0

I'm trying to understand the proof from this post: Show a certain group is contained in a Sylow p-group.

But certain things I cant decipher. Specifically, I'm looking at the top-rated answer.

If these types of posts are inappropriate i apologize, i just figured it would be better than arguing in the comments.

I'll let $P_i$ denote the parts of the proof in order.

$P_1$:

Assume $|G|=p^{n}m,(p,m)=1$

Q1: What does $(p,m) = 1$ mean? Is 1 the identity? Why are we looking at pairs of (p,m) anyways?

$P_2$:

I assume you already know the following Sylow theorems: Sylow p-sbgps. exist in G , they all are conjugate in G and their number equals 1(mod p): let K be a p-sbgp. and let P:={P≤G∣P is a Sylow p-sbgp. of G}. As noted above, |P|≡1(mod p).

Everything here is clear.

$P_3$:

Let now K act on P by conjugation: k∈K,P∈P⟹k⋅P→k−1Pk=:Pk .

Q2: Whats the intuition behind looking at conjugation as a potential solution?

Q3: Why is $k*P = k^{-1}Pk=P^k$? I have no idea how we reach any of these equalities. I can't tell if they are just defined in the group action or they follow from each other.

$P_4$:

If there is an orbit with one single element , say $O_Q$={Q},for some Q∈P, then $Q^k$=Q∀k∈K⟹QK=KQ⟹QK is a p-subgroup of G

Q4: If $Q^k = Q$, doesn't that imply that Q has order k? Why would it have order k?

Q5: Does does $Q^k = Q$ imply $QK=KQ$ We defined the group action by left multiplication. How would right multiplication even work with this group action, and why is it equal to left multiplication.

$P_5$:

so if K is not contained in any Sylow p-sbgp then |QK|>|Q|=$p^n$, which is absurd, so that all the orbit have size a power of p, but this means |P|≡0(mod p), which of course is also absurd since, as we mentioned above, |P|≡1(mod p)⟹ it must be that K is contained in some Sylow p-sbgp.

I have no idea what's happening here but I'm assuming its because i didn't understand the previous statements. I'll come back to this.

I apologize if some of the questions are trivial, I'm quite sleep deprived.Help would be appreciated.

user1729
  • 31,015
user569685
  • 597
  • 1
    For $P_1$, $(p,m)$ refers to the GCD of the numbers $p$ and $m$ so $(p,m) = 1$ is just saying that the GCD of $p$ and $m$ is $1$, i.e. they are relatively prime, or in this case, $m$ contains no factors of $p$. – Justin Young Sep 03 '20 at 18:14

1 Answers1

1

For the first two, i can certainly tell you that $(p,m)=1$ means that $p$ and $m$ are relatively prime. Secondly, I'm not sure what's going on with $P^k$, but it looks like it might just be a notation.

As to the intuition behind looking at the conjugation action, it is an artifice that just works in this case.

You seem to get lost around $Q4$. The action is conjugation, not left multiplication. Note that, in my guess we are using $P^k:=k^{-1}Pk$. So that almost immediately, $Q^k=Q\implies kQ=Qk$.