By the Sylow theorems, G has a subgroup of order 4. Call this subgroup H. If G permutes the letters abcd, H consists of abcd, badc, cdab, and dcba. These are the double transpositions. (A 4 cycle is impossible, since that would be an odd permutation.) So H is isomorphic to Z2*Z2.
The elements not in H are the eight 3 cycles. Since H is the only sylow subgroup it is normal in G, and the factor group is Z3. Are there any other normal subgroups?
Z3 cannot be normal, since there are 4 instances, all conjugate. If Z2 is normal then its single generator, x, satisfies yx/y = x for every y in A4. A double transposition does not commute with a 3 cycle, hence Z2 cannot be normal. Finally a group of order 6 would be normal, if one existed, but a double transposition and a 3 cycle conspire to build all of A4. Thus the only normal subgroup is Z2Z2, with factor group Z3.
Let G = An for n ≥ 5. We will show that G is simple.
First show that all the permutations in G can be generated by the three cycles that permute 1 2 and k, for every k between 3 and n. Given k and l, pull k back to position 2, then push k out to position l (moving l into position 1), then rotate l into position k. This swaps 1 and 2, as it swaps k and l.
Assume any even permutation and pull it back to start. If 1 and 2 aren't in the first 2 positions, make it so in just two steps. If 3 isn't in place, swap it with the element in position 3, thus 3 is now where it belongs. Do this for 4 and 5 and so on through n. Since all permutations are even, 1 and 2 will be in position when we're done.
Of course the selection of 1 and 2 in the above was arbitrary. The 3 cycles tied to any two positions will generate G.
Suppose H is a normal subgroup of G. Let H contain q, the 3 cycle on abc, and let x be the involution that swaps a with b and c with k. Now xq/x leaves c where it is and cycles a b and k. This must appear in H, and similarly for every k, thus generating G. If H is a proper normal subgroup, it does not contain an isolated 3 cycle.
Suppose q is in H and q includes a cycle of length r, for r ≥ 4. Let x be the three cycle that shifts the first three members of the r cycle. By normality, (xq/x)/q belongs to H. The r cycle has been turned into a 3 cycle. Other cycles within q have been pushed forward and pulled back, and are no longer present. Thus H contains an isolated 3 cycle, and H is all of G. Henceforth all cycles in the permutations of H have length 2 or 3.
Next suppose H has q with at least two disjoint three cycles, and let x permute the first two members of one cycle and the first member of the other. Now (x*q/x)/q is a 5 cycle in H, and by the above, H = G.
Next suppose q contains a three cycle and at least one two cycle. Since q2 is a three cycle, H = G.
Next suppose q contains an even number of two cycles. If x is a three cycle that permutes a pair and a member of another pair, then w = xq/x/q exchanges one pair for another, and leves everything else alone. Since n > 4, let z be a three cycle that permutes two corresponding entries in these pairs and another element not in either pair. Now zw/z/w is a three cycle in H, and H = G.
In each case H = G, so G is simple. The same proof shows A∞ is simple.
If Sn has another normal subgroup H, intersect H with An, giving a group that is normal in Sn, and in An, which is impossible, since An is simple. Actually there is a catch; H might intersect An in the identity element. But that means x in H cannot be applied twice, for that would be an even permutation in An. Every element of H is an involution. If there are two involutions, combine them to get something in An, hence H is a single involution, which we will call t. If x comes from An, a parity argument shows xt/x cannot be the identity permutation, hence xt/x = t, and xt = tx. This means t commutes with An. Write t as an odd number of transpositions, and it is easy to find even permutations that don't commute with t. That takes care of H normal in Sn, at least for n ≥ 5, or n = ∞.
S2 is simple, and S3 has A3 as normal subgroup. Let H be the normal subgroup of A4, the double transpositions, and verify that H is normal in S4. Then set H and A4 aside and look for something else. If K is normal in S4 it is not wholly contained in A4, hence it intersects A4 in H. It contains 8 elements, and might be the dihedral group D4, but then this sylow subgroup would be unique. And all groups of order 2 or 4 belong to a sylow subgroup, so everything else, the other 16 elements, would be 3 cycles. There are only 8 such cycles, building for instances of Z3, hence there are no other normal subgroups of S4.
As a corollary, Sn cannot have a subgroup of index 2, other than An, because such a subgroup would be normal in Sn.