[proofplan] The proof has two parts. First, because $p$ divides $|G|$, a Sylow $p$-subgroup has order divisible by $p$, so it is not the trivial subgroup. Since a [simple group](/page/Simple%20Group) has no normal subgroups other than $\{e\}$ and itself, a normal Sylow $p$-subgroup must therefore be all of $G$. For the consequence, if a non-abelian finite simple group had such a normal Sylow subgroup, the first part would make the whole group a finite nonidentity $p$-group; the nontrivial-centre theorem for finite $p$-groups then forces the centre to be all of $G$, making $G$ abelian, a contradiction. [/proofplan] [step:Show that the Sylow subgroup is nontrivial] Let $a$ be the positive integer such that $p^a$ is the largest power of $p$ dividing $|G|$. Since $p \mid |G|$, we have $a \geq 1$. By definition of a Sylow $p$-subgroup, $P$ has order \begin{align*} |P| = p^a. \end{align*} Thus $|P| \geq p > 1$, so $P \ne \{e\}$, where $e \in G$ denotes the identity element. [/step] [step:Use simplicity to force the normal Sylow subgroup to be the whole group] Since $P \trianglelefteq G$ by hypothesis and $G$ is simple, the only possibilities for the [normal subgroup](/page/Normal%20Subgroup) $P$ are \begin{align*} P = \{e\} \end{align*} or \begin{align*} P = G. \end{align*} The first possibility is excluded by the preceding step. Therefore $P=G$. [guided] We now use exactly the defining property of a simple group. A group $G$ is simple when $G$ is nonidentity and its only normal subgroups are the trivial subgroup $\{e\}$ and the whole group $G$. The subgroup $P$ is assumed to satisfy $P \trianglelefteq G$, so simplicity leaves only two cases: \begin{align*} P = \{e\} \end{align*} or \begin{align*} P = G. \end{align*} The first case cannot occur because $P$ is a Sylow $p$-subgroup and $p \mid |G|$. More explicitly, if $p^a$ is the largest power of $p$ dividing $|G|$, then $a \geq 1$ and $|P|=p^a>1$. Hence $P$ contains more than the identity element $e$, so $P \ne \{e\}$. The only remaining possibility is therefore $P=G$. [/guided] [/step] [step:Derive the obstruction for non-abelian finite simple groups] Suppose, for contradiction, that $G$ is a non-abelian finite simple group and that $G$ has a normal Sylow $p$-subgroup $P \le G$ for some prime divisor $p$ of $|G|$. By the result just proved, $P=G$. Hence $G$ is a finite nonidentity $p$-group. Define the centre $Z(G) \le G$ by \begin{align*} Z(G) := \{z \in G : zx = xz \text{ for every } x \in G\}. \end{align*} By the nontrivial-centre theorem for finite $p$-groups, $Z(G)$ is nontrivial. The centre is normal in $G$: for every $g \in G$ and every $z \in Z(G)$, the element $gzg^{-1}$ commutes with every $x \in G$, because \begin{align*} (gzg^{-1})x = g z (g^{-1}xg) g^{-1} = g (g^{-1}xg) z g^{-1} = x(gzg^{-1}). \end{align*} Thus $gzg^{-1} \in Z(G)$, so $Z(G) \trianglelefteq G$. Since $G$ is simple and $Z(G) \ne \{e\}$, we must have $Z(G)=G$. Therefore every element of $G$ commutes with every other element of $G$, so $G$ is abelian. This contradicts the hypothesis that $G$ is non-abelian. Hence a non-abelian finite simple group has no normal Sylow subgroup. [/step]