Induction Method: My Question About Theorem in Finite Groups

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SUMMARY

The discussion centers on the application of the induction method in group theory, specifically regarding finite groups and normal subgroups. The theorem states that if a finite group G has a normal subgroup H and satisfies an H-some statement, then G is solvable. The user questions the validity of applying induction when H is not necessarily a subgroup of another normal subgroup N of G, despite N satisfying the H-some statement. The user also presents a hypothetical theorem involving Sylow p-subgroups and their conjugacy in H, further exploring the implications of these conditions on solvability.

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  • Understanding of finite group theory
  • Familiarity with normal subgroups and their properties
  • Knowledge of Sylow theorems and Sylow p-subgroups
  • Proficiency in mathematical induction methods in group theory
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Mathematicians, particularly those specializing in group theory, students studying abstract algebra, and researchers exploring the solvability of finite groups will benefit from this discussion.

moont14263
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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.
 
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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 \in 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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