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Easy question on the Second Sylow Theorem

  1. Jun 22, 2011 #1
    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 [itex] \exists g \in G [/itex] such that [itex] H = gKg^{-1} [/itex]. 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 [itex] g \in G [/itex] such that [itex] gPg^{-1} = P [/itex] and that this needn't hold for all [itex] g \in G[/itex]. 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.
  2. jcsd
  3. Jun 22, 2011 #2
    Hi Kreizhn! :smile:

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

    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 pk, then the image under an automorphism will still have order pk.
    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)
  4. Jun 22, 2011 #3
    Excellent. Thanks.
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