Does the Grothendieck group detect the Picard group?

Let $$C$$ be a curve (=smooth projective curve) of genus $$g$$ over an algebraic closed field $$\mathbb{k}$$. It is well known that the Grothendieck group $$K_0(\operatorname{coh} C)$$ of the category of coherent sheaves on $$C$$ is the following (see, for example, Exercise 6.11 of Hartshorne): $$K_0(\operatorname{coh} C)\simeq \operatorname{Pic} C \times \mathbb{Z} \simeq \operatorname{Pic}_0 C \times \mathbb{Z}^2,$$ where $$\operatorname{Pic} C$$ is the Picard group of $$C$$ and $$\operatorname{Pic}_0 C$$ is the kernel of the degree map $$\mathrm{deg} : \operatorname{Pic} C\to \mathbb{Z}$$. In the case $$\mathbb{k} = \mathbb{C}$$, we also know that $$\operatorname{Pic}_0 C \simeq \mathbb{C}^g/\Lambda,$$ where $$\Lambda$$ is a $$\mathbb{Z}$$-submodule in $$\mathbb{C}^g$$ of rank $$2g$$.

My Questions:

1. For two curves $$C_1$$ and $$C_2$$, if $$K_0(\operatorname{coh} C_1) \simeq K_0(\operatorname{coh} C_2)$$ as abstract groups, then do we conclude that $$\operatorname{Pic}_0 C_1 \simeq \operatorname{Pic}_0 C_2$$ (or $$\operatorname{Pic} C_1 \simeq \operatorname{Pic} C_2$$) as abstract groups ? Here abstract groups means groups without additional structure, such as structure of topology, variety and so on.
2. Does an isomorphism $$\operatorname{Pic}_0 C_1 \simeq \operatorname{Pic}_0 C_2$$ of abstract groups contain some geometric information about $$C_1$$ and $$C_2$$?
• As an abstract group $Pic^0(C)$ is the same for every genus $g$ curve, because all complex tori of the same dimension are diffeomorphic? On the other hand, genus is recovered, as dimension of the $p$-torsion of this group (over $\mathbb{F}_p$) equals $2g$. Commented Jun 1, 2022 at 8:37

I don't know about 2., but the answer to 1. is yes.

More generally, suppose $$(S^1)^g\times\mathbb Z^k \cong (S^1)^h \times \mathbb Z^m$$ as abstract groups, then $$g=h, k=m$$ (in fact, you only need $$g=h$$ for question 1., because the abstract isomorphism type of $$\mathbb C^g/\Lambda$$ only depends on $$g$$, and because you already know $$k=m=2$$)

(Note that this gives part of the answer for 2., namely that $$C_1$$ and $$C_2$$ have the same genus)

Let's prove this statement.

First, if there is an isomorphism between the two groups, then their torsion parts are isomorphic. Their torsion parts are $$(\mathbb{Q/Z})^g$$ and $$(\mathbb{Q/Z})^h$$ respectively. But you can recover $$g$$ from $$(\mathbb{Q/Z})^g$$ in the following way : take the $$2$$-torsion (not $$2$$-power torsion, $$2$$-torsion) part as an $$\mathbb F_2$$-vector space, this has dimension $$g$$. (I chose $$2$$ but could have chosen any $$p$$, in fact for any integer $$n\geq 2$$, the $$n$$-torsion part is a free $$\mathbb Z/n$$-module of rank $$g$$, and these rings have the invariant basis property).

Therefore $$g=h$$ - this is all that's relevant to your question, but let me still prove $$k=m$$.

Note that to get $$k=m$$, we cannot rationalize and take dimensions because $$(S^1)^g\otimes \mathbb Q$$ is an uncounatbly dimensional rational vector space.

We now observe that we can mod out the divisible part - I don't know the standard name, but let me define it : $$\mathrm{Div}(A) = \{a\in A\mid \forall n \neq 0, \exists b \in A, nb = a\}$$. This is clearly a subgroup of $$A$$, and $$\mathrm{Div}(A\times B)\cong \mathrm{Div}(A)\times \mathrm{Div}(B)$$. In particular $$\mathrm{Div}((S^1)^g\times \mathbb Z^k) = (S^1)^g\times \{0\}$$, and so modding out $$A/\mathrm{Div}(A)$$ gives, in our case, an isomorphism $$\mathbb Z^k\cong \mathbb Z^m$$, and therefore $$k=m$$.