Matrices mod p, An Introduction

Introduction

The invertible matrices over Z_p, (or any other finite field for that matter), form a finite group, with the identity matrix acting as the identity element. If we are dealing with n by n matrices, p is assumed to be at least as large as n; otherwise things break down. (You'll see why in the next section.) What is the decomposition series for this group? Are there interesting subgroups to be had? What do the sylow subgroups look like? How about the group of automorphisms? These questions are very difficult to answer, even for modest values of n and p. I'm just exploring here, and I thought you might come along for the ride.

Determinant

The determinant is a group homomorphism from the invertible matrices into the nonzero elements mod p. In fact the map is onto. Put x in the upper left and place ones down the main diagonal. The determinant is x. We have created a reverse group homomorphism from x back to a matrix with determinant x. This makes the group a semidirect product. It is not a direct product however. Multiply on the left and you scale the first row by x; multiply on the right and you scale the first column by x. So the first step in the decomposition series pulls out the matrices with determinant 1, and yields a quotient group of Z_p^*. The latter group is cyclic, even for an arbitrary finite field, so I'm not going to spend any more time on that. Hereinafter, matrices will have determinant 1. This group is sometimes called the special linear group, with its own notation, but I think I'll just call it G.

Ring

In general, matrices could be drawn from a commutative ring R. A matrix is invertible iff its determinant is a unit. The invertible matrices form a group that maps onto the units of R via the determinant homomorphism. G is once again the kernel, i.e. those matrices with determinant 1.

The characteristic of R is assumed to be a power of p, where p ≥ n. An example is the integers mod p^k. In this case a ring homomorphism takes us back to the matrices mod p. Other rings may have characteristic p, whereupon the subring of integers mod p creates, as a subgroup, the matrices mod p. Therefore, understanding G mod p provides a solid foundation.

Trivial Case

When n = 1, there is but one matrix with determinant 1. Thus G is trivial. Henceforth n is assumed to be greater than 1.

Notation

As you move through the pages of this topic, you will encounter a lot of notation. If you aren't comfortable with the notation, things will get very confusing, very fast. Here is a summary of things to come.

