Unexpected output for the normalizer of V in A5?

Questions
1

My code works by finding the number of elements in the normalizer subgroup of a known Sylow p-subgroup of some group, so that I can find the number of Sylow p-subgroups of some group. For example:

A = GroupElements[SymmetricGroup[4]];
B = GroupElements[DihedralGroup[4]];
norm = {};
For[i = 1, i <= GroupOrder[SymmetricGroup[4]], i++, conj = {}; 
 For[j = 1, j <= GroupOrder[DihedralGroup[4]], j++, 
  AppendTo[conj, 
   PermutationProduct[A[[i]], B[[j]], InversePermutation[A[[i]]]]]]; 
 If[ContainsExactly[conj, B], AppendTo[norm, A[[i]]]]]
np = Length[A]/Length[Union[norm]]

The above code outputs 3, which is the correct answer to the question “how many Sylow 2-subgroups does S4 have?”. Replacing the groups also gives the correct solution for Sylow 3-subgroups in S4, and Sylow 3- and 5-subgroups in A5. However, when I switch to trying to find the number of Sylow 2-subgroups of A5, and I change the groups to A5 and D2 (or Z2 x Z2), I get the wrong answer with the following code:

Alt5 = GroupElements[AlternatingGroup[5]];
V = GroupElements[DihedralGroup[2]];
norm = {};
For[i = 1, i <= GroupOrder[AlternatingGroup[5]], i++, conj = {}; 
 For[j = 1, j <= GroupOrder[DihedralGroup[2]], j++, 
  AppendTo[conj, 
   PermutationProduct[Alt5[[i]], V[[j]], 
    InversePermutation[Alt5[[i]]]]]]; Print[conj]; 
 If[ContainsExactly[conj, V], AppendTo[norm, Alt5[[i]]]]]
np = Length[Alt5]/Length[Union[norm]]

This code outputs 15, when the correct answer should be 5.

2

Your use of D2 is not correct because this representation as cycles contains elements not in A5. You could instead use

d2 = PermutationGroup[{Cycles[{{1, 2}, {3, 4}}], Cycles[{{1, 3}, {2, 4}}]}];