I think I've got it.
The problem was to show for the general fifth-degree equation that the Galois Group of the resolvent sextic was just S5, the same as the Galois group of the quintic. I already talked about how you could generate the coefficients of the sextic by calculating the elementary symmetric polynomials of Dummits Functions. Which brought us to the question of: what is the Galois Group>
What I did is I wrote out all six Dummit functions in A,B,C,D and E (in a kind of shorthand, actually).
The "standard form" of Dummit's function is
1. AA(BE + CD) + BB(CA + DE) + CC(DB + EA)....
which is obviously based on the rotation ABCDE. Starting from this one, there are exactly five other "rotations" which work, and these are all of them:
1. ABCDE (= ACEBD)
2. ACBED (= ABDCE)
3. AECBD etc.
The terms in brackets are included to show that the alternate forms generate the same Dummit Function.
Then I took all possible 5-cycles on A,B,C,D and E and examined what there effect was on the Dummit Functions. (EDIT: Let me explain this a little better. First I'll shift A to B, B to C, etc. Obviously the first one is just going to map to a rotation of itself, which means for the sake of the associated Dummit Function, it goes nowhere. But the second one, ACBED, gets transformed to BDCAE...which is the same as No. 6 (or a reverse image, which comes to the same thing. Likewise you can plot where they all go. The first "orbit" is (26435). Then I tries a different five-cycle...actually, I chose a cycle of the 2nd function, because I wanted it to map to itself. The other five mapped to each other in a second five-cycle. I've listed all six five-cycles below:
And then of course you get the images of these under self-compounding...24 5-cycles altogether. (Just like S5.) The point is they didn't mushroom out into every possible 5-cycle in S6. So let's see what happens when you start multiplying them together? At first I started getting 3-cycles...actually, direct products of 3-cycles, like these:
And then I started getting new 5-cycles. But it was easy to verify that each new one was already within the orbits of my original 6. So far so good.
I was starting to think I was getting only even permutations...that would be interesting. Could the Galois Group of the resolvent be A5? I had already 24 5-cycles, and a whole bunch of 3-cycles, so unless I had some odd permutations, it looked like I might be heading there. So I went back to Dummit's Functions and just tried swapping A and B. And I got:
That's an odd permutation, so it can't be A5. I've got 5-cycles, 3-cycles, and odd permutations...if it's not S5, then God help me.
I know there are guys that would do a lot cleaner job of this than me, but there it is.