Finite subgroups of PGL(2,K)

TL;DR

This paper classifies finite subgroups of PGL(2,K) with order prime to the characteristic of K, providing a comprehensive understanding of their conjugacy classes.

Contribution

It offers a complete classification of finite subgroups of PGL(2,K) up to conjugacy for groups of order prime to char(K).

Findings

Classification of finite subgroups up to conjugacy

Identification of subgroup structures in PGL(2,K)

Results applicable to algebraic and geometric contexts

Abstract

We classify, up to conjugacy, the finite subgroups of PGL(2,K) of order prime to char(K).

Equations

$$\{(\pm 1,0,\pm \phi ),(\pm \phi ,\pm 1,0),(0,\pm \phi ,\pm 1)\}$$

$$A\cdot e_i=e_{i+1},B\cdot e_i={\zeta }^i{e}_i$$

$$A^r=B^r=I,BA=\zeta AB.$$

$${}^{\sigma }{\alpha }^{\prime }={\zeta }^{a(\sigma )}{\alpha }^{\prime }{}^{\sigma }{\beta }^{\prime }={\zeta }^{b(\sigma )}{\beta }^{\prime }for\; each\sigma \in \mathfrak{g}.$$

$$1\to {\mathbf{G}}_m\to {\mathrm{GL}}_r\to {\mathrm{PGL}}_r\to 1$$

$$H$$

$${}^{\sigma }{\alpha }^{\prime }=(-1{)}^{a(\sigma )}{\alpha }^{\prime }and{}^{\sigma }{\beta }^{\prime }=(-1{)}^{b(\sigma )}{\beta }^{\prime }for\; each\sigma \in \mathfrak{g}.$$

$$g^{-1}{}^{\sigma }g=i(a(\sigma ),b(\sigma )).$$

$$n_1:z\mapsto \frac{z+1}{z-1},n_2:z\mapsto \iota z,$$

$$n_2\cdot (\alpha ,\beta )=(\alpha ,-\alpha \beta ).$$

Full text

Introduction

The aim of this note is to describe, up to conjugacy, the finite subgroups of $\mathrm{PGL_{2}}(K)$, for an arbitrary field $K$. Throughout the paper, we consider only subgroups whose order is prime to the characteristic of $K$.

When $K=\mathbf{C}$, or more generally when $K$ is algebraically closed, the answer is well known: any such group is isomorphic to $\mathbf{Z}/r$, $D_{r}$ (the dihedral group), $\mathfrak{A}_{4}$, $\mathfrak{S}_{4}$ or $\mathfrak{A}_{5}$, and there is only one conjugacy class for each of these groups. If $K$ is arbitrary, the group $\mathrm{PGL_{2}}(K)$ is contained in $\mathrm{PGL_{2}}(\overline{K})$, so the subgroups of $\mathrm{PGL_{2}}(K)$ are among the previous list; it is not difficult to decide which subgroups occur for a given field $K$, see §1.

So the only question left is to describe the conjugacy classes in $\mathrm{PGL_{2}}(K)$ of the subgroups in the list. In §2 we give a general answer for subgroups of $G(K)$, for an algebraic group $G$, in terms of (non-abelian) Galois cohomology. We illustrate the method on one example in §3, and apply it to the case $G=\mathrm{PGL_{2}}$ in §4.

The motivation for looking at this question was to understand the appearance of the Brauer group in the case of $(\mathbf{Z}/2)^{2}$ considered in [B]. The result is somewhat disappointing, as it turns out that this case (which could be treated directly, as in [B]) is the only one where some second Galois cohomology group plays a role. At least our method explains this role, and hopefully may be useful in other situations.

1. The possible subgroups

We repeat that whenever we mention a finite group, we always assume that its order is prime to the characteristic of $K$. The following is classical (see [S2], 2.5).

