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All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore