Why $[E(K_q) : E_0(K_q)] ≤ 4$ and $E_0(K_q)$ is divisible by $2$ implies $E(K_q)[2^∞] = E[2]$?

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Let $E$ be an elliptic curve over $\Bbb{Q}$ with no non-trivial rational $2$-torsion point. We write $q$ for a odd prime of quadratic number field $K$. Let $K_q$ denote the $q$-adic completion of $K$. Suppose $dim_{\Bbb{F}_2}(E(\Bbb{Q}_{q})[2])=2$.

Lem 2.2. Page 3 of https://arxiv.org/pdf/1801.10536.pdf reads

Since $[E(K_q) : E_0(K_q)] ≤ 4$ and $E_0(K_q)$ is divisible by $2$ , $E(K_q)[2^∞] = E[2]$.

Why can we conclude $E(K_q)[2^∞] = E[2]$ ? In these conditions, it seems there is no contradiction even if $E$ has order $4$ or $8$ element for me. I may have forgotten some basic group theory or elliptic curve theoretical results. Thank you for your help.