Start with the identity matrix and place an x somewhere below. This generates a p cycle, so it implies 2x, 3x, 4x, and so on. Conjugate by a permutation matrix, specifically a transposition matrix, that swaps rows i and j, and columns i and j. Remember that the transposition matrix has to have a -1 somewhere, to keep the determinant 1. Something like this.
| 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 0 | -1 |
Use such a matrix to move x to a different column, or a different row. Notice that -1 is positioned so that x is not negated; but even if it was negated that really doesn't matter, because x implies the whole p cycle, including -x. In any case, x can be moved anywhere below the main diagonal, and all these matrices are at our disposal.
Write any matrix in S as the product of these elementary matrices. Read the entries from top to bottom, left to right, and multiply the corresponding elementary matrices together, in this order, to realize the given matrix in S.
But how do we get x alone?
If n = 2 then x is already alone, so assume n > 2. Thus p > 2, and 1 is different from -1, and 2 is nonzero. We'll need this below.
Assume y is in row two column one. Scale the first row and column, and the last row and column, by -1. (Do this by conjugating by a diagonal matrix with the corners set to -1). Now you have -y in row two column one, and when this is multiplied by the original matrix, you get 0 in that position. This kills off an entry, and may produce a simpler matrix that we can more easily manipulate.
Let's start with n = 3. Convert M to jordan canonical form. The eigen values are all 1, hence we do not need to extend the base field F to make this conversion. In other words, we can conjugate within the group G*. If the conjugating matrix has determinant 1/a, conjugate again by the diagonal [1,1,a]. Now you are conjugating within G. One matrix is in the normal closure of the other. Let's assume this has been done. If the subdiagonal is [1,0] then there is a single nonzero entry below the main diagonal, and we're done. So let M be a single jordan block, in other words, a 1j matrix. Its subdiagonal is [1,a]. Another matrix implied by M has subdiagonal [-1,b]. Multiply these together and find a matrix that is not the identity, and not 1j. (To demonstrate the latter, subtract the identity, square, and get 0.) This matrix brings in all of S, and that completes the proof for n = 3.
For larger n, convert to jordan form, as we did above. Assume M is a single jordan block, in other words, a 1j matrix. The entire subdiagonal is 1, except the bottom entry, which is tweaked to a. Negate column 1 and row n. Multiply this by M, and entry 2,1 goes away. The result is not the identity, and not a 1j matrix. (To demonstrate the latter, subtract the identity, raise to the n-2, and get 0.) Put this back into jordan form, with the bottom row once again scaled by a, and handle the case where M has more than one bloc.
By induction on n, each block can be reduced to a matrix with a main diagonal and one nonzero entry below. Put this in jordan form, and this block has a single 1 on its subdiagonal. Do this for all blocks, and M has a main diagonal of 1, and scattered ones along the subdiagonal. Let these ones alternate with zeros, starting at the left column. If n is at least 5, we have 1,0,1,0,… Negate rows 1 and 5, and columns 1 and 5, and then multiply by M. The first 1 (on the subdiagonal) goes away, while the second 1 becomes 2. This reduces the number of nonzero entries on the subdiagonal. Continue until there is but one nonzero entry on the subdiagonal. This brings in all of S as shown above. The only case remaining is n = 4, with a subdiagonal of [1,0,a]. Another implied matrix has subdiagonal [-1,0,b]. If b = -a, conjugate this matrix by the diagonal [1,1,s,1/s], thus scaling b by any square you like. Since p is at least 5, there are squares other than 1. Now b is not -a, and the product is not the identity matrix. Also, entry 2,1 goes away, hence the product has but one nonzero entry below the main diagonal, and meets our criteria. This implies all of S.
That completes the proof. Any matrix in S brings in all of S when you take the normal closure in G.
Move x to the lower left. Scale row n by y/x, and column n by x/y. At the same time, you have to scale some other row by x/y and some other column by y/x, so the conjugating diagonal matrix has a determinant of 1. This assumes n > 2. Thus x turns into y, and one p cycle, any p cycle, generates the sylow subgroup across the entire field. This int urn implies all sylow subgroups by conjugation.
