ok, the first thing to do, is find all elements x of S_{5}with x^{8}= e.

these are:

transpositions (of order 2)

double disjoint transpositions (of order 2)

4-cycles (of order 4)

(S_{5}has no elements of order 8).

now we need to count how many of each of these we have in S_{5}.

we have 10 transpositions, 15 double transpositions, and 30 4-cycles. note that every sylow 2-subgroup must have the same kinds of cycle types in it (since sylow subgroups are conjugate).

note as well, that we can only have 3,5 or 15 sylow 2-subgroups, since these are the odd divisors of 120/8 (i am not counting 1 since we obviously have more than one sylow 2-subgroup, since there are more than 7 elements of order 2 and 4). the 4-cycles also must belong to the sylow 2-subgroups in inverse-pairs (we have 15 such pairs).

clearly, one such subgroup is {e, (1 3), (2 4), (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 4)(2 3), (1 2)(3 4)}.

so every sylow 2-subgroup of S_{5}contains 2 transpositions, 3 double transpositions, and 2 4-cycles. furthermore, the 2 transpositions must be disjoint, since if they weren't the group would contain a 3-cycle, of order 3.

since we have 15 disjoint pairs of 2-cycles, 3 or 5 sylow 2-subgroups would not be enough to contain them all, so we must have 15 sylow 2-subgroups.

so each sylow 2-subgroup contains a unique 4-cycle inverse-pair: (a b c d) and (a d c b), and thus the disjoint double 2-cycle (a c)(b d).

but if g(1 2 3 4)g^{-1}= (a b c d) then either g(1 3)g^{-1}= (a c) or g(1 3)g^{-1}= (b d), so in any case our sylow 2-subgroup contains:

{e,(a b c d),(a c)(b d),(a d c b),(a c),(b d)} and thus must contain (a b c d)(a c) = (a d)(b c) and (a b c d)(b d) = (a b)(c d). we can summarize this as:

every 2-sylow subgroup of S_{5}is isomorphic to D_{4}, and is uniquely determined by any 4-cycle it contains.

now knock yourself out.