Proving Sylow p-Subgroup is Normal in Finite Group

[ehrenfest]

[SOLVED] Sylow p-subgroups

Homework Statement

Let G be a finite group and let primes p and q \neq p divide |G|. Prove that if G has precisely one proper Sylow p-subgroup, it is a normal subgroup, so G is simple.

EDIT: that should say "G is not simple"

Homework Equations

The Attempt at a Solution

I don't see the point of q. If G has precisely one proper Sylow p-subgroup, then you can conjugate with all the elements of the group and you cannot get of the subgroup or else you would have another Sylow-p-subgroup, right? So, it must be normal, right?

 

[ehrenfest]

anyone?

 

[NateTG]

I don't see the point of q. If G has precisely one proper Sylow p-subgroup, then you can conjugate with all the elements of the group and you cannot get of the subgroup or else you would have another Sylow-p-subgroup, right? So, it must be normal, right?

Yes. (Is it possible that you misread the question - if there is a normal subgroup, the group can't be simple.)

 

[ehrenfest]

See the EDIT. But am I right about the part about q being unnecessary? You can induce the existence of q by the fact there is a proper Sylow p-subgroup, right?

 

[morphism]

If |G| doesn't have another prime divisor, then we can't say for sure that G isn't simple (e.g. if |G|=p^2).

 

[ehrenfest]

If |G| doesn't have another prime divisor, then we can't say for sure that G isn't simple (e.g. if |G|=p^2).

But it must have another prime divisor if it has a proper p-Sylow subgroup.

 

[morphism]

Ah, you're right. Missed the word proper.

 

[NateTG]

See the EDIT. But am I right about the part about q being unnecessary? You can induce the existence of q by the fact there is a proper Sylow p-subgroup, right?

Yep.