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Finite group G=ST

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data
    let G be a finite group, and let S and T be nonempty subsets.Prove either G=ST={st|s is in S, t is in T} or |G|>=|S|+|T|


    2. Relevant equations



    3. The attempt at a solution

    So it is to prove G=ST, if |G|<|S|+|T|, which means also the intersection of S and T is nonempty (Note two smaller subsets can also have nonempty intersection, so |G|<|S|+|T| should have more properties than nonemptiness, but I fail to find one ). [tex]ST\subset G[/tex] is obvious. I want to prove the other direction by saying any g in G can be represented as g=st for some s,t, or [tex]s_{i}T[/tex] covers G.
    This is what I 've done:
    Let x be an element in both S and T. then x^2 is in ST. But x^2 can be outside S and T, So it is possible that [tex]x^{3}\notin ST[/tex], which seems to become useless...
    Any hint would be appreciated....
     
    Last edited: Oct 28, 2008
  2. jcsd
  3. Oct 29, 2008 #2

    Dick

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    This is really an annoying question. Did you get it? I didn't.
     
  4. Oct 30, 2008 #3
    No. The only thing I proved is that it is true when G is abelian. But..when G is nonabelian, it is a headache.
     
  5. Oct 30, 2008 #4

    morphism

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    Let S={s_1,...,s_|S|} and T={t_1,...,t_|T|}. Suppose that there is some x in G but not in ST. Then x != s_i t_j. In particular, (s_i)^(-1) x is never in T. Use this to deduce that |S| <= |G|-|T|.
     
  6. Oct 30, 2008 #5
    Ah, yea, it is true!
    I only tried to make a 1-1 correspondence from those elements g in both S ant T to something outside S and T and got stuck since g^(-1)x can possibly drop in S again.
    Thanks!
     
  7. Oct 30, 2008 #6

    Dick

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    Nice!
     
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