$P(𝔽_3)$的射影变换数|PGL(2,3)|=24

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$P(\mathbb F_2)=\{[1,0],[0,1],[1,1]\}$的自同构是任意置换

${\rm PSL}(2,2) ≅S_3$	[1,0]	[0,1]	[1,1]
$\pmatrix{1\\&1}$	[1,0]	[0,1]	[1,1]
$\pmatrix{&1\\1}$	[0,1]	[1,0]	[1,1]
$\pmatrix{1&1\\&1}$	[1,0]	[1,1]	[0,1]
$\pmatrix{&1\\1&1}$	[0,1]	[1,1]	[1,0]
$\pmatrix{1&1\\1}$	[1,1]	[1,0]	[0,1]
$\pmatrix{1\\1&1}$	[1,1]	[0,1]	[1,0]

$\mathbb P(\mathbb F_3)=\{[0,1],[1,0],[1,1],[1,2]\}$有多少射影变换?

The order of PGL(2, 3) is ${(3^2-1)(3^2-3^1)\over3-1}=24$.
The order of PSL(2, 3) is 12

Wikipedia
The order of ${\rm PGL}(n, q)$ is$${(q^n - 1)(q^n - q)(q^n - q^2) ⋅⋅⋅ (q^n - q^{n-1})\over q - 1}= q^{n^2-1} - O(q^{n^2-3}),$$which corresponds to the order of ${\rm GL}(n, q)$, divided by $q-1$ for projectivization; see q-analog for discussion of such formulas. Note that the degree is $n^2-1$, which agrees with the dimension as an algebraic group. The "O" is for big O notation, meaning "terms involving lower order". This also equals the order of ${\rm SL}(n, q)$; there dividing by $q-1$ is due to the determinant.

The order of ${\rm PSL}(n, q)$ is the above, divided by $|{\rm SZ}(n, q)|$, the number of scalar matrices with determinant $1$ – or equivalently dividing by $|F^×/(F^×)^n|$, the number of classes of element that have no $n$th root, or equivalently, dividing by the number of $n$th roots of unity in $\mathbb F_q$.

These are equal because they are the kernel and cokernel of the endomorphism $F^{\times }{\overset {x^{n}}{\to }}F^{\times };$ formally, $|μ_n| ⋅ |(F^×)^n| = |F^×|.$ More abstractly, the first realizes $\rm PSL$ as $\rm SL/SZ$, while the second realizes $\rm PSL$ as the kernel of ${\rm PGL} → F^×/(F^×)^n$.

${\rm GL}(n,q)/{\rm SL}(n,q)\cong F^\times$
$${\rm PGL}(n,q)={\rm GL}(n, q)/\mathrm Z({\rm GL}(n, q)),\quad\mathrm Z({\rm GL}(n, q))=\Set{\lambda I|\lambda\in F^\times}\cong F^\times$$
$${\rm PSL}(n, F) ={\rm SL}(n, F)/{\rm Z}({\rm SL}(n, F)),\quad{\rm Z}({\rm SL}(n, F))=\Set{\lambda I|\lambda\in F,\lambda^n=1}$$

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参考IntrodMagma用在线Magma计算

1. for x in SL(2,3) do
2. v1:=x*Matrix(GF(3),2,1,[0,1]);
3. v2:=x*Matrix(GF(3),2,1,[1,0]);
4. v3:=x*Matrix(GF(3),2,1,[1,1]);
5. v4:=x*Matrix(GF(3),2,1,[1,2]);
6. printf "$\\pmatrix{%o&%o\\\\%o&%o}$|[%o,%o]|[%o,%o]|[%o,%o]|[%o,%o]\n",x[1,1],x[1,2],x[2,1],x[2,2],v1[1,1],v1[2,1],v2[1,1],v2[2,1],v3[1,1],v3[2,1],v4[1,1],v4[2,1];
7. end for

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GL(2,3)有$(3^2-1)(3^2-3^1)=48$个元素.
SL(2,3)有$24$个元素但不同构于$S_4$！

