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Complement of a subgroup

  1. Oct 28, 2012 #1
    Let G be a finite group. Suppose that every element of order 2 of G has a complement in G, then G has no element of order 4.

    Proof. Let x be an element of G of order 4. By hypothesis, G=<x^{2}> K and < x^{2}> [itex]\cap[/itex]K=1 for some subgroup K of G. Clearly, G=< x> K and < x>[itex]\cap[/itex] K=1$, but |G|=|< x^{2}>||K|<|< x >||K|=|G|, a contradiction. Therefore G has no element of order 4.

    Is above true? Thanks in advance.
     
  2. jcsd
  3. Oct 29, 2012 #2
    It looks right to me, but to understand your last inequality I had to look up and verify the fact that the order of a product is the product of the orders divided by the order of the intersection. So I'd recommend putting that in there. My algebra prof recommended writing proofs so that someone 3 weeks behind could understand.
     
  4. Oct 29, 2012 #3
    Notice that [itex]|HK|=\frac{|H||K|}{|H\cap K|}[/itex], for any subgroups [itex]H[/itex] and [itex]K[/itex] of [itex]G[/itex]. Sorry, I could not add it there, but thank you very much.
     
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