4 Reduction of classification of finite subgroups of $\operatorname{PGL}_2(\bar{\mathbb {F}}_p)$ to classification of finite subgroups of $\operatorname{PSL}_2(\bar{\mathbb {F}}_p)$

4.1 Over an algebraically closed field $\operatorname{PSL}_n(F)$ is isomorphic to the projective $\operatorname{PGL}_n(F)$

When $F$ is algebraically closed and $\textrm{char}(F) \neq 2$ we can construct an isomorphism between the projective special linear group and the projective general linear group.

Definition 4.1

✓

Let $\varphi : \operatorname{SL}_n(R) \rightarrow \operatorname{PGL}_n(R)$ be the injection of $\operatorname{PSL}_n(R)$ into $\operatorname{PGL}_n(R)$ defined by

\[ S \mapsto i(S) \; (R^\times I) \]

where $i : \operatorname{SL}_n(F) \hookrightarrow \operatorname{GL}_n(F)$ is the natural injection of the special linear group into the general linear group.

We prove a useful fact about elements that belong to the center of $\operatorname{GL}_n(R)$:

Lemma 4.2

Let $R$ be a commutative ring, then $G \in GL_n(F)$ belongs to center of $\operatorname{GL}_n(R)$, $Z(\operatorname{GL}_n(R))$ if and only if $G = r \cdot I$ where $r \in R^\times $.

Proof ▶

Suppose $G \in GL_n(F)$ belongs to $Z(\operatorname{GL}_n(F))$ then for all $H \in \operatorname{GL}_n(F)$ we have that $G H = H G$. We will find it sufficient to only consider the case where $H$ is a transvection matrices. Let $1 \leq i {\lt} j \leq n$, then the transvection matrices are of the form $T_{ij} = I + E_{ij}$ where $E_{ij}$ is the standard basis matrix given by

\[ E_{{ij}_{kl}} = \begin{cases} 1 & \text{if $i = k$ and $l = j$}\\ 0 & \text{otherwise} \end{cases} \]

Given $T_{ij} G = (I + E_{ij}) G = G T_{ij} (I + E_{ij})$, and addition is commutative we can use the cancellation law to yield that

\[ E_{ij} G = G E_{ij} \]

But $G$ only commutes with $E_{ij}$ for all $i \neq j$ if $G = r \cdot I$ for some $r \in R^\times $.
Suppose $G = r \cdot I$ for some $r \in R^\times $ then it is clear that for all $H \in \operatorname{GL}_n(F)$ that $r \cdot I H = r \cdot H = H \cdot r = H (r \cdot I)$

Lemma 4.3

Let $R$ be a non-trivial commutative ring, then $Z(\operatorname{SL}_n(R)) \subseteq \ker (\varphi )$.

Proof ▶

If $S \in Z(\operatorname{SL}_n(R)) \leq \operatorname{SL}_n(F)$ then $S = \omega I$ where $\omega $ is a primitive root of unity.

Because $\varphi = \pi _{Z(\operatorname{GL}_n(F))} \circ i$, the kernel of $\varphi $ is $i^{-1}(Z(\operatorname{GL}_n(F)))$, where we recall that $i : \operatorname{SL}_n(R) \hookrightarrow \operatorname{GL}_n(F)$ is the injection of $SL_n(F)$ into $\operatorname{GL}_n(F)$.

But given $i(S) = i(\omega \cdot I) = \omega \cdot I$ is of the form $r \cdot I$ where $r \in R^\times $ by 4.2 it follows that $S \in \ker \varphi $, as desired.

Definition 4.4

✓

Given $Z(\operatorname{SL}_n(F)) \ker \varphi $ as shown in 4.3, by the universal property there exists a unique homomorphism $\bar{\varphi } : \operatorname{PSL}_n(F) \rightarrow \operatorname{PGL}_n(F)$ which is the lift of $\varphi $. Where $\varphi = \bar{\varphi } \circ \pi _{Z(\operatorname{SL}_n(F))}$ and $\pi _{Z(\operatorname{SL}_n(F))} : \operatorname{SL}_n(F) \rightarrow \operatorname{PSL}_n(F)$ is the canonical homomorphism from the group into its quotient.