n: the matrices under consideration are n by n. n is at least 2.
p: all the entries in each matrix are taken mod p. Remember that p is prime, and it is at least as large as n.
F: In some cases I can prove a theorem for an arbitrary finite field. F is the finite field, and it has characteristic p. The additive group is d copies of Z_p running in parallel, where d is the dimension of F over p. The multiplicative group of F is cyclic. Even if I am working mod p, I may still talk about F, with the understanding that F is the finite field of order p.
F^*: the multiplicative group of F.
w: the order of F. This is some power of p. As mentioned earlier, F^* is a cycle of length w-1.
Primitive: any element that generates F^*. There is a primitive element for each integer coprime to w-1. I will often use e for a primitive element.
K: a finite extension of F that is needed to split a polynomial, often the characteristic polynomial of a matrix. Thus K holds all the eigen values.
Real: in the context of the finite field extension K/F, the members of F are real. I guess it's easier to say "real eigen values", rather than "eigen values that all belong to F". Hope this isn't too confusing.
[Main] Diagonal: the entries in the matrix that run from upper left to lower right.
Diagonal Space: The group of diagonal matrices under multiplication. This is an abelian group. Each entry is independent of the others, generating its own cycle of length w-1. The structure is similar to a vector space, hence I call it diagonal space. It isn't stricly a vector space, but I think the notation helps more than it hurts. Place a primitive element e in various places along the main diagonal to find a basis for diagonal space. Take logs base e to make the group additive. It's dimension is n, or n-1 if we insist that the determinant equal 1.
Constant Diagonal: a diagonal matrix whose nonzero entries are all the same. This is a subspace of dimension 1 within diagonal space.
Subdiagonal: the diagonal line just below the main diagonal. It contains n-1 entries.
Superdiagonal: the diagonal line just above the main diagonal. It contains n-1 entries.
Stripe: a diagonal line in the matrix. The main diagonal has index 0. The superdiagonal has index 1. The subdiagonal has index -1. The lower left entry has index -(n-1). Multiply two stripes together and you add their indexes. Separate two matrices into stripes, and multiply, and the highest stripe of the product has index equal to the sum of the indexes of the highest stripes in the two matrices. If this index is less than -(n-1), then the product is 0.
Constant Stripe: a matrix where each stripe is constant. when multiplying two such matrices, separate them into stripes and multiply as described above. The product of two stripes is another stripe with all entries the same. The sum of two such stripes is also constant. Put this all together and the product is constant stripe. Therefore the constant stripe matrices form a subring within the ring of matrices. Furthermore, since any two stripes commute, this is a commutative ring. It is isomorphic to the ring of truncated rational expressions in an indeterminant x. The constant diagonal matrices are the constants. If the stripe of index i has entries of l, include the term lxⁱ. Add and multiply polynomials in the usual way, but cut off the result at xⁿ and x^-n. This is not a division ring. When n = 2, the polynomial x^-1+1+x corresponds to the matrix [1,1|1,1], and has no inverse. But if you restrict to the lower triangular matrices, with diagonal nonzero or with diagonal equal to 1, every matrix is invertible, and you have an abelian group.
Antidiagonal: the diagonal line that runs from upper right to lower left. It contains n entries.
Composite Matrix: a matrix where each entry is another matrix. If the outer matrix is n by n, and each submatrix is d by d, then this can be viewed as one large nd by nd matrix. Verify that addition and multiplication are the same either way. The two rings are isomorphic.
G: all the n by n matrices over F with determinant 1.
G^*: all the n by n matrices with nonzero determinant. As shown above, G is the kernel of G^*, with F^* as quotient.
G^±: the n by n matrices with determinant 1 or -1.
G₃: the group G when n = 3. The subscript defines n. I may also talk about G₅^*, or G₂^±.
O: the orthonormal n by n matrices with determinant 1.
O′: a certain subgroup of O based on even permutations. Not surprisingly, O′ has index 2 in O. This group will be described later.
O^±: all orthonormal n by n matrices. These will have determinant 1 or -1. Thus O is a normal subgroup of index 2 within O^±.
O₃: the group O when n = 3.
Q: the quaternions over F with norm 1.
Q^±: the quaternions over F with norm 1 or -1.
C: the center of G, or O, or Q, or whatever group we are talking about. I will often write G/C, for G mod its center. In fact, G/C is a simple group, with a few exceptions when p = 2. (I'll prove this later.) This is similar to the alternating groups, which are simple once you get past A₄.
U: the matrices of G that consist of standard unit vectors. Each row has one nonzero entry, which is 1 or -1. If a column has two nonzero entries, then the rows containing these two nonzero entries are linearly dependent, and the determinant is 0. Therefore each column also looks like a standard unit vector along one of the axes.
U^±: the group of matrices that consist of standard unit vectors along the axes. The determinant of such a matrix is 1 or -1. Thus U is a normal subgroup of index 2 within U^±. U lives in O lives in G, and U^± lives in O^± lives in G^±.
E: the matrices of G that consist of extended standard unit vectors. Each row has one nonzero entry. Verify that each column also has one nonzero entry.
S: the standard p sylow subgroup of G. This comprises the lower triangular matrices of G with ones down the main diagonal. (I'll prove this later.) Since p does not divide the index of G within G^*, S is also the standard sylow subgroup for G^*.
N: the normalizer of S. This comprises all the lower triangular matrices of G, or of G^*. (I'll prove this later.)
Atom: a matrix with ones down the main diagonal and one 1 below. Atoms can be multiplied together to build a stripe, thus forming an abelian group. Sometimes an atom contains something other than 1 below the main diagonal, especially if F goes beyond p. The atom 1 implies 2 3 4 etc, but nothing else.
Consider a 3 by 3 example. Let X have 1 in positiotn 2,1 and let Y have 1 in position 3,2. Let Z have 1 in position 3,1. Note that YX = XYZ. Atoms in the same stripe commute, except for adjacent atoms, which spin off the atom below.
φ: an automorphism of G, or O, or whatever group we are considering.
Conjugal: an automorphism that can be accomplished through similarity. If Z is an invertible matrix, then ZG/Z is a conjugal automorphism on G. Note that det(Z) need not equal 1. Thus ZG/Z need not be an inner automorphism. But every inner automorphism is conjugal. Thus the inner automorphisms form a subgroup of the conjugal automorphisms, which are a subgroup of all automorphisms. All the automorphisms of G₂ are conjugal. (I'll prove this later.)
1j: a matrix is 1j if it consists of one jordan block. Convert to jordan form, and you have ones down the main diagonal and ones down the subdiagonal.
J: the 1j matrix in jordan form. This has ones down the main diagonal and ones down the subdiagonal.
2j: a matrix is 2j if it consists of two jordan blocks. Convert to jordan form, and the subdiagonal has one zero, and all the rest ones.
mj: a matrix is mj if it consists of more than one jordan block. Convert to jordan form, and the subdiagonal has at least one zero somewhere.