Proposition 1.1**.**

$1)$* $\mathrm{PGL_{2}}(K)$ contains $\mathbf{Z}/r$ and $D_{r}\,$111We denote by $D_{r}$ the dihedral group with $2r$ elements. if and only if $K$ contains $\zeta+\zeta^{-1}$ for some primitive $r$-th root of unity $\zeta$.*

$2)$* $\mathrm{PGL_{2}}(K)$ contains $\mathfrak{A}_{4}$ and $\mathfrak{S}_{4}$ if and only $-1$ is the sum of two squares in $K$.*

$3)$* $\mathrm{PGL_{2}}(K)$ contains $\mathfrak{A}_{5}$ if and only if $-1$ is the sum of two squares and $5$ is a square in $K$.*

Proof : One way to prove this is to use the isomorphism \mathrm{PGL_{2}}(K)\vbox{\hbox to22.76228pt{\hfill{\scriptstyle\sim}\hfill} \nointerlineskip\hbox to22.76228pt{{\hfill\longrightarrow\hfill}} } $\mathrm{SO}(K,q)$, where $q$ is the quadratic form $q(x,y,z)=x^{2}+yz$ on $K^{3}$ ([D], II.9). If a group $H$ embeds into $\mathrm{SO}(K,q)$, we have a faithful representation $\rho$ of $H$ in $K^{3}$, which preserves an indefinite quadratic form.

\bullet\ Case $H=\mathbf{Z}/r$ : let $g$ be a generator; the existence of $q$ forces the eigenvalues of $\rho(g)$ in $\overline{K}$ to be of the form $(\zeta,\zeta^{-1},1)$, with $\zeta$ a primitive $r$-th root of 1. This implies $\zeta+\zeta^{-1}\in K$. Conversely, if $\lambda:=\zeta+\zeta^{-1}$ is in $K$, the homography $\displaystyle z\mapsto{(\lambda+1)z-1\over z+1}$ is an element of order $r$ of $\mathrm{PGL_{2}}(K)$.

\bullet\ Case $H=D_{r}$ : by the previous case, if $D_{r}\subset\mathrm{PGL_{2}}(K)$, $\lambda:=\zeta+\zeta^{-1}$ is in $K$. Conversely if $\lambda\in K$, the homographies ${z\mapsto 1/z}$ and $\displaystyle z\mapsto{(\lambda+1)z-1\over z+1}$ generate a subgroup of $\mathrm{PGL_{2}}(K)$ isomorphic to $D_{r}$.

\bullet\ Cases $H=\mathfrak{A}_{4},\mathfrak{S}_{4}$ or $\mathfrak{A}_{5}\,$. The representation $\rho$ must be irreducible. Each of the groups $\mathfrak{A}_{4}$ and $\mathfrak{S}_{4}$ has exactly one irreducible 3-dimensional representation with trivial determinant, which is defined over the prime field; the only invariant quadratic form (up to a scalar) is the standard form $q_{0}(x,y,z)=x^{2}+y^{2}+z^{2}$. Thus $\mathfrak{A}_{4}$ and $\mathfrak{S}_{4}$ are contained in $\mathrm{PGL_{2}}(K)$ if and only if $q_{0}$ is equivalent to $\lambda q$ for some $\lambda\in K^{*}$, which means that $q_{0}$ represents 0.

Since $\mathfrak{A}_{5}$ contains elements of order 5, the condition $\sqrt{5}\in K$ is necessary. Suppose this is the case, and put $\varphi=\frac{1}{2}(1+\sqrt{5})$; the subgroup of $\mathrm{SO}(K,q_{0})$ preserving the icosahedron with vertices

[TABLE]

is isomorphic to $\mathfrak{A}_{5}$. It follows as above that $\mathfrak{A}_{5}$ embeds in $\mathrm{SO}(K,q)$ if and only if $q_{0}$ represents 0.∎

2. Some Galois cohomology

2.1.

In this section we consider an algebraic group $G$ over $K$, and a subgroup $H\subset G(K)$. We choose a separable closure $K_{s}$ of $K$, and put $\mathfrak{g}:=\mathrm{Gal}(K_{s}/K)$. We are interested in the set of embeddings $H\hookrightarrow G(K)$ which are conjugate in $G(K_{s})$ to the natural inclusion $i:H\hookrightarrow G(K)$, modulo conjugacy by an element of $G(K)$. We denote this (pointed) set by $\mathop{\rm Emb}\nolimits_{i}(H,G(K))$.