Lemma 4.5

The homomorphism $\bar{\varphi }$ is injective.

Proof ▶

To show $\bar{\varphi }$ is injective we must show that $\ker \bar{\varphi } \leq \bot _{\operatorname{PSL}_n(F)}$ where $\bot _{\operatorname{PSL}_n(F)}$ is the trivial subgroup of $\operatorname{PSL}_n(F)$.

Let $[S] \in \operatorname{PSL}_n(F)$ and suppose $[S] \in \ker \bar{\varphi }$. If $[S] \in \ker \bar{\varphi }$ then $\bar{\varphi } ([S]) = [1]_{\operatorname{PGL}_n(F)}$. But on the other hand, $\bar{\varphi } ([S]) = \varphi (s)$ and so $\varphi (S) = 1_{\operatorname{PGL}_n(F)}$

and thus $S \in Z(\operatorname{GL}_n(F))$, from 4.2 it follows that $s = r \cdot I$ for some $r \in R^\times $. But given the restriction of $S \in \operatorname{SL}_n(F)$ we know that

\begin{equation*} \det (S) = \det (r \cdot I) = r^n = 1 \implies \text{$r$ is a $n$ \textsuperscript{th} root of unity} \end{equation*}

Therefore, given elements of $Z(\operatorname{SL}_n(F))$ are those matrices of the form $\omega \cdot I$ where $\omega $ is a $n$^th root of unity, we can conclude that $[S] = [1]_{\operatorname{PSL}_n(F)}$ and thus $\ker \bar{\varphi } \leq \bot _{\operatorname{PSL}_n(F)}$ as required.

Which shows that the homomorphism $\bar{\varphi }$ is injective.

Before we can show that $\bar{\varphi }$ is surjective we need the following lemma which allows us to find a suitable representative for an arbitrary element of $\operatorname{PGL}_n(F)$.

Lemma 4.6

If $F$ is an algebraically closed field then for every $G \in \operatorname{GL}_n(F)$ there exists a nonzero constant $\alpha \in F^\times $ and an element $S \in \operatorname{SL}_n(F)$ such that

\begin{equation*} G = \alpha \cdot S \end{equation*}

Proof ▶

Let $G \in \operatorname{GL}_n(R)$ then define

\begin{equation*} P(X) := X^n - \det (G) \end{equation*}

By assumption $F$ is algebraically closed and $\det (G) \in F^\times $ thus there exists a root $\alpha \in F^\times $ such that

\begin{equation*} \alpha ^n - \det (G) = 0 \iff \alpha = \sqrt[n]{\det (G)} \end{equation*}

Let $S = \frac{1}{\alpha } \cdot G$, by construction $S \in \operatorname{SL}_n(F)$ as

\begin{equation*} \det (S) = \left(\frac{1}{\alpha }\right) \cdot \det (G) = \frac{1}{\det (G)} \det (G) = 1 \end{equation*}

Lemma 4.7

The map $\bar{\varphi }$ is surjective.

Proof ▶

Let $G \; (F^\times I) = [G] \in \operatorname{PGL}_n(F)$, then $G \in \operatorname{GL}_n(F)$ we can find a representative of $[G']$, that lies within the special linear group. Given elements of the special linear group are matrices with determinant equal to one, we must scale $G$ to a suitable factor to yield a representative which lies within $\operatorname{SL}_n(F)$. Suppose $\det (G) \ne 1$ and let

\[ P(X) := X^n - \det (G) \in F[X] \]

By assumption, $F$ is algebraically closed so there exists a root $\alpha \ne 0\in F$ such that

\[ \alpha ^n - \det (G) = 0 \iff \alpha ^n = \det (G) \]

We can define

\[ G' := \frac{1}{\alpha } \cdot G \quad \text{where} \quad \det (G') = \frac{1}{\alpha ^n} \det (G) = 1. \]

Thus $G' \in \operatorname{SL}_n(F) \leq \operatorname{GL}_n(F)$ and given $G' = \frac{1}{\alpha } G$ we have that $G' \; (F^\times I) = G \; (F^\times I)$.