${\rm SL}(2,3)$	[0,1]	[1,0]	[1,1]	[1,2]
$\pmatrix{1&0\\0&1}$	[0,1]	[1,0]	[1,1]	[1,2]
$\pmatrix{0&1\\2&0}$	[1,0]	[0,2]	[1,2]	[2,2]
$\pmatrix{0&1\\2&2}$	[1,2]	[0,2]	[1,1]	[2,0]
$\pmatrix{0&1\\2&1}$	[1,1]	[0,2]	[1,0]	[2,1]
$\pmatrix{1&1\\0&1}$	[1,1]	[1,0]	[2,1]	[0,2]
$\pmatrix{2&1\\2&0}$	[1,0]	[2,2]	[0,2]	[1,2]
$\pmatrix{2&0\\2&2}$	[0,2]	[2,2]	[2,1]	[2,0]
$\pmatrix{2&2\\2&1}$	[2,1]	[2,2]	[1,0]	[0,1]
$\pmatrix{1&2\\0&1}$	[2,1]	[1,0]	[0,1]	[2,2]
$\pmatrix{1&1\\2&0}$	[1,0]	[1,2]	[2,2]	[0,2]
$\pmatrix{1&2\\2&2}$	[2,2]	[1,2]	[0,1]	[2,0]
$\pmatrix{1&0\\2&1}$	[0,1]	[1,2]	[1,0]	[1,1]
$\pmatrix{2&0\\0&2}$	[0,2]	[2,0]	[2,2]	[2,1]
$\pmatrix{0&2\\1&0}$	[2,0]	[0,1]	[2,1]	[1,1]
$\pmatrix{0&2\\1&1}$	[2,1]	[0,1]	[2,2]	[1,0]
$\pmatrix{0&2\\1&2}$	[2,2]	[0,1]	[2,0]	[1,2]
$\pmatrix{2&2\\0&2}$	[2,2]	[2,0]	[1,2]	[0,1]
$\pmatrix{1&2\\1&0}$	[2,0]	[1,1]	[0,1]	[2,1]
$\pmatrix{1&0\\1&1}$	[0,1]	[1,1]	[1,2]	[1,0]
$\pmatrix{1&1\\1&2}$	[1,2]	[1,1]	[2,0]	[0,2]
$\pmatrix{2&1\\0&2}$	[1,2]	[2,0]	[0,2]	[1,1]
$\pmatrix{2&2\\1&0}$	[2,0]	[2,1]	[1,1]	[0,1]
$\pmatrix{2&1\\1&1}$	[1,1]	[2,1]	[0,2]	[1,0]
$\pmatrix{2&0\\1&2}$	[0,2]	[2,1]	[2,0]	[2,2]

Automorphisms
The Automorphisms group of $\mathrm{SL}(2,3)$ is isomorphic to the symmetric group of degree four.
The inner automorphism group, which is the quotient of $\mathrm{SL}(2,3)$ by its center, and is also the projective special linear group $\mathrm{PSL}(2,3)$, is isomorphic to the alternating group of degree four.
根据蓝色句可列出$\mathrm{PSL}(2,3)$的元素

1. t:=[];
2. for x in SL(2,3) do
3. if not -x in t then
4. Append(~t,x);
5. v1:=x*Matrix(GF(3),2,1,[0,1]);
6. v2:=x*Matrix(GF(3),2,1,[1,0]);
7. v3:=x*Matrix(GF(3),2,1,[1,1]);
8. v4:=x*Matrix(GF(3),2,1,[1,2]);
9. printf "$\\pmatrix{%o&%o\\\\%o&%o},",x[1,1],x[1,2],x[2,1],x[2,2];
10. printf "\\pmatrix{%o&%o\\\\%o&%o}$|",-x[1,1],-x[1,2],-x[2,1],-x[2,2];
11. printf "[%o,%o]|[%o,%o]|[%o,%o]|[%o,%o]\n",v1[1,1],v1[2,1],v2[1,1],v2[2,1],v3[1,1],v3[2,1],v4[1,1],v4[2,1];
12. end if;
13. end for