We will use the standard conventions for non-abelian cohomology, as explained for instance in [S3], ch. I, §5. We will also use the notation of [S3] for Galois cohomology: if $G$ is an algebraic group over $K$, we put $\mathrm{H}^{i}(K,G):=\mathrm{H}^{i}(\mathfrak{g},G(K_{s}))$.

Proposition 2.2**.**

Let $Z$ be the centralizer of $H$ in $G(K_{s})$. The pointed set $\mathop{\rm Emb}\nolimits_{i}(H,G(K))$ is canonically isomorphic to the kernel of the natural map $\mathrm{H}^{1}(K,Z)\rightarrow\mathrm{H}^{1}(K,G)$.

Proof : Let $X\subset G(K_{s})$ be the subset of elements $g$ such that $g^{-1}\,{}^{\sigma}\!g\in Z$ for all $\sigma\in\mathfrak{g}$. The group $G(K)$ (resp. $Z$) acts on $X$ by left (resp. right) multiplication. By [S3], ch. I, 5.4, cor. 1, the kernel of $\mathrm{H}^{1}(K,Z)\rightarrow\mathrm{H}^{1}(K,G)$ is identified with the (left) quotient by $G(K)$ of the subset of $\mathfrak{g}$-invariant elements in $G(K_{s})/Z$; but this subset is by definition $X/Z$, so we can identify our kernel to the double quotient $G(K)\backslash X/Z$.

For every $g\in X$, the conjugate embedding $gig^{-1}$ belongs to $\mathop{\rm Emb}\nolimits_{i}(H,G(K))$. Any element $j\in\mathop{\rm Emb}\nolimits_{i}(H,G(K))$ is of the form $gig^{-1}$ for some $g\in G(K_{s})$; for $\sigma\in\mathfrak{g}$, the element $\,{}^{\sigma}\!g$ again conjugates $i$ to $j$, hence $g^{-1}\,{}^{\sigma}\!g\in Z$ and $g\in X$. Thus the map $g\mapsto gig^{-1}$ from $X$ to $\mathop{\rm Emb}\nolimits_{i}(H,G(K))$ is surjective. Two elements $g$ and $g^{\prime}$ of $X$ give the same element in $\mathop{\rm Emb}\nolimits_{i}(H,G(K))$ if and only if $g^{\prime}$ belongs to the double coset $G(K)gZ$. Therefore the above map induces a canonical bijection G(K)\backslash X/Z\vbox{\hbox to22.76228pt{\hfill{\scriptstyle\sim}\hfill} \nointerlineskip\hbox to22.76228pt{{\hfill\longrightarrow\hfill}} }\mathop{\rm Emb}\nolimits_{i}(H,G(K)).∎

2.3.

Let us write down the correspondence explicitly: a class in our kernel is represented by a 1-cocycle $\mathfrak{g}\rightarrow Z$ which becomes a coboundary in $G$, hence is of the form $\sigma\mapsto g^{-1}\,{}^{\sigma}\!g$ for some $g\in X$; we associate to this class the embedding $gig^{-1}$.

2.4.

We are actually more interested in the set $\mathop{\rm Conj}\nolimits(H,G(K))$ of subgroups of $G(K)$ which are conjugate to $H$ in $G(K_{s})$, modulo conjugacy by $G(K)$. Associating to an embedding its image defines a surjective map $im:\mathop{\rm Emb}\nolimits_{i}(H,G(K))\rightarrow\mathop{\rm Conj}\nolimits(H,G(K))$. The normalizer $N$ of $H$ in $G(K_{s})$ acts on $H$ by automorphisms, hence also on $\mathop{\rm Emb}\nolimits_{i}(H,G(K))$. Two embeddings with the same image differ by an automorphism of $H$, which must be induced by an element of $N$ if the embeddings are conjugate under $G(K_{s})$. It follows that $im$ induces an isomorphism \mathop{\rm Emb}\nolimits_{i}(H,G(K))/N\vbox{\hbox to22.76228pt{\hfill{\scriptstyle\sim}\hfill} \nointerlineskip\hbox to22.76228pt{{\hfill\longrightarrow\hfill}} }\mathop{\rm Conj}\nolimits(H,G(K)).

2.5.

