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After the discussion in the previous elliptic curve question, we know that the elliptic curve $y^2+4xy=x^3-6x^2-24x+144$ (Given by @YongHaoNg), which can be transform into $C:y^2=x^3-20x^2+108x$, has rank $1$, and the only non-identity torsion point in $C$ is $(0,0)$, which is of order two. Now I'm working on finding the generators of the infinite subgroup in $C(\mathbb Q)$, the group of all rational solutions of $C$, which is isomorphic to $\mathbb Z$.

We know that $$\Gamma=C(\mathbb Q)\cong \mathbb Z_2\times \mathbb Z$$ So I define the subgroup $S=\{\mathcal O, (0,0)\}$ of $\Gamma$ to be the subgroup of all rational points of finite order. I also define $\Gamma_1$ to be the subgroup of $\Gamma$ isomorphic to the $\mathbb Z$ part.

My approach is: We first find out the index $\Gamma:\Gamma_1$, which I believe is $2$, and then since $(6,12)$ is in $\Gamma$, by group theory $(6,12)$ is either in $\Gamma_1$ or in $(0,0)+\Gamma_1$, in which case $(6,12)-(0,0)=(18,-36)$ is in $\Gamma_1$.

Now let $Q_0$ be the generator of $\Gamma_1$, in the book "Rational Points on Elliptic Curves" by Silverman and Tate, page 49, it has been shown that any point $(x,y)$ in $\Gamma$ is of the form $x=m/e^2,y=n/e^3$ for some integers $m,n,e$ with $\gcd(m,e)=\gcd(n,e)=1$. If $P,Q$ are two points in $\Gamma$ such that $x(P)=m_1/e_1^2,x(Q)=m_2/e_2^2$ with $e_1,e_2$ having nontrivial common divisor, then $x(P+Q)$ mustn't be an integer.

Combining the few facts above, we know that there is an integer $n$ such that $nQ=(6,12)$, if $(6,12)\in \Gamma_1$, or $nQ=(18,-36)$ if $(18,-36)$ is in $\Gamma_1$, therefore I could show that $Q_0$ is also an $\textbf{integer point}$. Finally, I will use my result, which says that $\Gamma$ is generated by all points in $\Gamma$ having height not larger than $568000$, then I just need to check $1136000$ cases by computer, $Q_0$ is ready to be found!

But I'm skeptical at one point, this problem may sounds silly, but I should keep asking until I can convince myself. Given $\Gamma\cong \mathbb Z_2\times \mathbb Z$ is true, is it really true that the index $\Gamma:\Gamma_1$ is equal to $2$? I'm having this problem because I knew that although $\mathbb Z$ is properly contains $2\mathbb Z$, but they two are actually isomorphic, so I'm wondering can $\Gamma:\Gamma_1$ be something larger than $2$?

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Your index is $2$.

If we have an abelian group decomposition $G=T\oplus L$, then the first isomorphism theorem applied to projection on $T$ yields $G/L\simeq T$, and $[G:L]= \vert T\vert$ (provided that $T$ is finite).

Now take $G=\Gamma.$ Since $G$ is a finitely generated abelian group, we have a decomposition $\Gamma=T\oplus \Gamma_1$, where $T$ is the torsion subgroup of $G$ (the subgroup of points of finite order) and $\Gamma_1$ is a free abelian group of finite rank (the rank of the elliptic curve). In your example, $T$ has order two.

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  • $\begingroup$ Thanks for your verification! $\endgroup$ – kelvin hong 方 Nov 21 '19 at 0:27

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