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Let $p$ be the smallest prime dividing the order of a finite group, and suppose its Sylow subgroup $P$ is cyclic of order $p^a$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, of order $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$, this quotient has order prime to $p$. Its remaining possible prime divisors are smaller than $p$ and cannot divide $|G|$. Thus $N_G(P)=C_G(P)$, and the Burnside transfer theorem gives a normal p-complement.

Every nontrivial normal subgroup of a faithful primitive group of prime degree is transitive and contains a Sylow $p$-subgroup $P$. Its normalizer in $A_p$ has order $pq$. For a proper normal subgroup $N$, the Frattini argument gives $G=N{N}_G(P)$; unless $G=C_p$, its $P$-normalizer is the full order-$pq$ normalizer, while $N_N(P)=P$. The Burnside transfer theorem gives a characteristic normal $p$-complement in $N$. This complement is normal in $G$ and cannot be transitive because its order is prime to $p$, so it is trivial. Thus $N=P$ and $|G|=pq$.

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.

For the transfer homomorphism, choose representatives $T$ of the right cosets $H\backslash G$. For $t\in T$ and $g\in G$, write $tg=h(t,g)t_g$, where $h(t,g)\in H$ and $t_g\in T$. The permutation $t\mapsto t_g$ represents right multiplication on the cosets. DefineThe product has no order ambiguity in the abelianization $H/H^{\prime }$. Replace each representative $t$ by $a_{t}t$, with $a_t\in H$. The new factor is $a_{t}h(t,g)a_{t_g}^{-1}$. In the Abelian quotient, the extra products cancel because $t\mapsto t_g$ permutes $T$. Thus the transfer is independent of the transversal. Moreover,Multiplying and using the same permutation of representatives gives $V(g_1{g}_2)=V(g_1)V(g_2)$. This proves that group transfer is a well-defined group homomorphism.

For the Burnside transfer theorem, let $P\subseteq Z(N_G(P))$ be a Sylow $p$-subgroup. In particular $P$ is Abelian. We first prove the fusion fact that two elements of $P$ which are conjugate in $G$ must be equal. Suppose $u\in P$ and $v=g^{-1}ug\in P$. Both $P$ and $g^{-1}Pg$ centralize $v$, and they are Sylow subgroups of $C_G(v)$: they have the maximal $p$-order possible even in $G$. By Sylow theorems inside this centralizer, choose $h\in C_G(v)$ with $h^{-1}{g}^{-1}Pgh=P$. Thus $gh\in N_G(P)$ andBut $gh$ centralizes $P$ by hypothesis, so $v=u$.

Evaluate transfer to $P$ on $u\in P$. Partition the right cosets of $P$ into cycles under multiplication by $u$. For a cycle of length $r$ beginning at $Pt$, the product of its transfer factors telescopes to $t{u}^r{t}^{-1}\in P$. It is conjugate to $u^r\in P$, so the fusion fact makes it equal to $u^r$. The cycle lengths sum to $m=[G:P]$. Since $P$ is Abelian, all factors commute andThe integer $m$ is prime to $p$, and this power map is an automorphism of the finite Abelian $p$-group $P$: choose an inverse exponent modulo the exponent of $P$. Therefore $V_{G,P}:G\to P$ is surjective. Its kernel $K$ is normal and the first isomorphism theorem givesThus $K$ is the required normal p-complement; it also has trivial intersection with $P$.

Now suppose $P$ is cyclic of order $p^a$, where $p$ is the smallest prime divisor of $|G|$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, whose order is $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$ and $P$ is Sylow in $N_G(P)$, the quotient has order prime to $p$. Any remaining prime factor would divide $p-1$ and hence be smaller than $p$, but would also divide $|G|$, an impossibility. Thus $N_G(P)=C_G(P)$ and $P\subseteq Z(N_G(P))$. The Burnside transfer theorem proves that a cyclic Sylow subgroup at the least prime divisor gives a normal complement.

Apply this to a finite nonabelian simple group. If $p^3\nmid |G|$, its Sylow $p$-subgroup has order $p$ or $p^2$. A cyclic Sylow subgroup would give a normal complement; simplicity would then force either a trivial complement and a $p$-group, or an improper complement, neither possible for a nonabelian simple group. Groups of order $p^2$ are Abelian: their nontrivial group center, supplied by the class equation, is either the whole group or has cyclic quotient, which also forces the group Abelian. Thus the only remaining Sylow possibility is $P\cong C_p^2$.

The conjugation quotient $N_G(P)/C_G(P)$ now embeds in $G{L}_2(p)$ and is still prime to $p$. Its possible prime divisors come fromIf $p$ is odd, every prime divisor of $p-1$ is smaller than $p$, and every prime divisor of the even number $p+1$ is at most $(p+1)/2<p$. Since $p$ was the least prime divisor of $|G|$, the quotient must be trivial. Then the Burnside transfer theorem again contradicts simplicity. If $p=2$, the only nontrivial odd divisor of $|G{L}_2(2)|=6$ is three. Trivial conjugation would again give a normal complement, so three must divide $|G|$; the Sylow subgroup has order four. We have proved the least-prime divisibility constraint for a finite simple group:

For infinitely many instances of the latter alternative only, takeThese are simple, as permitted in the question (and also covered by the next solution). For $q=5^{2m+1}$, the group order is $q(q^2-1)/2$. Since $q\equiv 5(\mathrm{mod}8)$, $q-1$ is divisible by four but not eight and $q+1$ by two but not four. Hence the order is divisible by four but not eight. Also $q\not\equiv 0(\mathrm{mod}3)$, so $3\mid q^2-1$. Their smallest prime divisor is two andTheir orders strictly increase with $m$, so they give infinitely many distinct examples.

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