# Induction method

1. Apr 18, 2012

### moont14263

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. Apr 18, 2012

### Stephen Tashi

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"?

3. Apr 18, 2012

### moont14263

Here is an example of what I am talking about.

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 ?.