| a-sb | b |
| -s2b+s(a-d)+c | sb+d |
In characteristic 2, the lower left is always a square, for any s. For odd p, suppose the lower left is never a square. There are (p-1)/2 nonsquares, and each admits at most 2 values of s. So some s leads to a square in the lower left. If this square is 0, then the matrix is upper triangular, and we're done. So we have a nonzero square in the lower left.
The next trick is to conjugate by a diagonal matrix [s,0|0,t]. Naturally st = 1. Conjugation leaves a and d the same, multiplies b by s2, and multiplies c by t2. So b and c can trade off squares. By the above, the lower left is a square. Move the square to the upper right, and the lower left becomes 1.
Assume our matrix has been transformed into [a,1|c,d]. (I can put 1 in the upper right, as easily as the lower left.) Conjugate by [1,0|s,1]. This gives the matrix at the top of the page with b set to 1.
| a-s | 1 |
| -s2+s(a-d)+c | s+d |
Multiply this on the right by [a,1|c,d], and the upper right entry is s-(a+d). We want this to be 0, thus building a lower triangular matrix. Set s to the trace, a+d, and after substituting for s, and setting c = ad-y (where y is the determinant), the product matrix looks like this.
| -y | 0 |
| -2y(a+d) | -y |
Of course y = 1, but even if we did not know y was 1, it is definitely nonzero mod p. This matrix has real eigen values, namely -y. If the lower left is nonzero, then we are not in the center, and all the matrices with real eigen values come rolling in, along with everything else. We can stop here, unless p = 2 or a+d = 0.
Suppose a+d = 0. The base matrix is now [a,1|c,-a], with trace 0 and determinant y. Conjugate by [1,0|s,1] as we did before, and to simplify things, replace s with t+a. The result is [-t,1|-t2-y,t]. When t = -a we are back to our original matrix [a,1|c,-a]. But now we see that any t is fair game.
Multiply two such matrices together, based on t and u, where t ≠ u. The upper right is nonzero. How about the lower left? The expression simplifies to (u-t)*(y-ut). Select any u and t so that their product is y, and the result is upper triangular. That completes the case a+d = 0, unless this can only be done when u = t. Write t2 = y, and there are but 2 solutions. As long as there are more than 2 nonzero values in our base field, there is no trouble. We only need look at p = 2 and p = 3. Let's look at these now.
When p = 2, G = S3. I'll let you verify this yourself.
When p = 3, G mod ±1 has order 12, and can't possibly be simple. With 4 sylow subgroups of order 3, there are 8 elements of order 3. The remaining 4 elements form the single, normal sylow subgroup of order 4.
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Multiply these by each other, negating the result any time you wish, and this subgroup is Z2Z2.
Let the entire group of order 12 act on its 4 sylow subgroups by conjugation. The sylow theorems state that each sylow group moves the others around. So 3 elements stabilize the standard sylow subgroup, and 6 do not. The diagonal involution turns the lower triangular group into upper triangular, so the stabilizer has order 3, and is only the group itself. The upper triangular sylow group is only stabilized by itself. Thus the only element that fixes all four sylow subgroups is 1. Our group embeds in S4, and has order 12, hence a normal subgroup of S4. The decomposition series for S4 leaves us no choice; G mod its center is A4.
Don't confuse G, of order 24, with S4. G has a center ±1, and S4 does not.
As shown above, we can always put 1 in the corner. Let an arbitrary matrix be [a,b|1,d]. Conjugate by a diagonal matrix to get [a,u|v,d], where uv = b. Multiply the first by the second, and the lower left entry is a+dv. Assume both a and d are nonzero. Choose v so that the lower left corner of the product is 0. Thus v = a/d, and u = bd/a. (I'm using + and - interchangeably here.) The upper right entry is au+bd, and this too comes out 0. The matrix is diagonal. If the eigen values are distinct, then we're done. The eigen values are a2+bv and d2+u. After substituting, these become a2+ab/d and d2+bd/a. Suppose they are equal.
a2 + d2 + b(a/d + d/a) = 0
(a2+d2) * (1 + b/ad) = 0
With determinant 1, the second factor is 1/ad. Clear the denominator, and a2+d2 = 0. This is (a+d)2, hence the trace is once again 0. We've already handled this case, so let's move on.
The last case is where a or d is 0. Since the determinant is 1, we can call this [a,1|1,0], or [a,u|v,0] where uv = 1. Multiply these together and get this.
| a2+v | au |
| a | u |
If a is nonzero, then we are not in the center. Choose any v other than a2, (which we can do if the field is more than Z2), and the diagonal entries are both nonzero. Such a matrix brings in everything, so let's consider a = 0. The above matrix becomes diagonal, with v and u as eigen values. Let u and v be distinct, and the result is not in the center. That completes the proof.
For any field of characteristic 2, other than Z2, G/C is simple. Unlike the odd primes, this has no connection to the quaternions. G is nonabelian, and the quaternions are abelian.
Start with the finite field of order 4 and build G. In characteristic 2, there is no center. So G has order 60, and we're back to A5.
When p = 7 the order is 168, and G/C is isomorphic to the unique simple group of order 168. Larger base fields lead to simple groups that are like no other.