This is taken from Judson's Abstract Algebra textbook.

I will notate the inverse of h as h` and the inverse of g as g`.

"Let H be a subset of a group G. Then H is a subgroup of G if and only if H is non-empty and whenever g, h are elements of H then gh` is in H."

Simple enough statement. The proof of it confuses me.

"Let H be a non-empty subset of G. Then H contains some element g. So gg`=e is in H. If g is in H, then eg`=g` is also in H. Finally, let g, h be elements of H. We must show that their product is also in H. However, g(h`)`=gh is in H. Hence, H is indeed a subgroup of G. Conversely, if g and h are in H, we want to show gh`is in H. Since h is in H, its inverse h`must also be in H. Because of the closure of the group operation, gh` is in H."

This proof confuses me. It seems to just throw around random statements. I know what a subgroup is - it is subset of a group that is closed, contains the identity element, and is invertible. I don't understand how the above proof shows these things.

Another issue with this writing is that we are given "H is a non-empty subset that contains an element 'g.'" We are given no information other than "H is a subset." How do we conclude that g`is also an element of this set?

Also, this statement bothers me:

"Finally, let g, h be elements of H. We must show that their product is also in H. However, g(h`)`=gh is in H."

I understand that we show that the product of g and h is in H to show that H is closed under the operation. I do not understand why we are working with "g(h`)`". Where did this "(h`)`" come from? The "however" also implies some kind of unexpected result.

Is this theorem trying to state the following?

gg` is in H

hh` is in H

g, g`, h, h`, e are in h

This implies that gh is in H.

The last statement is great, but nowhere did we state that H was a subgroup. gh must be an element of the group (G,+), but this proof that H is a subgroup seems to assume that H is a subgroup.