Let $A$ be an Abelian group. Define $A[2^n] = \{a \in A \mid 2^na = 0\}$ for a positive integer $n$.
Then, let $A[2^\infty] = \bigcup_{n\geq 1} A[2^n]$.
In general, $A[2]=0 \nRightarrow A[2^\infty]=0$.
However, this paper (https://arxiv.org/pdf/1003.4393v2.pdf), which deals with the case where $A$ is the Tate-Shafarevich group $Sha(E_n/\mathbb{Q})$ of an elliptic curve $E_n: y^2 = x^3 - n^2x$, states in the proof of example $G$ that $Sha(E_n/\mathbb{Q})[2] = 0 \implies Sha(E_n/\mathbb{Q})[2^\infty] = 0$.
Why does the implication hold in this specific case, or have I misunderstood some fundamental fact about Abelian groups?
If $a \in A[2^\infty]\backslash\{0\}$, then $a \in A[2^n]$ for some $n \ge 1$. So $2^na=0$; in particular, $a'=2^{n-1}a$ is such that $a' \neq 0$ and $2a'=0$ (if $2^{n-1}a=0$, just replace $n-1$ for $n-2$ and so on).