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${\rm PSL}(2,3)≅A_4$	[0,1]	[1,0]	[1,1]	[1,2]
$\pmatrix{1&0\\0&1},\pmatrix{2&0\\0&2}$	[0,1]	[1,0]	[1,1]	[1,2]
$\pmatrix{0&1\\2&0},\pmatrix{0&2\\1&0}$	[1,0]	[0,2]	[1,2]	[2,2]
$\pmatrix{0&1\\2&2},\pmatrix{0&2\\1&1}$	[1,2]	[0,2]	[1,1]	[2,0]
$\pmatrix{0&1\\2&1},\pmatrix{0&2\\1&2}$	[1,1]	[0,2]	[1,0]	[2,1]
$\pmatrix{1&1\\0&1},\pmatrix{2&2\\0&2}$	[1,1]	[1,0]	[2,1]	[0,2]
$\pmatrix{2&1\\2&0},\pmatrix{1&2\\1&0}$	[1,0]	[2,2]	[0,2]	[1,2]
$\pmatrix{2&0\\2&2},\pmatrix{1&0\\1&1}$	[0,2]	[2,2]	[2,1]	[2,0]
$\pmatrix{2&2\\2&1},\pmatrix{1&1\\1&2}$	[2,1]	[2,2]	[1,0]	[0,1]
$\pmatrix{1&2\\0&1},\pmatrix{2&1\\0&2}$	[2,1]	[1,0]	[0,1]	[2,2]
$\pmatrix{1&1\\2&0},\pmatrix{2&2\\1&0}$	[1,0]	[1,2]	[2,2]	[0,2]
$\pmatrix{1&2\\2&2},\pmatrix{2&1\\1&1}$	[2,2]	[1,2]	[0,1]	[2,0]
$\pmatrix{1&0\\2&1},\pmatrix{2&0\\1&2}$	[0,1]	[1,2]	[1,0]	[1,1]

把上述代码中的SL改为GL就得到PGL(2,3)的元素

${\rm PGL}(2,3)≅S_4$	[0,1]	[1,0]	[1,1]	[1,2]
$\pmatrix{1&0\\0&1},\pmatrix{2&0\\0&2}$	[0,1]	[1,0]	[1,1]	[1,2]
$\pmatrix{2&1\\2&0},\pmatrix{1&2\\1&0}$	[1,0]	[2,2]	[0,2]	[1,2]
$\pmatrix{1&1\\1&0},\pmatrix{2&2\\2&0}$	[1,0]	[1,1]	[2,1]	[0,1]
$\pmatrix{0&2\\1&2},\pmatrix{0&1\\2&1}$	[2,2]	[0,1]	[2,0]	[1,2]
$\pmatrix{2&0\\0&1},\pmatrix{1&0\\0&2}$	[0,1]	[2,0]	[2,1]	[2,2]
$\pmatrix{1&2\\2&0},\pmatrix{2&1\\1&0}$	[2,0]	[1,2]	[0,2]	[2,2]
$\pmatrix{2&2\\1&0},\pmatrix{1&1\\2&0}$	[2,0]	[2,1]	[1,1]	[0,1]
$\pmatrix{0&1\\1&2},\pmatrix{0&2\\2&1}$	[1,2]	[0,1]	[1,0]	[2,2]
$\pmatrix{1&0\\1&2},\pmatrix{2&0\\2&1}$	[0,2]	[1,1]	[1,0]	[1,2]
$\pmatrix{2&1\\0&1},\pmatrix{1&2\\0&2}$	[1,1]	[2,0]	[0,1]	[1,2]
$\pmatrix{1&1\\0&1},\pmatrix{2&2\\0&2}$	[1,1]	[1,0]	[2,1]	[0,2]
$\pmatrix{0&2\\2&0},\pmatrix{0&1\\1&0}$	[2,0]	[0,2]	[2,2]	[1,2]
$\pmatrix{2&0\\2&2},\pmatrix{1&0\\1&1}$	[0,2]	[2,2]	[2,1]	[2,0]
$\pmatrix{1&2\\2&2},\pmatrix{2&1\\1&1}$	[2,2]	[1,2]	[0,1]	[2,0]
$\pmatrix{2&2\\1&2},\pmatrix{1&1\\2&1}$	[2,2]	[2,1]	[1,0]	[0,2]
$\pmatrix{0&1\\2&2},\pmatrix{0&2\\1&1}$	[1,2]	[0,2]	[1,1]	[2,0]
$\pmatrix{1&0\\2&1},\pmatrix{2&0\\1&2}$	[0,1]	[1,2]	[1,0]	[1,1]
$\pmatrix{2&1\\0&2},\pmatrix{1&2\\0&1}$	[1,2]	[2,0]	[0,2]	[1,1]
$\pmatrix{1&1\\0&2},\pmatrix{2&2\\0&1}$	[1,2]	[1,0]	[2,2]	[0,1]
$\pmatrix{0&2\\1&0},\pmatrix{0&1\\2&0}$	[2,0]	[0,1]	[2,1]	[1,1]
$\pmatrix{2&0\\1&1},\pmatrix{1&0\\2&2}$	[0,1]	[2,1]	[2,2]	[2,0]
$\pmatrix{1&2\\1&1},\pmatrix{2&1\\2&2}$	[2,1]	[1,1]	[0,2]	[2,0]
$\pmatrix{2&2\\2&1},\pmatrix{1&1\\1&2}$	[2,1]	[2,2]	[1,0]	[0,1]
$\pmatrix{0&1\\1&1},\pmatrix{0&2\\2&2}$	[1,1]	[0,1]	[1,2]	[2,0]

