(a) Let be a finite group. Prove that if is a nonempty subset of G that is closed under , then is a subgroup of .

(b) Show that the result in (a) is false if is an infinite group.

My attempt for for part (a):

Given:

,

,

is closed under ,

and is a finite group.

Proof:

If we can show satisfies properties; namely, that it is closed under ; it is associative; it has an identity, and each of its element has an inverse, then H is a subgroup of .

Since is closed under , then if , then . Let . Since , , , so is associative, and as well.

For all , there is an inverse .

Here run into difficulties: Without knowing how to show , I could go nowhere.

For part (b), I find it hard to believe that an infinite group cannot have a subgroup.