Let us translate this in cohomological terms. Let $\mathrm{H}^{1}(K,Z)_{0}$ denote the kernel of the map $\mathrm{H}^{1}(K,Z)\rightarrow\mathrm{H}^{1}(K,G)$. An element $n$ of $N$ acts on $\mathop{\rm Emb}\nolimits_{i}(H,G(K))$ by $j\mapsto j\kern 2.0pt{\scriptstyle\circ}\kern 2.0pt\mathrm{int}(n^{-1})$; if $j=gig^{-1}$, this amounts to replace $g$ by $gn$, hence the 1-cocycle $\varphi:\sigma\mapsto g^{-1}\,{}^{\sigma}\!g$ by $n^{-1}\varphi\,{}^{\sigma}\!n$. This formula defines an action of $N$ on $\mathrm{H}^{1}(K,Z)$ which preserves $\mathrm{H}^{1}(K,Z)_{0}$; the map $g\mapsto gHg^{-1}$ induces an isomorphism of pointed sets \mathrm{H}^{1}(K,Z)_{0}/N\vbox{\hbox to22.76228pt{\hfill{\scriptstyle\sim}\hfill} \nointerlineskip\hbox to22.76228pt{{\hfill\longrightarrow\hfill}} }\mathop{\rm Conj}\nolimits(H,G(K)).

3. An example

3.1.

In this section we fix an integer $r\geq 2$, prime to $\mathrm{char}(K)$, and we assume that $K$ contains a primitive $r$-th root of unity $\zeta$. We consider the matrices $A,B\in\mathrm{M}_{r}(K)$ defined on the canonical basis $(e_{1},\ldots,e_{r})$ of $K^{r}$ by

[TABLE]

for $1\leq i\leq r$, with the convention $e_{r+1}=e_{1}$.

The matrices $A$ and $B$ generate the $K$-algebra $\mathrm{M}_{r}(K)$, with the relations

[TABLE]

Their classes $\bar{A},\bar{B}$ in $\mathrm{PGL}_{r}(K)$ commute; we consider the embedding $i:(\mathbf{Z}/r)^{2}\hookrightarrow\mathrm{PGL}_{r}(K)$ which maps the two basis vectors to $\bar{A}$ and $\bar{B}$. The image $H$ of $i$ is its own centralizer; in particular, $H$ is a maximal commutative subgroup of $\mathrm{PGL}_{r}(K)$.

By the Kummer exact sequence (and the choice of $\zeta$), the group $\mathrm{H}^{1}(K,\mathbf{Z}/r)$ is identified with $K^{*}/K^{*r}$; the pointed set $\mathrm{H}^{1}(K,\mathrm{PGL}_{r})$ can be viewed as the set of isomorphism classes of central simple $K$-algebras of dimension $r^{2}$ ([S1], X.5).

Lemma 3.2**.**

Let $\alpha,\beta\in K^{*}$, and let $\bar{\alpha},\bar{\beta}$ be their images in $K^{*}/K^{*r}$. The map $\mathrm{H}^{1}(i):\mathrm{H}^{1}(K,\mathbf{Z}/r)^{2}\rightarrow\mathrm{H}^{1}(K,\mathrm{PGL}_{r})$ associates to $(\bar{\alpha},\bar{\beta})$ the class of the cyclic $K$-algebra $A_{\alpha,\beta}$ generated by two variables $x,y$ with the relations $x^{r}=\alpha$, $y^{r}=\beta$, $yx=\zeta xy$.

Proof : We choose $\alpha^{\prime},\beta^{\prime}$ in $K_{s}$ with $\alpha^{\prime r}=\alpha$ and $\beta^{\prime r}=\beta$. The Kummer isomorphism associates to $(\alpha,\beta)$ the homomorphism $(a,b):\mathfrak{g}\rightarrow(\mathbf{Z}/r)^{2}$ defined by

[TABLE]

Its image in $\mathrm{H}^{1}(K,\mathrm{PGL}_{r}(K_{s}))$ is the class of the 1-cocycle $\sigma\mapsto\bar{A}^{a(\sigma)}\bar{B}^{b(\sigma)}$.

