Question

... every Riemann surface of genus $1$ appears as a complex one-parameter subgroup of $G$?

Answer 1

No. In a connected complex Lie group all compact complex subgroups must be in the center, that is, in the kernel of the adjoint representation, because in a complex general linear group there is no nontrivial compact connected complex subgroup. And in a connected abelian complex Lie group there can be only countably many compact one-parameter complex subgroups.

Edit: So in fact in a complex Lie group there are only countably many compact connected one-parameter complex subgroups (not just up to isomorphism).

Edit: Just to summarize what I have learned from this exercise: Let CCC="compact connected complex".

(1) (I already knew this.) A CCC subgroup of $GL_n(\mathbb C)$ must be trivial. Compact implies contained in a conjugate of $U_n$; complex then implies $0$-dimensional; connected then implies trivial. (Alternatively, a holomorphic map from a CCC manifold to $\mathbb C$ (therefore to $GL_n(\mathbb C)$) must be constant.)

(2) (IAKT) A CCC group is always abelian. This follows from (1) using the adjoint representation. Therefore it must be a compact complex torus group, the quotient of a complex vector space by a full lattice (i.e. a subgroup generated by an $\mathbb R$-basis).

(3) (This had never occurred to me before, but it's obvious by the same argument.) In a connected complex Lie group a CCC subgroup is always in the center. Thus a complex Lie group $G$ has a maximal CCC subgroup in the strongest sense -- one which contains all others -- namely the (unique) maximal compact subgroup of the connected component of the center of the connected component of $G$.

(4) Therefore there cannot be a nonconstant family of CCC subgroups of a complex Lie group. (3) reduces this statement to the case where the ambient group is abelian, and that case is clear.

I don't know what you mean by infinite-dimensional Lie group, exactly, but it seems to me that the arguments above can be adapted to show that in the group of all complex automorphisms of a connected complex manifold there is a CCC containing all others (again contained in the center of the component) so that again there are no families.

Answer 2

Let me show that such a group, if it exists, can't be algebraic. Any complex algebraic group $G$ is an extension of an abelian variety $A$ by an affine group $H$, i.e. we have an exact sequence $1\to H\to G\to A\to 0$ (Chevalley's theorem). Suppose $G$ had every elliptic curve as a subgroup. All of those would project isomorphically to $A$, since none of them cam intersect $H$. So we may just as well restrict ourselves to the case $G=A$. But $A$ has only countably many Lie subgroups.

upd: here is a proof in the general case: A subgroup of $G$ which is an elliptic curve is contained in a maximal (compact) torus; moreover, since all tori are conjugate, the isomorphism classes of the elliptic curves they contain are the same. Now we use the above argument: every torus contains countably many (closed) Lie subrgoups.