Suppose that H is a subgroup of G, and fix elements x and y of H. I want to show that xy-1 is in H. Since y is in H, H contains its inverse y-1. Now by closure, the product xy-1 is in H.The sentence "I want to show..." is saying what I'm going to do. Then I do it. At the end I could have added "This finishes the first part of the proof," to say what I had just done; when I was writing this, I thought that this part was short enough that I didn't need to do that.
The solution continues
Conversely, suppose that H is a nonempty subset of G such that xy-1 is in H for all x,y in H.I probably should have added "I want to show that H is a subgroup of G."
Then there were a bunch of details verifying the subgroup axioms for H, which I'll skip here. My solution ended like this:
Thus H is closed, so H is a subgroup.In the clause "so H is a subgroup," I'm reminding you of the point of this part of the solution, and thus pointing out that the proof is over.