= Solution
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>. Define
$$
\boxed{V_{G,H}(g)=\prod_{t\in T}h(t,g)\pmod{H'}.}
$$
The product has no order ambiguity in the <abelianization> $H/H'$. Replace each representative $t$ by $a_tt$, 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,
$$
h(t,g_1g_2)=h(t,g_1)h(t_{g_1},g_2).
$$
Multiplying and using the same <permutation> of representatives gives $V(g_1g_2)=V(g_1)V(g_2)$. 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 $gh\in N_G(P)$ and
$$
(gh)^{-1}u(gh)=h^{-1}vh=v.
$$
But $gh$ 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 $tu^rt^{-1}\in P$. 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 and
$$
V_{G,P}(u)=u^m.
$$
The 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> gives
$$
\boxed{[G:K]=|P|,\qquad |K|=[G:P].}
$$
Thus $K$ 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 $\operatorname{Aut}(P)$, 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 \b[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 $GL_2(p)$ and is still prime to $p$. Its possible prime divisors come from
$$
|GL_2(p)|=(p^2-1)(p^2-p)=p(p-1)^2(p+1).
$$
If $p$ 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 $|GL_2(2)|=6$ 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>:
$$
\boxed{p^3\mid |G|\quad\text{or}\quad p=2\text{ and }12\mid |G|.}
$$
For infinitely many instances of the latter alternative only, take
$$
G_m=PSL_2(5^{2m+1}),\qquad m\geq0.
$$
These 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 $q\equiv5\pmod8$, $q-1$ 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 $q\not\equiv0\pmod3$, so $3\mid q^2-1$. Their smallest prime divisor is two and
$$
\boxed{12\mid |G_m|,\qquad8\nmid |G_m|.}
$$
Their orders strictly increase with $m$, so they give infinitely many distinct examples.
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