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Let $p$ be the smallest prime divisor of the order of a finite nonabelian simple group. A cyclic Sylow subgroup would give a normal p-complement, so if $p^3$ does not divide the order, the Sylow subgroup must be $C_p^2$. Conjugation embeds its normalizer modulo centralizer into $G{L}_2(p)$, and this quotient has order prime to $p$. For odd $p$, every other prime divisor of $|G{L}_2(p)|=p(p-1{)}^2(p+1)$ is smaller than $p$, again forcing trivial conjugation and a contradiction to the Burnside transfer theorem. For $p=2$, nontrivial conjugation must have order three, giving $12\mid |G|$. The groups $PS{L}_2(5^{2m+1})$ give infinitely many examples with exactly two factors of two in their orders.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

Put $Q=G/H$ and write $M^H$ for the invariant submodule. The five-term exact sequence in group cohomology associated with the Lyndon–Hochschild–Serre spectral sequence isThe inflation map in group cohomology composes a cocycle on $Q$ with the quotient homomorphism $G\to Q$. The restriction map in group cohomology restricts a cocycle from $G$ to $H$. The quotient action on the middle term is, for $q=gH$ and a one-cocycle $f$,This is independent of the lift and of the representative at the level of cohomology. Finally, the transgression in group cohomology extends a $Q$-invariant class on $H$ to a one-cochain on $G$; its coboundary is $H$-basic and descends to the two-cocycle on $Q$ representing $d_2[f]$. Changing the extension changes that cocycle by a group coboundary.

For the application, choose free generators $x_1,\dots ,x_n$ of $F$ and normal generators $r_1,\dots ,r_n$ of $R$. Since a finite nonabelian simple group $K$ is a perfect group, its abelianization is zero. The five-term sequence for $1\to R\to F\to K\to 1$ with trivial coefficients containsThe first term is zero because $K$ is finite, and the last term is zero because a free group has cohomological dimension one. It remains to prove that restriction is surjective.

The invariant submodule of homomorphisms $R\to \mathbb{Z}$ is exactlyThe images of the relators $r_1,\dots ,r_n$ generate $R/[F,R]$, so such a homomorphism $\lambda$ is determined by the integer vector $b=(\lambda (r_1),\dots ,\lambda (r_n){)}^T$. Let $A$ be the relator exponent-sum matrix. This square integer matrix presents $K^{\mathrm{ab}}$, which is zero, so $A$ is a unimodular matrix. There is therefore an integer vector $v$ satisfying $Av=b$. Define $\psi :F\to \mathbb{Z}$ by assigning to $x_i$ the $i$th entry of $v$. The definition of $A$ gives $\psi (r_j)=\lambda (r_j)$ for every $j$. Since the relator images generate $R/[F,R]$, the restriction of $\psi$ to $R$ equals $\lambda$. Restriction is surjective, exactness now gives

A finite nonabelian simple group embeds in $A_{n_p}$, where $n_p$ is its number of Sylow subgroups for a prime dividing its order. It is not a p-group by the nontrivial-centre property. Thus its Sylow subgroups are proper and cannot be normal, so $n_p>1$. The conjugation action is nontrivial, and its normal kernel of a group homomorphism must therefore be trivial. The sign of this faithful permutation action must be trivial, since a nontrivial sign homomorphism would inject the group into a group of order two. Lagrange's theorem consequently gives $|G|\mid n_p!/2$.