Matrices mod p, Inner Automorphisms

Automorphisms

The inner automorphisms of a group G are naturally isomorphic to G mod its center. So the group of automorphisms is not much different than the group we're already exploring. This doesn't include outer automorphisms however, which cannot be realized by conjugation. For instance, let S be the lower triangular matrices with ones down the main diagonal. In the 2 by 2 case, as simple as it gets, the group is essentially a p cycle. An automorphism carries [1,0|1,1] to [1,0|x,1] for any nonzero x. But this isn't an inner automorphism. Since S is abelian, S mod its center is trivial, and there are no inner automorphisms.

Outer automorphisms will be addressed later; for now let's focus on inner automorphisms. The group of inner automorphisms may not be much different from the matrix group itself, but it might act, in interesting ways, on various subgroups. Let's have a look.

Sylow Subgroups

G moves sylow subgroups to sylow subgroups by conjugation, and it acts transitively. Something in G takes any sylow subgroup to any other sylow subgroup.

By the third sylow theorem, a q sylow subgroup H fixes itself, and moves all others through orbits divisible by q. That's how we know the number of q sylow subgroups is 1 mod q.

Let an element z have order r, where r is prime, and let z (actually the group generated by z) act on q sylow subgroups. H is fixed by z iff z is in the normalizer of H. If r does not divide the size of the normalizer then this can't happen. Nothing is fixed, and all orbits are of size r.

Yes, this really does happen. Let q be an odd prime greater than n that divides p-1. Let z have order p. Or perhaps |z| is a prime greater than n, and dividing p+1, or p2+p+1, or one of the other factors. Now z acts on q sylow subgroups, but |z| does not divide the normalizer, so z pushes them around in orbits of size |z|.

In the same way, z, whose prime order exceeds n and divides one of the other factors, pushes p sylow subgroups around in orbits of size |z|, because |z| does not divide the normalizer of a p sylow subgroup.

Conjugacy Classes

Although G moves any sylow subgroup to any other sylow subgroup, it does not act transitively on p cycles. Let n = 3, and build two different p cycles, each in jordan form. Let the first have a one on the subdiagonal, and let the second have two ones on the sub diagonal. Since these are distinct jordan matrices, no amount of conjugation will take you from one to the other.

Each p cycle has eigen values of 1, and can be converted to jordan form without extending the field. Therefore, the conjugacy classes of the p cycles within G* are the jordan matrices with ones on the main diagonal. How many conjugacy classes are there? Separate each matrix into jordan blocks. These have various sizes, and these sizes sum to n. Conjugacy classes correspond to sets of integers that sum to n. This is a notoriously difficult problem in combinatorics.

You can tell at a glance if a standard p cycle is diagonalizable. The eigen values remain 1, so the diagonal equivalent is the identity matrix. Yet every conjugate of I is I. So a p cycle in standard form is diagonalizable iff it is the identity matrix. At the other extreme, M corresponds to a single jordan block of size n, a 1j matrix, iff (M-1)n-1 is nonzero, iff the lower left corner of the product is nonzero, iff everything on the subdiagonal is nonzero. This is going to happen most of the time. Other conjugacy classes may take a little more work to determine, if anybody cares.

If q is a prime greater than n, that divides p-1, the standard q sylow subgroup is diagonal. Like the p cycles above, each matrix is already in jordan form. Conjugate by a permutation matrix to move eigen values about. Thus the conjugacy classes correspond to sets of eigen values - another difficult problem in combinatorics.

Moving q Cycles About

Let a diagonal matrix M live in the standard q sylow subgroup, for q > n. Conjugate M by Z and assume the matrix lies in the same sylow subgroup. In other words, ZMY is diagonal. The entries (eigen values) are the same, just permuted. What can we say about Z?

I'm going to make an assumption that is usually true, especially for large q. The eigen values are distinct. Now move forward with a proof that is similar to what we saw before. Take the dot product of the ith row of Z with the jth column of Y, for j ≠ i. The result is 0 of course, since Y and Z are inverses. Let the terms of the dot product be t1 through tn. Now scale the columns of Z by the entries in M. Take the dot product again, and a linear combination of t1 through tn is equal to 0. The coefficients are the entries in M. Now step up to M2. The conjugate of M2 is the square of the conjugate, and is diagonal. So apply the above to M2. A linear combination of t1 through tn is 0, with coefficients taken from M2. Continue this process up to Mn. Write this as a system of n linear equations in t1 through tn. The matrix is vandermonde. Since eigen values are distinct, the matrix is nonsingular. The only input vector that yields 0 is 0. Therefore all the terms of the dot product are 0. The nonzero entries in Zi completely miss the nonzero entries in Yj. As we saw before, this proves Z lies in E. Conversely, any Z in E will map M into a q cycle belonging to the same sylow subgroup.

For M as above, the odds of selecting a matrix Z that keeps M in the standard sylow subgroup are approximately 1 over the index of E in G. This is approximately 1/p to the n2-n. Not very likely!