Now let us recall how we associate to the algebra $A_{\alpha,\beta}$ a cohomology class $[A_{\alpha,\beta}]$ in $\mathrm{H}^{1}(K,\mathrm{PGL}_{r})$ (loc. cit.). We choose an isomorphism of $K_{s}$-algebras u:\mathrm{M}_{r}(K_{s})\vbox{\hbox to22.76228pt{\hfill{\scriptstyle\sim}\hfill} \nointerlineskip\hbox to22.76228pt{{\hfill\longrightarrow\hfill}} }A_{\alpha,\beta}\otimes_{K}K_{s}. For each $\sigma\in\mathfrak{g}$, $u^{-1}\,{}^{\sigma}\!u$ is an automorphism of $\mathrm{M}_{r}(K_{s})$, hence of the form $\mathrm{int}(g_{\sigma})$ for some $g_{\sigma}$ in $\mathrm{PGL}_{r}(K_{s})$. Then $[A_{\alpha,\beta}]$ is the class of the 1-cocycle $\sigma\mapsto g_{\sigma}$.

In our case we define $u$ on the generators $A,B$ by $u(A)=\beta^{\prime}y^{-1}$, $u(B)=\alpha^{\prime-1}x$. Then the automorphism $u^{-1}\,{}^{\sigma}\!u$ multiplies $A$ by $\zeta^{b(\sigma)}$ and $B$ by $\zeta^{-a(\sigma)}$, which gives $g_{\sigma}=\bar{A}^{a(\sigma)}\bar{B}^{b(\sigma)}$ as above.∎

3.3.

The exact sequence

[TABLE]

gives rise to a coboundary homomorphism \partial_{r}:\mathrm{H}^{1}(K,\mathrm{PGL}_{r})\rightarrow$$\mathrm{H}^{2}(K,\mathbf{G}_{m})=\mathop{\rm Br}\nolimits(K) which is injective (loc. cit.). The class $\partial_{r}[A_{\alpha,\beta}]\in\mathop{\rm Br}\nolimits(K)$ is the symbol $(\alpha,\beta)_{r}$; it depends only on the classes of $\alpha$ and $\beta$ (mod. $K^{*r}$). The map $(\ ,\ )_{r}:(K^{*}/K^{*r})^{2}\rightarrow\mathop{\rm Br}\nolimits(K)$ is bilinear and alternating. Since $\partial_{r}$ is injective, we find:

Proposition 3.4**.**

The set $\mathop{\rm Emb}\nolimits_{i}((\mathbf{Z}/r)^{2},\mathrm{PGL}_{r}(K))$ is isomorphic to the set of couples $(\alpha,\beta)$ in $(K^{*}/K^{*r})^{2}$ such that $(\alpha,\beta)_{r}=0$.∎

We will describe the correspondence more explicitely in the case $r=2$ in the next section.

4. Conjugacy classes in $\mathrm{PGL_{2}}(K)$

Proposition 4.1**.**

Assume that $K$ is separably closed. Two finite subgroups of $\mathrm{PGL_{2}}(K)$ which are isomorphic $($and of order prime to $\mathrm{char}(K))$ are conjugate.

*Proof *: Again this is certainly well-known; we give a quick proof for completeness. The possible subgroups are those which appear in Proposition 1.1.

An element of order $r$ of $\mathrm{PGL_{2}}(K)$ comes from a diagonalizable element of $\mathrm{GL}_{2}(K)$, hence is conjugate to the homothety $z\mapsto\zeta z$ for some $\zeta\in\boldsymbol{\mu}_{r}(K)\,$222As usual we denote by $\boldsymbol{\mu}_{r}(K)$ the group of $r$-th roots of unity in $K$.; thus a cyclic subgroup of order $r$ of $\mathrm{PGL_{2}}(K)$ is conjugate to the group $H_{r}$ of homotheties $z\mapsto\lambda z,\ \lambda\in\boldsymbol{\mu}_{r}(K)$.

There is only one group $D_{r}$ containing $H_{r}$, namely the subgroup generated by $H_{r}$ and the involution $z\mapsto 1/z$; it follows that all dihedral subgroups of order $2r$ are conjugate to this subgroup.