But what about n = 2? Conjugate by [s,0|0,t] and x is multiplied by t2. All the square multiples of x are accessible. In characteristic 2 we are done; so let p be odd. Multiply two of these together and get x times (a2+b2). A rather technical lemma proves that two squares sum to a nonsquare, so that covers everything.
The normal closure includes matrices with [2,s|t,0] in the upper left and ones down the main diagonal, where st = -1. (Verify the eigen values are all 1.) Multiply this by [1,0|x,1] and get [2+sx,s|t,0]. The upper left can be anything we like, as long as the adjacent elements have a product of -1. In particular, the upper left can be 1.
Multiply [1,s|t,0] by [1,u|v,0] and get [1+sv,u|t,tu]. Premultiply by an elementary matrix to subtract s times the second row from the first, giving [sv,0|t,tu]. Subtract a multiple of the first row from the second to get [sv,0|0,tu]. Since s and v are unrelated, any eigen value is fair game. The "other" eigen value takes up the slack, giving a determinant of 1.
Apply the above reasoning to the second and third entries, and build a diagonal matrix whose entries are all ones, except for the second entry, which is arbitrary, and the third entry, which is the inverse of the second. Do the same for the third and fourth entries, and the fourth and fifth, and so on. These diagonal matrices can be multiplied to produce any diagonal matrix in G.
The abelian diagonal group combines with the sylow subgroup S to generate the normalizer N, i.e. all the lower triangular matrices. In summary, one p cycle is enough to bring in all of N, and all the conjugates of N follow from there. This includes every matrix with eigen values in F, as these are similar to a lower triangular by Schur's theorem.
Conversely, start with any matrix whose eigen values lie in F, and convert it to jordan canonical form. It is now diagonal with scattered ones on the subdiagonal. (Again, the entry on the floor might be a instead of 1.) Let the diagonal have order k. This is the least common multiple of the orders of the eigen values. If k is 1 then all the eigen values are 1. We either have 1 or a p cycle. Only the identity matrix is similar to the identity matrix, so if the initial matrix was nontrivial, then we have a p cycle, and everything follows from there.
Next assume k is not 1. There is at least one eigen value that is not one;yet they all lie in F. If there is a jordan block with a subdiagonal, let d be the diagonal and let y be the subdiagonal, and rais d+y to the k. We only need look at the top two rows of this block. Since d and y commute, use the binomial theorem. The diagonal turns to 1, and the subdiagonal is kdk-1y. This is 0 only if k is 0. Yet k is w-1, or a factor thereof. Our jordan matrix, raised to the k, has something below the main diagonal, and is a p cycle.
Finally assume the matrix is diagonal. If it is constant then it is in the center of G. That is a normal subgroup, and I'll assume this has been pulled out. We're really looking at G mod its center. See the previous section for the characterization of the center of G.
Permute the entries so that the same eigen values are grouped together. Thus the eigen value at the upper left is not equal to the eigen value at the lower right. Call these s and t respectively. Premultiply by an elementary matrix that adds the first row to the last. Then post multiply by the inverse, which subtracts the last column from the first. The lower left corner is now s-t, which is nonzero. I'll call it x.
Raise this matrix to the w-1. The diagonal becomes 1, thus we have 1 or a p cycle. It all depends on the lower left corner. Things don't commute, so we can't use the binomial theorem, but we can expand (d+x)w-1 by the distributive property, and throw away any term with two instances of x, as these are all 0. This leaves the sum of dixdw-1-i, as i runs from 0 to w-1. This in turn is the sum of tixsw-1-i. Pull x out, and the remaining factor is (sw-tw)/(s-t). (Since s ≠ t, this is well defined.) Since sw = s, and tw = t, the expression is 1. Therefore x survives, and the matrix is a p cycle.
In summary, a normal subgroup of G mod its center contains every matrix with eigen values in F, or it contains none of them.
We may want to apply this to G mod its center. Start with a normal subgroup in G/C that contains a matrix with real eigen values. Real eigen values remain real when you multiply by any diagonal matrix in the center, so this is well defined. Lift to a normal subgroup of G. This contains the aforementioned matrix, and becomes all of G. Hence the normal subgroup downstairs is G/C. If we can get from an arbitrary matrix to a matrix with real eigen values, then G/C is simple.