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Another group theorem

  1. Apr 27, 2007 #1


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    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.
  2. jcsd
  3. Apr 27, 2007 #2


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    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).
  4. Apr 27, 2007 #3


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    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.
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