The theorem for the two-step subgroup test says:
The subset H of a group G is a subgroup IFF the binary operation of 2 ordered pairs of elements of H are in G and for each element in H, there each exists an inverse that is in H.
Going back to the definition of Subgroup:
A subset H of a group G is a subgroup if H itself is a group under the binary operation of H,
'IF' part for the two-step subgroup test:
If H is a subgroup then H itself is a group and for it to be a group it must have already satisfied the 4 properties of the Group axiom. And we are done for 'If' part.
'Only If' part for the two-step subgroup test:
If the binary operation of two ordered pairs (a,b) are in H and the inverse of a is in H for all 'a' and 'b' in H, then H is a subgroup of G.
But the requirement for a subset H to be a subgroup is more than that! In fact, we require 2 more. How does one reconcile this?