Therefore, $\varphi (G') = i(G') (F^\times I) = G' (F^\times I) = G (F^\times I)$.

Lemma 4.8

The map $\bar{\varphi }$ is bijective

Proof ▶

We have shown that $\bar{\varphi }$ is injective in 4.5 and have shown that $\bar{\varphi }$ is surjective in 4.7. Therefore,$\bar{\varphi }$ defines a bijection from $\operatorname{PSL}_n(F)$ to $\operatorname{PGL}_n(F)$.

Theorem 4.9

If $F$ is an algebraically closed field, then the map $\bar{\varphi } : \operatorname{PSL}_n(F) \rightarrow \operatorname{PGL}_n(F)$ defines a group isomorphism between $\operatorname{PSL}_n(F)$ and $\operatorname{PGL}_n(F)$.

Proof ▶

The map $\bar{\varphi }$ was shown to be a bijection in 4.8 and given $\bar{\varphi }$ is mulitplicative as it was defined to be the lift of the homomorphism $\varphi $, we can conclude that $\bar{\varphi }$ defines a group isomorphism between $\operatorname{PSL}_n(F)$ and $ºPGL_n(F)$

This isomorphism will be essential to the classification of finite subgroups of $\operatorname{PGL}_2(\bar{\mathbb {F}}_p)$, as we only need understand a the classification of subgroups of $\operatorname{PSL}_2(F)$ structure to reach our desired result.

4.2 Christopher Butler’s exposition

Following from the isomorphism defined in the previous section, we can now proceed to classify the finite subgroups of $\operatorname{PGL}_2(\bar{\mathbb {F}}_p)$ by classifying the finite subgroups of $\operatorname{PSL}_2(\bar{\mathbb {F}}_p)$. In turn, one can begin classifying the finite subgroups of $\operatorname{PSL}_2(\bar{\mathbb {F}}_p)$ by classifying the finite subgroups of $\operatorname{SL}_2(\bar{\mathbb {F}}_p)$ and then considering what happens after quotienting by the center, $Z(\operatorname{SL}_2(F)) = \langle -I\rangle $.

We now turn our attention to the more general setting when $F$ is an arbitrary field that is algebraically closed, as this will turn out to be sufficient for our purposes.

Given $|\langle -I \rangle | = 2$ when $\textrm{char} F \ne 2$ and $\langle -I\rangle = \bot $ when $\textrm{char} F = 2$. When a finite subgroup of $\operatorname{SL}_2(F)$ is sent through the canonical mapping $\pi _{Z(\operatorname{SL}_2(F))} : \operatorname{SL}_2(F) \rightarrow \operatorname{PSL}_2(F)$ the resulting subgroup will at most shrink by a factor of two or remain intact should the center not be contained within the subgroup.

We now proceed to classify all finite subgroups of $\operatorname{SL}_2(F)$ when $F$ is algebraically closed field. From now on, we follow Christopher Butler’s exposition of Dickson’s classification of finite subgroups of $\operatorname{SL}_2(F)$ over an algebraically closed field $F$. Christopher has been kind enough to provide the TeX code so I could prepare this blueprint which crucially hinges on the result which his exposition covers.

Classification of finite subgroups of PGL

4 Reduction of classification of finite subgroups of \(\operatorname{PGL}_2(\bar{\mathbb {F}}_p)\) to classification of finite subgroups of \(\operatorname{PSL}_2(\bar{\mathbb {F}}_p)\)

4.1 Over an algebraically closed field \(\operatorname{PSL}_n(F)\) is isomorphic to the projective \(\operatorname{PGL}_n(F)\)

4.2 Christopher Butler’s exposition