted the list here, and it is taken from B. Huppert Endliche Gruppen. Suppose $q \notin \{7,9\}$. Then the dihedral groups of order containing $u$ are maximal, as the dihedral group arising of point (8) in the link have smaller order than $q \pm 1$, also if (7) gives rise to dihedral subgroups it has at most the same order, and for case (4), (5), (6) the possible dihedral groups are also smaller, for example the largest dihedral group in $A_5$ is $D_{10}$ of order $10$, see here, and this is the one whose order is not divisible by four, hence the one containing $u$ must have larger order as $q \ne 9$. Case (4) and (5) I guess could also be handled by a detailed case analysis. Case (1) and (2) do not give rise to any dihedral groups if $p \ne 2$.

If $p = 9$, then by (5) and (6) we have $S_4$ and $A_5$ as subgroups, I do not see how these are related to the dihedral groups and their maximality immediately, but at least we cannot draw the same conclusions. If $p = 7$ by (5) we have that $S_4$ is a subgroup.

As a last note. The essential facts used are that for a dihdral group its center contains the unique involution of its cyclic group if its order is divisible by four, and is trivial otherwise. See for example here, Theorem 4.2. Also the partitioning of $PSL(2, q)$, and the properties of the mentioend groups $U$ and $S$, namely that their normalizers are dihedral, these could all be found in B. Huppert, Endliche Gruppen I, chapter II.8.