If P is normal in H, and H normal in G, then P is normal in G

[mathgirl313]

Homework Statement

Now, G is a finite group, and P is a Sylow P-Subgroup of G, and H is a subgroup of G with P≤H≤G. So if P is normal in H, and H is normal in G, then P is normal in G.

Homework Equations

The Attempt at a Solution

I know I want to show that gpg^-1 is contained in P but that's about as far as I can get. I know gPg^-1 is going to be another Sylow P-Subgroup, but other than that I'm kind of stuck. When I try going about showing that gpg^-1 q, where q is an element in p, I have trouble showing the two elements are in P. I get as far as q=hph^-1 since q is in P and i think my best bet for showing it is contained in P is for closure...but that's all I get. I know I need to use the P is Sylow P-Subgroup since the property normally isn't transitive, but I'm not sure how to.

Thanks for any help.

 

[Dick]

All sylow p subgroups are conjugate, right?

 

[mathgirl313]

Ya, gPg^-1 is going to be a conjugate to P, but why does that help me show that it is P? Couldn't it be another Sylow P-Subgroup?

 

[Dick]

Ya, gPg^-1 is going to be a conjugate to P, but why does that help me show that it is P? Couldn't it be another Sylow P-Subgroup?

If it's another Sylow P-Subgroup then that subgroup is contained in H. What next?

 

[mathgirl313]

If it's another Sylow P-Subgroup then that subgroup is contained in H. What next?

Why is it necessarily going to be contained in H? Just because of orders and conjugate? Are you going to get that P is unique and so it's normal?

 

[Dick]

Why is it necessarily going to be contained in H? Just because of orders and conjugate? Are you going to get that P is unique and so it's normal?

Yes, I am. gHg^(-1)=H and P is contained in H.

 

[mathgirl313]

Yes, I am. gHg^(-1)=H and P is contained in H.

So is gPg^(-1)=P because of that?

I'm sorry, I've always been dreadful at showing that something is a normal subgroup.

 

[Dick]

So is gPg^(-1)=P because of that?

I'm sorry, I've always been dreadful at showing that something is a normal subgroup.

No, that's not the reason. P and gPg^(-1) are two p-sylow subgroups of H. Think about that.

 

[mathgirl313]

No, that's not the reason. P and gPg^(-1) are two p-sylow subgroups of H. Think about that.

Since they're normal in H they have to be the same subgroup.

 

[Dick]

Since they're normal in H they have to be the same subgroup.

There's a little more to it than that. P is normal in H. Why is gPg^(-1)?

 

[mathgirl313]

There's a little more to it than that. P is normal in H. Why is gPg^(-1)?

Why is it normal? I can only think that since P is a Sylow P-Subgroup of H, and P is normal in H. But a p-Sylow group is normal if and only if the number of Sylow p-subgroups is one (for that
). So P is the unique Sylow p-subgroup. So P is the unique Sylow p-subgroup in G as well, and as such it's normal.

 

[Dick]

Why is it normal? I can only think that since P is a Sylow P-Subgroup of H, and P is normal in H. But a p-Sylow group is normal if and only if the number of Sylow p-subgroups is one (for that
). So P is the unique Sylow p-subgroup. So P is the unique Sylow p-subgroup in G as well, and as such it's normal.

Yeah, that does it. Since P and gPg^(-1) are both sylow p-subgroups of H. Then they are conjugate in H. So gPg^(-1)=hPh^(-1) where h is in H. Since P is normal in H, gPg^(-1)=P.

 

[mathgirl313]

Yeah, that does it. Since P and gPg^(-1) are both sylow p-subgroups of H. Then they are conjugate in H. So gPg^(-1)=hPh^(-1) where h is in H. Since P is normal in H, gPg^(-1)=P.

Awesome! Thank you do much for walking me through it!