# All Rational Points of a particular Elliptic Curve

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 $$2$$?

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.