For the three remaining groups, we use again the isomorphism \mathrm{PGL_{2}}(K)\vbox{\hbox to22.76228pt{\hfill{\scriptstyle\sim}\hfill} \nointerlineskip\hbox to22.76228pt{{\hfill\longrightarrow\hfill}} }\mathrm{SO}_{3}(K). The groups $\mathfrak{A}_{4}$ and $\mathfrak{S}_{4}$ have exactly one irreducible representation of dimension 3 with trivial determinant, while $\mathfrak{A}_{5}$ has two such representations which differ by an outer automorphism: this is elementary in characteristic 0, and the general case follows by [I], ch. 15. Therefore two isomorphic subgroups $H$ and $H^{\prime}$ of $\mathrm{SO}_{3}(K)$ of this type are conjugate in $\mathrm{GL}_{3}(K)$. The only quadratic forms preserved by $H$ or $H^{\prime}$ are the multiple of the standard form; thus the element $g$ of $\mathrm{GL}_{3}(K)$ which conjugates $H$ to $H^{\prime}$ must satisfy ${}^{t}\!g\,g=\lambda I$ for some $\lambda\in K$. Replacing $g$ by $\pm\mu g$, with $\mu^{2}=\lambda^{-1}$, we have $g\in\mathrm{SO}_{3}(K)$, hence our assertion.∎

Recall that the determinant induces a homomorphism $\ \overline{\mathop{\rm det}\nolimits}:$ $\mathrm{PGL_{2}}(K)\rightarrow K^{*}/K^{*2}$.

Theorem 4.2**.**

$1)\ \mathrm{PGL_{2}}(K)$* contains only one conjugacy class of subgroups isomorphic to $\mathbf{Z}/r\ (r>2)$, $\mathfrak{A}_{4}$, $\mathfrak{S}_{4}$ or $\mathfrak{A}_{5}$.*

$2)$* The conjugacy classes of cyclic subgroups of order $2$ of $\mathrm{PGL_{2}}(K)$ are parametrized by $K^{*}/K^{*2}$: to $\alpha\in K^{*}\ \mathrm{(mod.}\,K^{*2})$ corresponds the involution $z\mapsto\alpha/z$.*

$3)$* The homomorphism $\overline{\mathop{\rm det}\nolimits}:\mathrm{PGL_{2}}(K)\rightarrow K^{*}/K^{*2}$ induces a bijective correspondence between:*

$\bullet$* conjugacy classes of subgroups of $\mathrm{PGL_{2}}(K)$ isomorphic to $(\mathbf{Z}/2)^{2}$;*

$\bullet$* subgroups $G\subset K^{*}/K^{*2}$ of order $\leq 4$, such that $(-\alpha,-\beta)_{2}=0$ for all $\alpha,\beta$ in $G$ (see (3.3)).*

$4)$* Assume that $\boldsymbol{\mu}_{r}(K)$ has order $r$. The conjugacy classes of subgroups $D_{r}$ of $\mathrm{PGL_{2}}(K)$ are parametrized by $K^{*}/K^{*2}\boldsymbol{\mu}_{r}(K)$. The subgroup corresponding to $\alpha\in K^{*}\ {\rm(mod.}\ K^{*2}\boldsymbol{\mu}_{r}(K)\,)$ consists of the homographies $z\mapsto\zeta z$ and $z\mapsto\alpha\eta/z$, for $\zeta,\eta\in\boldsymbol{\mu}_{r}(K)$.*

Proof : Using Proposition 4.1 we can apply the method of §3. We give the list of the subgroups of $\mathrm{PGL_{2}}(K_{s})$ and their centralizers:

[TABLE]

In case 1), we have $\mathrm{H}^{1}(K,Z)=\{1\}$ (using $\mathrm{H}^{1}(K,\mathbf{G}_{m})$ $=\{1\}$). The result follows from (2.5).

