Every classical Collatz cycle contains a Syracuse-periodic point

Let $C$ be the classical Collatz map and $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ the accelerated Syracuse map. Every positive cycle of $C$ contains a point which is genuinely periodic for $T$:

Mathematical role. This is the transfer principle that makes the accelerated theory applicable to the original problem. Every sharp statement about Collatz cycles — the inequality $3^{a}<2^{K}$, the bound $2^{K}m^{a}\le(3m+1)^{a}$ on a cycle's minimum, the resulting estimate $m \lesssim a\,3^{a-1}$, and the exclusion of short accelerated periods — is a statement about $T$, because $T$ carries the arithmetic of the problem without the intervening halvings. None of it says anything about a hypothetical classical cycle until one knows that such a cycle contains a $T$-periodic point. That is what this provides, together with the fact that the point is reached from $x$ by the classical map, so the two cycles genuinely share an orbit.

The argument is a finiteness one. A classical cycle contains an odd number $y$, at which the accelerated dynamics can start. Every Syracuse iterate of $y$ is again a classical iterate of $x$, since one $T$-step is a run of $C$-steps; and the classical orbit of a periodic point is finite, being determined by the residue of the index modulo $p$. So the sequence $T^{t}(y)$ takes only finitely many values and must repeat: $T^{s}(y)=T^{t}(y)$ for some $s<t$. The point $z=T^{s}(y)$ then satisfies $T^{\,t-s}(z)=z$ with $t-s>0$.

What it does and does not give. Combined with the exclusion of accelerated periods $1$ through $5$, it shows that any positive classical cycle whose accelerated period is at most $5$ must contain $1$, hence is the trivial cycle. It does not bound that accelerated period, so it does not by itself exclude classical cycles; supplying such a bound is the remaining content of the cycle problem.

Formalization note. The conclusion asserts $z$ is reached from $x$ in $k$ classical steps, rather than that $z$ lies in a set, so no orbit set need be defined. The period $a$ produced is positive but not claimed minimal.