Sylow theorems state that sylow p subgroups of a group G are conjugate. Often I see argumentation that if there are n sylow p subgroups in G then we can define a group action on it by conjugation and hence create a homomorphism from G into Symmetric group or order n. Please provide a proof that this homomorphism is legitimate and why conjugation by any element on a sylow p subgroup takes you to another sylow p subgroup?
[Math] Why does group action by conjugation on sylow subgroups define a homomorphism into the symmetric group
group-theorysylow-theory
Best Answer
Every action of a group $G$ on a set $S$ defines a homomorphism from $G$ to $\operatorname{Sym}(S)$ (by $g\mapsto(s\mapsto \varphi_g(s):=\mathcal{A}(g,s)$), and viceversa (by $(g,s)\mapsto \mathcal{A}(g,s):=\varphi_g(s)$), and the action of $G$ on $\operatorname{Syl}_p(G)$ by conjugation doesn't make an exception to this general fact about actions. Furthermore, given two equicardinal sets, say $X$ and $Y$, and a bijection $\alpha$ between them, it's easy to build up an isomorphism between $\operatorname{Sym}(X)$ and $\operatorname{Sym}(Y)$, and thence the above mentioned homomorphism can be thought of from $G$ to $S_{n_p}$ $($rather then to $\operatorname{Sym}(\operatorname{Syl}_p(G)))$, where $n_p:=|\operatorname{Syl}_p(G)|$.