Proving the Subgroup Theorem for Nonempty Subsets of a Group

[radou]

So, I need to prove that, if G is a group and X is a nonempty subset of G, then the subgroup <X> generated by the set X consists of all finite products a1^n1*a2^n2*... *at^nt, where ai is from X and ni are integers.

First of all, one needs to show that the set H of all such products is a subgroup of G. Let a, b be elements from H, where a = a1^n1*a2^n2*... *at^nt, b = a1^p1*a2^p2*...*at^pt. H < G iff ab^-1 is in H. We have
ab^-1 = a1^n1*a2^n2*... *at^nt * (at^pt)^-1*...(a1^p1)^-1. I don't see where this leads, perhaps I'm doing it the wrong way.

After showing that H < G, one should show that H is contained in every subgroup of G containing X, which is almost obvious.

One thing may have confused me, though. Does a1^n1*a2^n2*... *at^nt mean that X = {a1, ..., at}, or is the t not important, since it isn't specified in the theorem (it's only important that ai is from X)? In that case, the proof would be very easy.

Thanks in advance.

 

[StatusX]

The a_i are just any elements from X. There can be elements from X not in the product, and elements appearing more than once. The only thing you have to notice to see it's a subgroup is that the product of two elements of that form is another element of the same form (ie, a product of powers of elements from X).

 

[radou]

The a_i are just any elements from X. There can be elements from X not in the product, and elements appearing more than once. The only thing you have to notice to see it's a subgroup is that the product of two elements of that form is another element of the same form (ie, a product of powers of elements from X).

OK, then the proof is pretty easy. This was not worth of a whole new thread, but Hungerford gives me some trouble, and I want to make sure I understand things before I move on.