A few days ago I wrote about finite simple groups and said that I might write more about them. This post is a follow-on to that one.
In my earlier post I said that finite simple groups fell into five broad categories, and that three of these were easy to describe. This post will chip away at one of the categories I said was hard to describe briefly, namely groups of Lie type or classical groups. These are finite groups that are analogous to (continuous) Lie groups.
General linear group: GL(n, q)
Given a field F, finite or infinite, GL(n, F) is the general linear group over F, consisting of non-singular n by n matrices with entries from the field F. When F is the real or complex numbers, the general linear groups are Lie groups.
If F is a finite field, we get finite groups. For every prime power q there is a unique finite field of order q (up to isomorphism) and so we usually denote the field by its number of elements q. We could define the groups in this post over any field, but we’re primarily interested in finite groups.
Special linear group: SL(n, q)
The general linear groups GL(n, F) are not usually not simple groups until we whittle them down some. Our first step will be to take the subgroup of matrices with determinant 1. These are called the special linear groups SL(n, F).
Projective special linear group: PSL(n, q)
The next step is to identify matrices that are a scalar multiple of each other. This gives us the projective special linear groups PSL(n, q). More formally, we will look at the quotient group of SL(n, q) by its center.
PSL(n, q) = SL(n, q) / Z( SL(n, q) ).
The notation Z here stands for “center of.” This is standard notation, and has nothing to do with the use of Z to denote integers. (The notation comes from zentrum, the German word for center.) In our case the center is the set of multiples of the identity matrix. The examples below will show what it means to mod out by the center.
The groups PSL(n, q) are simple unless n =2 and q = 2 or 3.
Note: The groups denoted An(q) in the previous post correspond to PSL(n+1, q) here.
Example: PSL(2, 5)
Let’s look at the case n = 2 and q = 5, i.e. 2 by 2 matrices whose entries are integers mod 5. This is the smallest example of PSL(n, q) which is a simple group.
We denote these matrices by
and the requirement of having determinant 1 means ad – bc = 1.
How many matrices are in SL(2, 5)? A rough estimate would be 125 because we have three degrees of freedom (four variables minus constraining equation) and each degree of freedom can take on five possible values. It’s not quite that simple because you can’t always fix three variables and solve for the fourth. The exact number of elements turns out to be 120.
Now to move from SL(2, 5) to PSL(2, 5) we have to look at the center of SL(2, 5). These are the diagonal matrices in SL(2, 5) with constant entries along the diagonal. This implies a = d and b = c = 0. Since the determinant is 1, we have a² = 1, and so a = 1 or a = 4. Or more intuitively we could say a = 1 or a = -1 since 4 and -1 are the same mod 5. So we consider a matrix and its negative to be the same matrix. Since SL(2, 5) had 120 elements and we’ve identified elements in pairs, PSL(2, 5) has 60 elements.
The orders of GL(n, q), SL(n, q), and PSL(n, q) in general are given below .
Example: PSL(3, 5)
If we set n = 3 then we’re looking at 3 by 3 matrices. SL(3, 5) is the set of 3 by 3 matrices with elements from the integers mod 5 and determinant 1.
The center of SL(3, 5) is the set of multiples of the identity matrix with determinant 1. If the diagonal elements are a, then the determinant condition says a³ = 1. If you cube the numbers 0, 1, 2, 3, 4 and take the remainders by 5, you’ll see that 1 is the only cube root of 1 mod 5. So the center is just the identity matrix, and modding out by the group identity does nothing. So PSL(3, 5) = SL(3, 5).
Example: PSL(3, 7)
SL(3, 7) is the set of 3 by 3 matrices with entries taken from the integers mod 7, and with determinant 1, much like the example above. This time, however, the center will not be trivial. As before, if the diagonal elements are a, then a³ = 1. But this equation has more solutions mod 7 than it had mod 5. You can see by brute force that the possible values of a are 1, 2, and 4. This means that for a given matrix M in SL(3, 7), we consider M, 2M, and 4M to be the same element.
Example: PSL(n, 2)
The general linear group GL(n, 2) is the set of non-singular n by n matrices with binary entries. Non-singular matrices are those whose determinant is not 0, and over the field with 2 elements, the determinant can only be 0 or 1. So GL(n, 2) = SL(n, 2). There are only two multiples of the identity matrix with binary elements, the zero matrix and the identity matrix, and so the only multiple of the matrix in SL(n, 2) is the identity itself, and we have GL(n, 2) = SL(n, 2) = PSL(n, 2).
Example: PSL(2, 25)
The examples above had q equal to a prime, but q can be a power of a prime, like 25.
The field with pn elements consists of polynomials of degree n-1 with coefficients from the integers mod p. Addition works as expected, but multiplication is different. You select an irreducible polynomial of degree n, and multiplication is carried out modulo this irreducible polynomial. For the field with 25 elements we could choose x² + x + 1 as our irreducible polynomial.
SL(2, 25) is the set of 2 by 2 matrices with entries from the field described above, with determinant 1. The center of SL(2, 25) consists of the identity matrix times a where a² = 1. This equation has two solutions: a = 1 and a = 3x + 3. So to form PSL(2, 25) from SL(2, 25) we identify each matrix M with (3x + 3)M.
More group theory posts
 Denote the order of a group G by |G|. Then we have the following.
Source: The Finite Simple Groups by Robert A. Wilson
2 thoughts on “Finite simple group example: PSL(q)”
Why does “diagonal” imply the diagonal elements are the same?
Good point. I should not have said diagonal matrices but multiples of the identity matrix. I’ll edit the post.