We can approach this another way:
Consider the homomorphism .
If is ANY subgroup of , then is also a subgroup of . Restricting the homomorphism sgn to this subgroup yields a homomorphism from to {-1,1}.
Since there are only two possibilities for the image of this homomorphism, we have either:
, which implies that is a subgroup of (why?), or:
, which implies that is of index 2 in .
Now if , the first possibility leads to . So if , we must have that .
Now is thus a subgroup of of order 6. There are two possibilities:
a) This subgroup is cyclic, but has no elements of order 6, which leaves us with:
b) .
However, if has a subgroup isomorphic to , this subgroup would contain 3 elements of order 2. However, the (only) 3 elements of order 2 in , namely:
(1 2)(3 4), (1 3)(2 4), (1 4)(2 3) generate a subgroup of order 4, and has no such subgroup (it cannot, for 4 does not divide 6).