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Induction method

  1. Apr 18, 2012 #1
    My question is about the induction method. This was in a theorem that I read.
    Let H be a normal subgroup of a finite group G. If G satisfies H-some statement then G is solvable.
    In the poof I have this.
    Let G be a counter example of minimal order. Let N be a proper normal subgroup of G. Since N satisfies the H-some statement then N is solvable by the induction in the order of G.


    Here is my question. H may not be a subgroup of N so, how did he apply the induction method?
     
  2. jcsd
  3. Apr 18, 2012 #2

    Stephen Tashi

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    moont14263,

    You should clarify your question. The expression "If G satisfies H-some statement then G is solvable" doesn't convey a coherent pattern for a mathematical thought. Does the "-" mean "and"?
     
  4. Apr 18, 2012 #3
    Here is an example of what I am talking about.
    I made up this theorem.

    Let H be a normal subgroup of a finite group G. If all Sylow p-subgroup P of G are conjugate in H then G is solvable.



    Conjugate in H means the set {P^{h}:h [itex]\in[/itex] H} contain all Sylow p-subgroup of G where P is a Sylow p-subgroup of G.


    Let G be a counter example of minimal order. Let N be a proper normal subgroup of G. I assume that all Sylow p-subgroup P of N are conjugate in H, "this is just an assumption ,it may not be true".then N is solvable by the induction in the order of G.


    Here is my question. H may not be a subgroup of N so, how did he apply the induction method in his theorem which has the same situation ?.
     
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