Disjoint Cycles

If M has distinct eigen values, and order q, for q > n and q ≠ p, then M belongs to exactly one q sylow subgroup. Suppose not, and swing M over to a diagonal matrix. (Extend F up to K if necessary to make this happen.) The eigen values are still distinct; but now they lie along the diagonal. Let U be the set of q cycles with distinct eigen values that live on the main diagonal. Let M generate B, one of the cycles in U. Conjugate by Z and every cycle in U stays within U if Z belongs to E, or moves outside of U if Z does not belong to E. Select Z so that conjugation by Z maps U onto V, the cycles in the "other" sylow group containing B. This means some cycle A in U maps onto B. Yet Z, outside of E, must map A to something outside of U. This is a contradiction. Thus B cannot lie in both U and V after all. When eigen values are distinct, M extends to a unique sylow subgroup.

Now assume Z has order q with distinct eigen values. Use the disjoint variation of the third sylow theorem to state that Z fixes the sylow group that contains it, and moves all the others about. It also moves the other q cycles with distinct eigen values about, since each of these uniquely determines a sylow subgroup.

Of course some other Z, with order coprime to E, will move the q sylow subgroups, and the cycles with distinct eigen values, about, since Z cannot live in anybody's normalizer.

The bottom line: hardly anything is fixed by conjugation.

p Cycles and Sylow Groups

Can we make a similar claim for p cycles? Not for all p cycles you understand, but for most of them? Put a p cycle in jordan form, and assume it consists of a single jordan block. In other words, this is a 1j matrix. We showed above that most p cycles fit this category. For instance, a lower triangular p cycle is 1j iff its subdiagonal has no zeros. And when p is large, the subdiagonal is almost certain to be devoid of zeros. So convert to jordan form, and M has ones on the main diagonal, and ones on the subdiagonal, and zeros elsewhere.

Assume ZMY is lower triangular. Since eigen values are preserved, we still have ones down the main diagonal, and the conjugate is part of the standard sylow subgroup along with M.

Separate M into two pieces, the diagonal and the subdiagonal, and conjugate each piece separately, as we did before. The conjugate of the identity is the identity, and that is obviously lower triangular. Conjugate the subdiagonal, and each row of Z is shifted to the left, then dotted with each column of Y. Let the last column of Y have entries t1 through tn. You can call this column vector t if you like. Since Y and Z are inverses, a linear combination of t1 through tn is 0, using the first row of Z as coefficients. Call this row vector r; thus r.t = 0. But now we have another linear equation; the coefficients of r are shifted left (with 0 put on the end), and applied to t, and the result is still 0.

If ZMY is lower triangular, then the same holds for ZMY squared. This is the same as the conjugate of M2. The subdiagonal is 2, and the diagonal below that is 1. Discard the identity, then separate M2 into two pieces, the subdiagonal and the subsubdiagonal. Now r is shifted left and doubled, and then added to r shifted left by 2. This is dotted with t to get 0. Yet we know that the dot product of r shifted left by 1 with t is already 0, so we can take this away. Therefore r, shifted left by 2, and dotted with t, is 0.

apply the same reasoning to M3. The diagonals contain 1, 3, 3, 1, moving down and to the left, according to pascal's triangle. This tells us r, shifted by 3, and dotted with t, yields 0. Continue this process through Mn-1. This builds a matrix of shifted versions of r, that when multiplied by t, yields a column vector of zeros. If the rightmost entry of r is nonzero, the matrix is antitriangular, and nonsingular. The only solution is 0, hence t = 0. In other words, the last column of Y is 0. Yet Y is invertible, a contradiction, hence the last entry of r is 0.

Do the same for the second row of Z, and the third, and the fourth, and so on, until the last column of Z, other than the lower right, is 0.

Remember that Y is the inverse of Z. The inverse is the adjoint, and the determinants that become the rightmost column of Y come out 0. Thus both matrices are 0 on the right, above the main diagonal.

Ignore the bottom row and apply this reasoning to the upper left n-1 by n-1 block. Again, both matrices come out 0 in column n-1 above the main diagonal. Continue all the way up to the top, and Z and Y are lower triangular.

If M is any 1j matrix in S, some Y conjugates it into jordan form. Naturally Z conjugates it back. As shown above, Z is lower triangular. Therefore Y is lower triangular.

Apply this twice, and a lower triangular matrix moves any 1j matrix in S to any other 1j matrix in S. Conversely, if Z makes such a move, and is not lower triangular, then conjugate both sides by L, so that the destination is in jordan form. Now LZ takes a 1j matrix in S to jordan form, and has to be lower triangular. Since L is lower triangular, multiply by L inverse on the left, and Z is lower triangular after all. In summary, Z moves a 1j matrix in S to another 1j matrix in S iff Z is lower triangular.

From here it's familiar territory. Suppose two sylow subgroups have a 1j matrix in common. Conjugate, so that one of the two sylow groups is S. Conjugate S by Z to realize the "other" sylow subgroup. Thus Z is not lower triangular. Let B be the 1j p cycle in common. now Z moves the cycle A onto the cycle B, and this is a contradiction, as we saw before. Each 1j matrix, and each implied p cycle, belongs to one and only one sylow subgroup.

apply the disjoint variation of the third sylow theorem, and M pushes all the p sylow subgroups (other than the one that contains it) about in orbits of size p. The 1j p cycles are also pushed about in orbits of size p, since they determine p sylow subgroups uniquely. Almost everything moves under conjugation.

Subgroups of G

All this propagates down to subgroups, such as O. Eigen values remain distinct, and these q cycles push each other about; there are just fewer of them. The jordan equivalent of a p cycle does not change, so 1j p cycles push each other about; there are just fewer of them.