Does the torsion subgroup of a torus have more than one complement?

Question

Let $A$ be a torus (compact connected abelian Lie group) and $T$ its torsion subgroup. Since $T$ is divisible, then $A = T \times H$ for some torsion-free subgroup $H$. Is $H$ unique? (I know in general in an abelian group $A$ it is not unique, but wondering if in Lie groups it is).

Answer 1

In general, if you have a group with a direct decomposition $G\times H$, the (semidirect) complements of $H$ are precisely the graph of homomorphisms $G\to H$.

In your case, assuming the torus is non-trivial, $H\simeq (\mathbf{Q}/\mathbf{Z})^n$ with $n\ge 1$, and $G$ is isomorphic to $\mathbf{Q}^{(c)}$ (direct sum of continuum copies of $\mathbf{Q}$.

So this provides as many direct complements as elements of $$\mathrm{Hom}(\mathbf{Q}^{(c)},(\mathbf{Q}/\mathbf{Z})^n)\simeq \mathrm{Hom}(\mathbf{Q},\mathbf{Q}/\mathbf{Z})^c;$$ since $\mathrm{Hom}(\mathbf{Q},\mathbf{Q}/\mathbf{Z})$ is obviously nontrivial [it actually has continuum cardinal itself], we deduce that the number of direct complements is continuum as well.