Easy question on the Second Sylow Theorem

[Kreizhn]

So Sylow's second theorem tells us that if G is a group, p a prime, and H,K are both p-Sylow subgroups, then $\exists g \in G$ such that $H = gKg^{-1}$. Now I have some questions about this, because I don't think I've ever properly understood the consequences of this theorem.

My issue I think is really the act of conjugation. Namely, this theorem states that we can always relate two p-Sylow subgroups via conjugation, but is it really more that conjugation is a transitive group action on the set of p-Sylow subgroups?

This may sound silly, but we always say that if there is precisely one p-Sylow subgroup, it must be normal right? But the way that I've always read this is that we can only be assured that there is a single element $g \in G$ such that $gPg^{-1} = P$ and that this needn't hold for all $g \in G$. So I guess what I'm asking is, does conjugation of a p-Sylow subgroup ALWAYS yield another p-Sylow subgroup? This would then clear up a lot of my confusion.

 

[micromass]

Hi Kreizhn!

So Sylow's second theorem tells us that if G is a group, p a prime, and H,K are both p-Sylow subgroups, then $\exists g \in G$ such that $H = gKg^{-1}$. Now I have some questions about this, because I don't think I've ever properly understood the consequences of this theorem.

My issue I think is really the act of conjugation. Namely, this theorem states that we can always relate two p-Sylow subgroups via conjugation, but is it really more that conjugation is a transitive group action on the set of p-Sylow subgroups?

Yes, that's exactly what the theorem says. It says that the conjugation action acts transitively on the p-Sylows.

This may sound silly, but we always say that if there is precisely one p-Sylow subgroup, it must be normal right? But the way that I've always read this is that we can only be assured that there is a single element $g \in G$ such that $gPg^{-1} = P$ and that this needn't hold for all $g \in G$. So I guess what I'm asking is, does conjugation of a p-Sylow subgroup ALWAYS yield another p-Sylow subgroup? This would then clear up a lot of my confusion.

Yes, the image of a p-Sylow under any automorphism (example a conjuection) is always a p-Sylow. This is true since automorphisms preserve the order of the group. That is, if a group has order p^k, then the image under an automorphism will still have order p^k.
This shows in general that if there is precisly one p-Sylow, then it is characteristic (i.e. its image under any automorphism is again the p-Sylow)

 

[Kreizhn]

Excellent. Thanks.