用Magma验证

1. K := GF(3);
2. G := GL(2,K);
3. V := VectorSpace(G);
4. PGL := OrbitImage(G,sub<V|V.1>);
5. "The order of PGL(2,3) is",#PGL;
6. Ker := OrbitKernel(G,sub<V|V.1>);
7. "The order of GL(2,3)/PGL(2,3) is",#Ker;
8. "The generator of GL(2,3)/PGL(2,3) is",Generators(Ker)

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输出
The order of PGL(2,3) is 24
The order of GL(2,3)/PGL(2,3) is 2
The generator of GL(2,3)/PGL(2,3) is $\pmatrix{2&0\\0&2}$

hbghlyj

${\rm PSL}(2,2)\cong{\rm PGL}(2,2)\cong{\rm SL}(2,2)\cong{\rm GL}(2,2)\cong S_3$ Groupprops
${\rm PSL}(2,3)\cong A_4$ Groupprops   ${\rm PGL}(2,3)\cong S_4$ Groupprops
${\rm PSL}(2,4)\cong{\rm PGL}(2,4)\cong{\rm SL}(2,4)\cong A_5$ Groupprops   ${\rm PGL}(2,4)\cong A_5\times C_3$ Groupprops
${\rm PSL}(2,5)\cong A_5$ Groupprops
${\rm PSL}(2,9)\cong A_6$ Groupprops

hbghlyj

PGL(2,2) is isomorphic to S3 We will prove in this article that PSL(2,2) is isomorphic to S3, that GL(2,2) is isomorphic to S3, and GL(F2) is isomorphic to S3. By definition, \(PSL(V)\) is equal to the quotient \(GL(V)/Z(V)\) where \(Z(V)\) is the centralizer of \(GL(V)\). In this case, the centralizer is just the identity element itself. So we have that \(GL(2,2) = PGL(2,2)\). Further, recall that

\begin{equation*}
S_3 = \{e,(12),(13),(23),(123),(132)\}.
\end{equation*}

Proof. Define the mapping \(\rho: S_3 \longrightarrow PGL(2,2)\) with the next elements

\[ e \longmapsto \mathbb{I}_2 = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, \ (123) \longmapsto \begin{pmatrix}
0 & 1 \\
1 & 1
\end{pmatrix}, \ (132) \longmapsto \begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix} \]

\[ (12) \longmapsto \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}, \ (12)(123) \longmapsto \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}, \ (12)(132) \longmapsto \begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix} \]

Note that \((12)(132) = (13)\), \((12)(123) = (23)\) and \((123)^2 = (132)\). The map is surjective as there exists an element \(S_3\) for any element of \(PGL(2,2)\). That the mapping is a group homomorphism is easy to see and we will give one example, and the rest is for the reader to verify:

\[ \rho(123)\rho(123) = \begin{pmatrix}
0 & 1 \\
1 & 1
\end{pmatrix} \begin{pmatrix}
0 & 1 \\
1 & 1
\end{pmatrix} = \begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix} = \rho(132) \]

Repeat this process for any element.

What is left is to prove injectivity, but we do see already in the mapping above that \(\ker(\rho) = \{\mathbb{I}_2\}\).

