This is a problem (3.9b on pg. 113) from "Rational Points on Elliptic Curves 2nd Edition" by Silverman and Tate.
Let $C : y^2 = x^3 + 5x$ and $\overline{C} : y^2 = x^3 -20x $. Let $\Gamma = C(\mathbb{Q})$ and $\overline{\Gamma} = \overline{C}(\mathbb{Q})$. In the textbook, they define the homomorphisms $\alpha : \Gamma\to \mathbb{Q}^*/{\mathbb{Q}^*}^2$ and $\overline{\alpha}: \overline{\Gamma}\to \mathbb{Q}^*/{\mathbb{Q}^*}^2$ by:
$$ \alpha((x,y)) = x \mod{{\mathbb{Q}^*}^2} $$ $$ \alpha((0,0)) = 5 \mod{{\mathbb{Q}^*}^2} $$ $$ \overline{\alpha}((x,y)) = x \mod{{\mathbb{Q}^*}^2} $$ $$ \overline{\alpha}((0,0)) = -20 \mod{{\mathbb{Q}^*}^2} $$ The methodology presented in the book tells us that the rank $r$ of $C$ is equal to: $$ 2^r = \frac{\left|\alpha(\Gamma)\right|\left|\overline{\alpha}(\overline{\Gamma})\right|}{4} $$ I have found that: $$ \alpha(\Gamma) = \{ 1,5 \} \subset \mathbb{Q}^*/{\mathbb{Q}^*}^2 $$ $$ \overline{\alpha}(\overline{\Gamma}) = \{ \pm 1, \pm 5 \} \subset \mathbb{Q}^*/{\mathbb{Q}^*}^2 $$ Therefore, the rank of $C$ is $1$. Now, for the second part of the problem it's asking us to find generators for $C(\mathbb{Q})/2C(\mathbb{Q})$.
$$ C(\mathbb{Q}) = \mathbb{Z}\times (\text{torsion subgroup}) $$ $$ C(\mathbb{Q})/2C(\mathbb{Q}) = \mathbb{Z}/2\mathbb{Z}\times (\text{half of torsion subgroup}) $$
The generators for "half of torsion subgroup" are easy to find via Nagell-Lutz. I was wondering how I can find a generator for $\mathbb{Z}/2\mathbb{Z}$ in order to finish the question.
I just figured it out. I will reference notation and results in the book without going into too much detail. But for anyone else reading the book, I hope this answer will be helpful. Following the notation in the book, there are homomorphisms $\psi : \overline{\Gamma} \to \Gamma$ and $\phi: \Gamma \to \overline{\Gamma}$ such that: $$ \psi(\phi(\Gamma)) = 2\Gamma $$ By a result in the book, we know that $\ker(\alpha) =\psi(\overline{\Gamma})$. Therefore, we simply have to find an element $(x,y)\in \Gamma$ with infinite order such that $\alpha((x,y))\neq 1$. In other words, this will ensure that $(x,y)\notin 2\Gamma$ and hence will indeed be the non-zero element in $\mathbb{Z}/2\mathbb{Z}$.
Explicitly:
Using the method developed in Chapter 3.6, we know that $5\in \alpha(\Gamma)$ if and only if there is a solution in $M,e,N$ with $e > 0$, $M\neq 0$ to the equation $N^2 = 5M^4 + e^4$, and moreover $$ (x_0,y_0) = \left( 5 \cdot\frac{M^2}{e^2}, 5\cdot \frac{MN}{e^3} \right) $$ will be an affine point on the curve $C$ satisfying $\alpha((x_0,y_0)) = 5$. Thus, with a little bit of luck, I found that $M=2,N=9,e=1$ works, and therefore we have a candidate $$ \left( 20,90 \right) \notin 2\Gamma $$ It suffices to check that this element has infinite order, which is easy by taking twice the point and checking that its coefficients are non-integer.