= Solution
Let $L$ be any degree-zero <line bundle> on the genus-one <smooth projective curve>. The twist $M=L\otimes\mathcal O(p_0)$ has degree one. The <Riemann-Roch theorem> gives
$$
h^0(M)-h^0(K_X\otimes M^{-1})=1.
$$
The second <line bundle> has degree $2g-2-1=-1$, so it has no nonzero section: the zero divisor of such a section would be effective of negative degree. Hence $h^0(M)=1$. A nonzero section has an effective zero <divisor> of degree one, which is a single point $p$. It is a $k$-rational point even over a general field, since a degree-one effective divisor has residue degree one. Thus $M\cong\mathcal O(p)$ and $L\cong\mathcal O(p-p_0)=\alpha(p)$.
This proves surjectivity; part (a) gives injectivity. Hence \b[$\alpha$ is a bijection], as in the <Abel-Jacobi map of a genus-one curve>. Define
$$
\boxed{p\oplus q=\alpha^{-1}\bigl(\alpha(p)\otimes\alpha(q)\bigr),\qquad \ominus p=\alpha^{-1}\bigl(\alpha(p)^{-1}\bigr).}
$$
Tensor product of <line bundles> is associative and commutative, so this is an <abelian group> law on $X(k)$. Its identity is $p_0$, since $\alpha(p_0)=\mathcal O_X$. Equivalently, $p\oplus q$ is the unique point $r$ with $\mathcal O(r-p_0)\cong\mathcal O(p+q-2p_0)$.
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