Explicit elements of projective special linear group

Question

How would one explicitly write the elements of $PSL(n, q)$? Say for $PSL(2,3)$. I know that it is isomorphic to $A_4$, but in this case I am trying to prove this fact, so I can't use those elements.

I know that $$PSL(n,q) = SL(n,q)/Z(SL(n,q))$$

where $Z$ denotes the centre of the group and $SL(n,q)$ is the group of $n \times n$ matrices with elements in $F_q$ having determinant $1$. Is there a (relatively) simple to actually write down the explicit elements of this group? I'm not sure how they will be written due to the quotient.

Answer 1

Here is an explicit isomorphism between the two groups using an intermediate group:

$$\begin{array}{|l|l|l|} \hline PSL(2,\mathbb{Z_3})& \ \ \ Homog(2,\mathbb{Z_3})& \ \ \ \ \ \ \ \ \ \ \ \ \ A_4\\ \hline \pmatrix{1&0\\0&1}&h_{1}(z)=z&\pmatrix{0 & 1 & 2 & \infty\\0 & 1 & 2 & \infty}\\ \hline \pmatrix{1&0\\1&1}&h_{2}(z)=\dfrac{z}{z+1}&\pmatrix{0 & 1 & 2 & \infty\\0 & 2 & \infty & 1}\\ \hline \pmatrix{1&1\\0&1}&h_{3}(z)=z+1&\pmatrix{0 & 1 & 2 & \infty\\1 & 2 & 0 & \infty}\\ \hline \pmatrix{1&0\\2&1}& h_{4}(z)=\dfrac{z}{2z+1}&\pmatrix{0 & 1 & 2 & \infty\\0 & \infty & 1 & 2}\\ \hline \pmatrix{1&1\\1&2}&h_{5}(z)=\dfrac{z+1}{z+2}&\pmatrix{0 & 1 & 2 & \infty\\ 2 & \infty & 0 & 1}\\ \hline \pmatrix{1&1\\2 & 0}&h_{6}(z)=\dfrac{z+1}{2z}&\pmatrix{0 & 1 & 2 & \infty\\ \infty & 1 & 0 & 2}\\ \hline \pmatrix{1&2\\0 & 1}&h_{7}(z)=z+2&\pmatrix{0 & 1 & 2 & \infty\\ 2 & 0 & 1 & \infty}\\ \hline \pmatrix{1&2\\1 & 0}&h_{8}(z)=\dfrac{z+2}{z}&\pmatrix{0 & 1 & 2 & \infty\\ \infty & 0 & 2 & 1}\\ \hline \pmatrix{2&1\\1 & 1}&h_{9}(z)=\dfrac{2z+1}{z+1}&\pmatrix{0 & 1 & 2 & \infty\\ 1 & 0 & \infty & 2}\\ \hline \pmatrix{0&2\\1 & 0}&h_{10}(z)=\dfrac{2}{z}&\pmatrix{0 & 1 & 2 & \infty\\ \infty & 2 & 1 & 0}\\ \hline \pmatrix{0&2\\1 & 1}&h_{11}(z)=\dfrac{2}{z+1}&\pmatrix{0 & 1 & 2 & \infty\\ 2 & 1 & \infty & 0}\\ \hline \pmatrix{0&1\\2 & 1}&h_{12}(z)=\dfrac{1}{2z+1}&\pmatrix{0 & 1 & 2 & \infty\\ 1 & \infty & 2 & 0}\\ \hline \end{array}$$

i.e., we are going to establish a combined isomorphism between 3 groups, the middle one making the connection between the two extreme ones:

• The group $PSL(2,\mathbb{Z_3})$ for matrix multiplication,

• The group of homographic functions on $\mathbb{Z_3}$ extended to $\mathbb{Z_3} \cup \{ \infty\}$ for the composition of functions (see below), and

• The group $A_4$ of even permutations on 4 objects $\mathbb{Z_3} \cup \{\infty\}$, for the composition of bijections.

This "Rosetta stone" deserves of course a detailed explanation.

