TITLE: Invariant Sylow subgroups and solvability of finite groups - Antonio Beltran.
QUESTION [1 upvotes]: This is a article which Antonio Beltran. I'm reading lemma 2.2. I see that:
"Lemma 2.2: Suppose that $A$ is a finite group acting coprimely on a finite group $G$, and let $C={\textbf{C}_G}(A)$. Then, for every prime $p$, $n_p^A(G) = \left| {C:{\textbf{N}_C}(P)} \right|$ for every $A-$invariant Sylow $p-$subgroup $P$ of $G$.
Proof:
Let $P$ be an $A-$invariant Sylow $p-$subgroup of $G$. Then follows
from the fact that the $A-$invariant Sylow subgroups of $G$ are all $C-$conjugate, so $v_p^A(G)$ is exactly the number of distinct $C-$conjugates of $P$, and this is exactly equal to $\left| {C:{\textbf{N}_C}(P)} \right|$."
I don't understand why "the $A-$invariant Sylow subgroups of $G$ are all $C-$conjugate".
https://www.researchgate.net/publication/291552840_Invariant_Sylow_subgroups_and_solvability_of_finite_groups

REPLY [0 votes]: Check out Theorem 3.23 in Finite Group Theory, by I.M. Isaacs.


Since $C$ acts on $Syl_p^A(G)$, the set of $A$-invariant Sylow $p$-subgroups, the length of the orbit of this action is the index of the stabilizer in $C$, and $Stab(P)=\{c \in C: P^c=P\}=N_G(P) \cap C=N_C(P)$. So $n_p^A(G)=|C:N_C(P)|$, where $P$ is an $A$-invariant Sylow $p$-subgroup.
{"set_name": "stack_exchange", "score": 1, "question_id": 3344372}