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Let S be the subset of group G that contains identity element 1?

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Let S be the subset of group G that contains identity element 1 such that left co sets aS with a in G, partition G .Probe that S is a subgroup of G.
    2. Relevant equations

    {hS : h belongs to G } is a partition of G.

    3. The attempt at a solution
    For h in S if I show that hS is S then that would imply that S is closed.
    Now hS is a partition of S and contains h since 1 is in S and h is in S also.Hence h belongs to S intersection hS.
    More over both these partition or two partition are disjoint or equal sets.
    Hence h=hS which says that S is closed.
     
  2. jcsd
  3. Sep 4, 2014 #2

    Fredrik

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    It was difficult to follow your argument, but I think you got it right. This is assuming that you meant to write S=hS where you wrote h=hS in the last line.

    My version:
    We will prove that S is closed under multiplication. Let ##a,b\in S## be arbitrary. Since ##ab\in aS##, it will be sufficient to prove that ##aS=S##. Since ##e\in S##, we have ##a=ae\in aS##. By assumption, we also have ##a\in S=eS##. Since aS and eS are both cells of the partition, they are either equal or disjoint. So the fact that they both contain ##a## implies that ##aS=eS=S##.
     
  4. Sep 5, 2014 #3
    The same I can show that it contains the inverses. WHy aren't they partitions of G?? As  ,h belongs to G , hS is a left coset. 
     
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