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Math Help - Proposition Regarding Cosets and Orders

  1. #1
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    Proposition Regarding Cosets and Orders

    Let A be a finite subset of a group G (not necessarily a subgroup). Denote by A^2 the set \{a_1 a_2 | a_1, a_2 \in A\}.

    Prove that |A^2| = |A| \iff the following is true:
    A equals a left coset aH for some subgroup H \leq G and some
    element a \in G, and A also equals some right coset Hb, b \in G.

    The backward implication is easy enough since all that's required is a simple bijection, but I'm having problems proving the forward implication.

    Thanks in advance!
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  2. #2
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    I also realize that the professor went through quite a bit about group actions, would they be required to give a proof of this?
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  3. #3
    Senior Member roninpro's Avatar
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    I haven't worked this out completely, but I'll throw out some ideas.

    Let A=\{a_1, a_2,\ldots, a_n\} and write A^2=\bigcup_{a\in A} aA. Since a\in G, the operation x\mapsto ax is injective. Therefore, |aA|=|A| for every a\in A. Now, we are given that |A^2|=|A|. This implies that every term in the union is equal; that is, a_1A=a_2A=\ldots=a_nA.

    This seems to be coset-like behaviour, but I don't quite see to how get the conclusion.
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  4. #4
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    Solved it -- took much more effort than I thought was initially required, presumably because this question was adapted from a conference paper that's currently investigating groups with such properties.

    Basically the proof shows that the identity element is in a^{-1}A for any a \in A. Then A \subset a^{-1}A^{2}. From there, a few further arguments will show that a^{-1}A is a subgroup, and it is clear that A = a.a^{-1}A.
    Last edited by h2osprey; November 6th 2010 at 04:36 PM.
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