I am trying to solve this problem:
Let $G$ be a finite and simple group, and let $p$ be a prime number such that $p$ divides $|G|$. If $n_p(G)=n$ for $n>1$ (n_p denotes the number of p-Sylows) then $G$ is isomorphic to a subgroup of $\mathbb A_n$.
What I thought of was the following:
I consider the action of $G$ on $X$={H subgroup of G, H p-Sylow} by conjugation. Then there is a morphism $$\psi:G \to S(X)$$ $$g \to gSg^{-1} \space \forall S \in X,$$
where $S(X)$ is the set of all bijective functions $f:X \to X$. First of all, I would like to affirm $S(X) \cong \mathbb S_n$, but I don't know if this is true. Second, note that $Ker(\psi) \neq G$ because if not $n_p=1$ and $Ker(\psi)$ can't be a proper subgroup because if that was the case $Ker(\psi) \lhd G$, then we have $Ker(\psi)=\{e\}$.
By the first isomorphism theorem, $G \cong G/Ker(\psi) \cong Im(\psi)$. Now, if I could show that $Im(\psi)$ is isomorphic to a subgroup of $\mathbb A_n$, then I would be done.
I would appreciate any hints or suggestions.
Hint: $A_n$ is the kernel of the signature morphism $$\epsilon:S_n\longrightarrow\lbrace-1,+1\rbrace$$ What can you say about the kernel of the composite $\epsilon\circ\psi$? ... Whence $\mathrm{Im}(\phi)\subset A_n$.