So we get \(S_3 \cong PGL(2,2)\).

hbghlyj

math.mit.edu/~dav/smallgroups.pdf
Lemma For any prime $p$, there are $p + 1$ elements in the projective line with the following form:
$$P^{\prime}\left(\mathbb{F}_{p}^{2}\right)=\left\{\left(\begin{array}{l}1 \\ 0\end{array}\right)\left(\begin{array}{l}1 \\ 1\end{array}\right) \cdots\left(\begin{array}{c}1 \\ (p-1)\end{array}\right)\left(\begin{array}{l}0 \\ 1\end{array}\right)\right\}$$
Description of $\operatorname{PGL}\left(2, \mathbb{F}_3\right)$ Because 3 is a prime number we know from the above lemma that $P^{\prime}\left(\mathbb{F}_3^2\right)$ has 4 elements. Similarly to $\mathrm{PGL}\left(2, \mathbb{F}_2\right)$ the action of $\mathrm{PGL}\left(2, \mathbb{F}_3\right)$ permutes these four elements creating an inclusion:
$$
\operatorname{PGL}\left(2, \mathbb{F}_3\right) \hookrightarrow \operatorname{Perm}\left(P^{\prime}\left(\mathbb{F}_3^2\right)\right) \cong S_4
$$
Since both $\mathrm{PGL}\left(2, \mathbb{F}_3\right)$ and $S_4$ have exactly 24 elements (from first lemma) and an inclusion exists, we know that we must have an isomorphism between the groups
$$
\mathrm{PGL}\left(2, \mathbb{F}_3\right) \cong S_4
$$
Description of $\mathrm{PGL}\left(2, \mathbb{F}_4\right)$ Now we will examine $\mathrm{PGL}\left(2, \mathbb{F}_3\right)$ which has $(4+1)(4)(4-1)=60$ elements. The projective line over $\mathbb{F}_4$ contains the following five elements:
$$
\operatorname{Perm}\left(P^{\prime}\left(\mathbb{F}_3^2\right)\right)=\left\{\left(\begin{array}{l}
1 \\
0
\end{array}\right)\left(\begin{array}{l}
1 \\
1
\end{array}\right)\left(\begin{array}{l}
1 \\
a
\end{array}\right)\left(\begin{array}{c}
0 \\
(1+a)
\end{array}\right)\left(\begin{array}{l}
0 \\
1
\end{array}\right)\right\}
$$
These five elements of the projective plane will be referred to as $\overline{0}, \overline{1}, \bar{a}, \overline{1+a}$, and $\bar{\infty}$ respectivefully in the following text. As in a previous lecture on finite fields, $a$ is the elment of $\mathbb{F}_4$ which is the root of $x^2+x+1$.

Similar to the above cases, we know that there exists an inclusion based on the fact that the projective linear group permutes the five elements of the projective line in some way:
$$
\operatorname{PGL}\left(2, \mathbb{F}_4\right) \hookrightarrow \operatorname{Perm}\left(P^{\prime}\left(\mathbb{F}_4^2\right)\right) \cong S_5
$$
However, in this case the orders of the two groups are different. We have $\left|S_5\right|=120$. As discussed above $\left|\mathrm{PGL}\left(2, \mathbb{F}_3\right)\right|=60$. Therefore, there cannot possibly be an isomorphism between groups. For any symmetric group, $S_n$ of order $2 k$, there is only one subgroup of order $k$ : The alternating group $A_n$. Therefore, due to the above inclusion, we must have the following isomorphism:
$$
\operatorname{PGL}\left(2, \mathbb{F}_4\right) \cong A_5
$$

hbghlyj

21Sheet1

4. (i) List the elements of $P G L\left(2, \mathbb{F}_2\right)$. What is the order of $P G L(2, \mathbb{F})$ if $|\mathbb{F}|=q$ ?
(ii) By considering the action of $P G L\left(2, \mathbb{F}_2\right)$ on $\mathbb{F}_2 \mathbb{P}^1$, show that $P G L\left(2, \mathbb{F}_2\right) \cong S_3$. Is $P G L\left(2, \mathbb{F}_3\right) \cong S_4$ ? Is $P G L\left(2, \mathbb{F}_5\right) \cong S_6 ?$