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For $H\le G$ of finite subgroup index, let $H^{\prime }=[H,H]$ and choose a right-coset transversal $T$ for the cosets $Ht$. For each $g\in G$ writeThe transfer homomorphism isThe product is taken in the abelianization, so its order is irrelevant. If another transversal has representative $a_{t}t$ for $Ht$, with $a_t\in H$, its factor is $a_{t}h(t,g)a_{t_g}^{-1}$. The map $t\mapsto t_g$ permutes $T$, so the extra factors cancel in $H/H^{\prime }$. This proves independence of the transversal. AlsoMultiplying over $t$ and again permuting the second indices proves $V(g_1{g}_2)=V(g_1)V(g_2)$. Thus it is indeed a group homomorphism.

The Burnside transfer theorem states that if a Sylow $p$-subgroup $P$ of a finite group satisfies $P\subseteq Z(N_G(P))$, then $G$ has a normal $p$-complement: a normal subgroup of order $|G|/|P|$. The hypothesis first makes $P$ abelian. It also makes fusion inside $P$ trivial. To prove this, suppose $u,v\in P$ and $u^g=v$, using $u^g=g^{-1}ug$. Both $P$ and $P^g$ lie in $C_G(v)$: the former because $P$ is abelian, and the latter because $v\in P^g$. They are Sylow subgroups of that centralizer. Choose $c\in C_G(v)$ with $(P^g{)}^c=P$. Then $gc\in N_G(P)$ and $u^{gc}=v^c=v$, while the central-normalizer hypothesis gives $u^{gc}=u$. Thus $u=v$.

Evaluate transfer to the abelian group $P$ on $u\in P$. In the action of $u$ on right cosets, a cycle of length $r$ contributeswith $t_r=t_0$. This is a conjugate of $u^r$ inside $P$ and hence equals $u^r$ by the fusion argument. Summing the cycle lengths therefore givesThe exponent is coprime to $p$, so this power map is an automorphism of the finite abelian $p$-group $P$. Transfer is consequently onto $P$, and its kernel is normal of subgroup index $|P|$ and of order prime to $p$. This proves the theorem. Notice also that a normal $p$-complement consists exactly of the elements of prime-to-$p$ order: they map trivially to the $p$-group quotient, while all elements of the complement have such order. It is therefore unique and a characteristic subgroup.

Now let $G\le A_p$ be primitive of degree $p=2q+1$, with $q$ prime. Every nontrivial normal subgroup of a faithful primitive group action is transitive: its orbits form a $G$-invariant block partition, and singleton orbits would make it trivial. Such a subgroup has order divisible by $p$. A Sylow $p$-subgroup $P$ has order $p$, since $p!$ contains only one factor $p$. The normalizer of a regular $p$-cycle in $S_p$ is the affine group $x\mapsto ax+b$ on ${\mathbb{F}}_p$, of order $p(p-1)$. Translations are even, whereas a generating multiplier gives a $(p-1)$-cycle and is odd. Its normalizer in $A_p$ therefore has order $p(p-1)/2=pq$.

Suppose $G$ is not simple and choose $1<N◃G$ proper. It is transitive, so we can choose $P\le N$. The Frattini argument gives $G=N{N}_G(P)$: for each $g$, the subgroup $P^g$ is a Sylow subgroup of $N$ and can be conjugated back to $P$ within $N$. Since $N_G(P)$ contains $P$ and lies in the order-$pq$ normalizer, it has order $p$ or $pq$. If its order were $p$, the Burnside theorem applied to $G$ would give a normal $p$-complement. A nontrivial such complement cannot be transitive because its order is prime to $p$, contradicting primitivity. It must be trivial, giving $G=P$, contrary to the choice of $N$. Thus $|N_G(P)|=pq$.

The intersection $N_N(P)$ again has order $p$ or $pq$. If it had order $pq$, the whole normalizer would lie in $N$ and the Frattini equality would give $G=N$. Hence $N_N(P)=P$. Applying Burnside transfer within $N$ gives its normal $p$-complement $K$. By the uniqueness argument, $K$ is a characteristic subgroup of $N$, and so normal in $G$. Primitivity again forces this prime-to-$p$ subgroup to be trivial. Thus $N=P$ is normal in $G$, and $G=N_G(P)$ has order $pq$. The even primitive groups of safe-prime degree conclusion is