Case 2): We apply Proposition 2.2, taking for $i(\bar{1})$ the involution $z\mapsto 1/z$. The centralizer $Z$ is the semi-direct product of a torus $\mathbf{G}_{m}$ and the subgroup $\mathbf{Z}/2$ generated by the involution $\iota:z\mapsto-z$. The pointed set $\mathrm{H}^{1}(K,\mathbf{G}_{m}\rxtimes\mathbf{Z}/2)$ is identified with $\mathrm{H}^{1}(K,\mathbf{Z}/2)\cong K^{*}/K^{*2}$. The map $\mathrm{H}^{1}(K,\mathbf{Z}/2)\rightarrow\mathrm{H}^{1}(K,\mathrm{PGL_{2}})$ is trivial, for instance because the injection $\mathbf{Z}/2\rightarrow\mathrm{PGL_{2}}$ factors through a torus $\mathbf{G}_{m}$. Hence the set of conjugacy classes of involutions of $\mathrm{PGL_{2}}(K)$ is identified with $K^{*}/K^{*2}$.

To describe the correspondence explicitely we follow (2.3). Let $\alpha\in K^{*}$, and let $\alpha^{\prime}\in K_{s}^{*}$ such that $\alpha^{\prime 2}=\alpha$; the class of $\alpha$ (mod. $K^{*2}$) corresponds to the class of the 1-cocycle $a:\mathfrak{g}\rightarrow\mathbf{Z}/2$ given by $\,{}^{\sigma}\!\alpha^{\prime}=(-1)^{a(\sigma)}\alpha^{\prime}$. In $\mathrm{PGL_{2}}(K_{s})$ we have $i(a(\sigma))=g^{-1}\,{}^{\sigma}\!g$, where $g$ is the homography $z\mapsto\alpha^{\prime}z$. Thus the subgroup corresponding to $\alpha$ is $gHg^{-1}$, which is generated by the involution $z\mapsto\alpha/z$.

Case 3): Let $i:(\mathbf{Z}/2)^{2}\hookrightarrow\mathrm{PGL_{2}}(K)$ be the embedding which maps the basis vectors $e_{1}$ and $e_{2}$ to the involutions $z\mapsto 1/z$ and $z\mapsto-z$. By Proposition 3.4 the set ${\mathop{\rm Emb}\nolimits_{i}((\mathbf{Z}/2)^{2},\mathrm{PGL_{2}}(K))}$ is canonically identified to the set of couples $(\alpha,\beta)$ in $(K^{*}/K^{*2})^{2}$ with $(\alpha,\beta)_{2}=0$.

Again we make the correspondence explicit following (2.3). Let $\alpha,\beta\in K^{*}$ with $(\alpha,\beta)_{2}=0$. This means that the conic $x^{2}-\alpha y^{2}-\beta z^{2}=0$ is isomorphic to $\mathbf{P}^{1}_{K}$, thus there exists $\lambda,\mu$ in $K$ with $\lambda^{2}-\alpha-\beta\mu^{2}=0$. We choose $\alpha^{\prime}$ and $\beta^{\prime}$ in $K_{s}$ such that $\alpha^{\prime 2}=\alpha$ and $\beta^{\prime 2}=\beta$; as above we define the homomorphisms $a$ and $b:\mathfrak{g}\rightarrow\mathbf{Z}/2$ by

[TABLE]

Put $\displaystyle\theta:={\beta^{\prime}\mu\over\lambda+\alpha^{\prime}}={\lambda-\alpha^{\prime}\over\beta^{\prime}\mu}$; let $g\in\mathrm{PGL_{2}}(K_{s})$ be the homography $\displaystyle z\mapsto\alpha^{\prime}{z-\theta\over z+\theta}$ . An easy computation gives

[TABLE]

Thus the embedding of $(\mathbf{Z}/2)^{2}$ associated to $(\alpha,\beta)$ is $gig^{-1}$; it maps $e_{1}$ to the homography $\displaystyle h_{1}:z\mapsto{\lambda u-\alpha\over z-\lambda}$ , and $e_{2}$ to $h_{2}:z\mapsto\alpha/z$ . Note that $\overline{\mathop{\rm det}\nolimits}(h_{1})=-\beta$ and $\overline{\mathop{\rm det}\nolimits}(h_{2})=-\alpha$.