Let us begin by recalling what is $PSL(2,\mathbb{Z_3})$ by using a kind of etymological analysis:

• (1) P is as "Projective", [see below],

• (2) SL2 as "Special Linear", $2$-dimensional, more precisely: "L2" is for $2 \times 2$ matrices and "S" (special) for determinant equal to 1 (mod. 3),

• (3) the entries of these matrices being in $\mathbb{Z_3}=\{\bar{0},\bar{1},\bar{2}\}$ with the usual addition and multiplication mod. 3, with the following rules: $3 \equiv 0$, $2 \equiv -1$, $2 \times 2 \equiv 1.$

About the term "projective": two matrices are in the same equivalence class (thus are considered as a same projective "object") iff they obey the equivalence rule:

$$\tag{*}\exists k \neq 0 \ \ \text{such that} \ \ \pmatrix{a&c\\b&d}=k \pmatrix{a'&c'\\b'&d'}.$$

In fact, the unique non trivial case is with $k \equiv 2$ (mod $3$).

Therefore, every equivalence class, i.e., every coset, has 2 elements.

For example, the two following matrices are considered as a same element:

$$\pmatrix{2&0\\1&2} \sim \pmatrix{1&0\\2&1},$$

because the first one is twice the second and reciprocally (only one of them is present in the first column of the table upwards).

Remark: The equivalence relationship (*) is the same as yours because the center of group $GL(2,\mathbb{Z_3})$ is the group with two elements:

$$I=\pmatrix{1&0\\0&1} \ \ \text{and} \ \ 2 I=\pmatrix{2&0\\0&2}$$

constituting the neutral class (the "coset" of $I$).

It's time now to explain the isomorphism between $PGL(2,\mathbb{Z_3})$ and $A_4$, that has been explicited in the table at the beginning.

Every matrix $M=\pmatrix{a&c\\b&d} \in PGL(2,\mathbb{Z_3})$, can be associated with a unique "homography":

$$z \rightarrow h_M(z)=\frac{az+c}{bz+d}$$

Why this uniqueness ? Because equivalence relationship (*) "up to a non-zero multiplicative factor $k$" is associated with a cancellation when a same factor $k$ is present in the numerator and in the denominator of a fractional expression.

Let us now consider the images of $(0,1,2,\infty)$ by $h_M.$

Remark: the presence of $\infty$ isn't a true surprise in such a context because, in general, projective geometry uses infinity almost constantly. Without giving a fully rigorous explanation, it suffices to say that the rules governing the special symbol $\infty$ are the same as the rules dealing with limits "when the variable tends to $\infty$" in real analysis.

Let us consider the example of $M=\pmatrix{1&1\\1&2}$, and its associated homography $h(z)=\frac{1z+1}{1z+2}=1-\frac{1}{z+2}$. Using the upsaid rules, we have:

$$\begin{cases}h(0)&=&1/2=1.2^{-1}=2,\\h(1)&=&"1/0"=\infty,\\h(2)&=&0/4=0,\\h(\infty)&=&1 \ - "1/\infty"=1-0=1. \end{cases}$$

and the fact is that this permutation $\pmatrix{0 & 1 & 2 & \infty\\2 & \infty & 0 & 1}$ on the 4 objects $(0,1,2,\infty)$ is even...

A "rationale" for this evenness : The only possible subgroup with 12 elements of the symmetrical group with $4!=24$ elements is the group of even permutations $A_4$ ($A_n$ is the only subgroup of $S_n$ of index $2$.).

Remark : for projective geometry and homographies on finite fields or rings, see (https://en.wikipedia.org/wiki/Projective_line_over_a_ring).

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Edit: using SAGE (calling sagecell.sagemath.org), one can have further informations, for example :

G=PSL(2,3);G

Answer : Permutation Group with generators [(2,3,4), (1,2)(3,4)]

With the request :

G.list()

one gets the representation

 [(),
 (1,3)(2,4),
 (1,4)(2,3),
 (1,2)(3,4),
 (2,4,3),
 (1,3,4),
 (1,4,2),
 (1,2,3),
 (2,3,4),
 (1,3,2),
 (1,4,3),
 (1,2,4)]

(under the form of permutations, presented as composition of cycles). A good and simple exercise is to establish a correspondence between letters $a,b,\cdots l$ and the last column of the array above. The following request :

G.cayley_table()

gives :

 *  a b c d e f g h i j k l
  +------------------------
 a| a b c d e f g h i j k l
 b| b c a f d e h i g l j k
 c| c a b e f d i g h k l j
 d| d g j a h k b e l c f i
 e| e i k c g l a f j b d h
 f| f h l b i j c d k a e g
 g| g j d k a h e l b i c f
 h| h l f j b i d k c g a e
 i| i k e l c g f j a h b d
 j| j d g h k a l b e f i c
 k| k e i g l c j a f d h b
 l| l f h i j b k c d e g a

If furthermore, you ask :

G.is_isomorphic(AlternatingGroup(4))

the answer is :

True...