# A simple finite group $G$ with $n$ p-Sylows is isomorphic to a subgroup of $\mathbb A_n$

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.

• $X$ is not a group, you are correct to have a doubt. But it does not need to be. The point is that $G$ permutes (by conjugation) the $n$ Sylow $p$-subgroups which allows us to identify it as a subgroup of $S_{n}.$ Also, you should probably amend the title: there is a big difference between "isomorphic to $A_{n}$" and "isomorphic to a subgroup of $A_{n}.$" – Geoff Robinson Dec 3 '14 at 18:45
• @GeoffRobinson Thanks for the correction and the explanation, if it occurs to you how could I complete the solution , I would appreciate any suggestions in the part I got stuck at. – user16924 Dec 3 '14 at 20:34

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 $\mathrm{Im}(\phi)\subset A_n$.
• Well, $1=Ker(\psi) \subset Ker(\epsilon \circ \psi) \subset Ker(\epsilon)=A_n$, but I don't see why this implies $Im(\psi) \subset A_n$ (sorry to ask this if it is too obvious). – user16924 Dec 3 '14 at 21:26
• If ${\rm Im}(\psi) \not\le A_n$ then $|{\rm Im}(\psi) : {\rm Im}(\psi) \cap A_n| = 2$, contradicting the simplicity of ${\rm Im}(\psi) \cong G$. – Derek Holt Dec 3 '14 at 21:59