Now we have to take into account the action of the normalizer $N$ of $H$ in $\mathrm{PGL_{2}}(K_{s})$. This is the subgroup $\mathfrak{S}_{4}$ generated by $H$ and the homographies

[TABLE]

where $\iota$ is a square root of $-1$. We apply the recipe of (2.5). Since $n_{1}\in\mathrm{PGL_{2}}(K)$, it acts on $\mathrm{H}^{1}(K,H)$ through its action on $H$, which permutes $e_{1}$ and $e_{2}$; thus it maps $(\alpha,\beta)\in(K^{*}/K^{*2})\times(K^{*}/K^{*2})$ to $(\beta,\alpha)$. The action of $n_{2}$ on $H$ fixes $e_{2}$ and exchanges $e_{1}$ with $e_{1}+e_{2}$; to get the action on $\mathrm{H}^{1}(K,H)$ we have to multiply by the class of the cocycle $\sigma\mapsto n_{2}^{-1}\,{}^{\sigma}\!n_{2}$, that is, \sigma\mapsto i\bigl{(}(\sigma(\iota)/\iota)\,e_{2}\bigr{)}. Hence $n_{2}$ acts on $\mathrm{H}^{1}(K,H)$ by

[TABLE]

Let $G_{\alpha,\beta}$ be the subgroup of $K^{*}/K^{*2}$ generated by $-\alpha$ and $-\beta$; it is the image of $H$ by the homomorphism $\overline{\mathop{\rm det}\nolimits}:\mathrm{PGL_{2}}(K)\rightarrow K^{*}/K^{*2}$. If $G_{\alpha,\beta}\cong(\mathbf{Z}/2)^{2}$, the orbit $N\cdot(\alpha,\beta)$ in $(K^{*}/K^{*2})\times(K^{*}/K^{*2})$ has 6 elements, which are the couples $(-x,-y)$ with $x,y\in G_{\alpha,\beta}$, $x\neq y$. If $G_{\alpha,\beta}\cong(\mathbf{Z}/2)$, the orbit has 3 elements, which are the couples $(-x,-y)$ with $x,y\in G_{\alpha,\beta}$, $(x,y)\neq(1,1)$. Finally if $G_{\alpha,\beta}$ is trivial the orbit consists only of $(-1,-1)$. Thus the conjugacy classes of subgroups $(\mathbf{Z}/2)^{2}$ in $\mathrm{PGL_{2}}(K)$ are parametrized by the subgroups $G\subset K^{*}/K^{*2}$ of order $\leq 4$, with the property $(-\alpha,-\beta)_{2}=0$ for each $\alpha,\beta$ in $G$.

Case 4): The group $D_{r}$ is generated by two elements $s,t$ with the relations $s^{2}=t^{r}=1$ and $sts=t^{-1}$. We choose a primitive $r$-th root of unity $\zeta$ and consider the embedding $i:D_{r}\hookrightarrow\mathrm{PGL_{2}}(K)$ such that $i(s)$ is the involution $z\mapsto 1/z$ and $i(t)$ the homothety $z\mapsto\zeta z$. The centralizer is $\mathbf{Z}/2$, generated by the involution $z\mapsto-z$. As in case 2) it follows that $\mathop{\rm Emb}\nolimits_{i}(D_{r},\mathrm{PGL_{2}}(K))$ is isomorphic to $\mathrm{H}^{1}(K,\mathbf{Z}/2)$. Also the previous argument shows that the embedding corresponding to $\alpha\in K^{*}$ is the conjugate of $i$ by the homography $z\mapsto\alpha^{\prime}z$, with $\alpha^{\prime 2}=\alpha$, so it maps $s$ to $z\mapsto\alpha/z$ and $t$ to $z\mapsto\zeta z$.

To complete the picture we have to take into account the action of the normalizer $N$ of $i(D_{r})$ in $\mathrm{PGL_{2}}(K_{s})$. This is the subgroup $D_{2r}$ generated by $i(s):z\mapsto 1/z$ and the homothety $n:z\mapsto\eta z$, where $\eta\in K_{s}$ is a primitive $2r$-th root of unity. The action of $i(s)$ is trivial, and $n$ acts by multiplication by the cocycle $\sigma\mapsto n^{-1}\,{}^{\sigma}\!n$, which corresponds to the class of $\eta^{2}$ in $K^{*}/K^{*2}$. Since $\eta^{2}$ generates $\boldsymbol{\mu}_{r}(K)$, the assertion 4) follows.∎