a)
1234
1243
1324
1342
1423
1423
2134
2143
2314
2341
2413
2431
3124
3142
3214
3241
3412
3421
4123
4132
4231
4213
4312
4321

it has 12,
A4=1/2 s4=12
Since A4 has cyclical groups to get elements in group 4, they exist in s4
b)
1234  i=0
2314  i=2
3124  i=2
The above each have even inversions, there is also the identity permutation along with H, allowing H to be alternating 4 group

c)
1342	1423	2143	2431	3241	3412	4132	4213	4321
	
1342H
1342*1234=	1342
1342*2314=	2143
1342*3124=	3241

remove these
1423	2431	3412	4132	4213	4321

next:
1423*1234=	1423
1423*2314=	2431
1423*3124=	3412

4132	4213	4321

next

4132*1234=	4132	
4132*2314=	4213
4132*3124=	4321

thus none left,

so 

s4=1342	u 1423 u 4132

there are 3 cosets, They have the same amount of cosets as right as they are the gh = hg. they cannot have the same, as when elimination happens with the identity set, it eliminates all of the first row.
d)
The cosets above is group of s4 into H, making them the same number of elements where none overlap, thus s4/h is an integer.
2)Legranges Theorum is defined as G=(G:H)*H,
where the left side, the partition of G is aH, is the same cardinality of H, thus the inverse.