Why Are There Only $2^t$ Points of Order $2$ in an Elliptic Curve Let $E(\mathbb{Q})$ be an elliptic curve over rationals. Mordell's theorem says that $E(\mathbb{Q}) \simeq \mathbb{Z}^r \oplus (\mathbb{Z}_{p_1^{\nu{_1}}} \oplus...\oplus \mathbb{Z}_{p_s^{\nu{_s}}})$ where $p_j$ is prime. I'm reading Silverman and Tate's book 'Rational Points on Elliptic Curves', and they 'proved' that the subgroup of points  $Q\in E(\mathbb{Q})$, such that $2Q=0$ (where $0$ is the point at infinity), that is, $$2Q=2(n_1P_1+n_2P_2+...+n_rP_r+...+m_1Q_1+...m_rQ_r)=0$$ where the $P_i$'s are infinite ordered points and $Q_j$'s are finite ordered points, basically they are the generators as what the Mordell's theorem says, has order $2^t$. Where $t=\# \text{of j's such that}\ p_j=2$. I think what the book did is to count the number of "integers" (sort of like scalars) $n_i$'s and $m_j$'s so that the equation above is true. But I do not really get why we end up with $2^t$. Can anybody help enlighten me as to why?
 A: If I understand your question correctly, what you are asking is just a fact of abstract algebra (a fact about abelian groups). Can you prove the following steps? Below, let $G$ be a finite abelian group (which in your case will be the torsion subgroup, i.e., $G\cong \mathbb{Z}/p_1^{\nu_1}\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/p_s^{\nu_s}\mathbb{Z}$ in your notation).


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*The number of non-trivial elements in $G/2G$ is equal to the number of elements in $G$ of order exactly $2$.

*If the order of $G$ is odd, then $G/2G$ is trivial.

*If $G$ is finite abelian, then there are subgroups $H$ and $J$, with $H$ of even order, and $J$ of odd order, such that $G \cong H \oplus J$. (Hint: cyclic decomposition theorem.)

*With $G,H,J$ as above, $G/2G \cong H/2H$.

*If $H = \mathbb{Z}/2^n\mathbb{Z}$ with $n\geq 1$, then $H/2H \cong \mathbb{Z}/2\mathbb{Z}$.

*If $H = \mathbb{Z}/2^{\nu_1}\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/2^{\nu_t}\mathbb{Z}$, then $H/2H \cong (\mathbb{Z}/2\mathbb{Z})^t$, and therefore, $|H/2H| = 2^t$.
