Let $G=\langle g_1,\dots g_n\rangle<\mathrm{PSL}_2(\mathbb{C})$ be discrete, i.e. a finitely generated Kleinian group. Let $H=\langle g^{\dagger}g\mid g\in G\rangle$, (the group generated by the $g^{\dagger}g$) where $g^{\dagger}$ is the conjugate transpose.
Under what conditions is $H$ discrete?
Edit: In light of comments from @Misha, below I am removing my claims about what I can prove and adding some motivation for the question.
In the spinor representation of the restricted Lorentz group, $\mathrm{PSL}_2(\mathbb{C})$ is faithfully represented as the group of orientation-preserving isometries of the hyperboloid model $\mathbb{I}$ for hyperbolic 3-space. In this construction, we identify the points on $\mathbb{I}$ with the Hermitian matrices of norm 1, up to identification of $\pm1$. Then for $(g,p)\in\mathrm{PSL}_2(\mathbb{C})\times\mathbb{I}$ the action can be defined by $g(p)=g^{\dagger}pg$. Skipping some details, the point $p=(1,0,0,0)\in\mathbb{I}\subset\mathbb{R}^4$ is identified with the identity matrix. So in this construction, the set $\{g^{\dagger}g\mid g\in G\}$ is identified with the orbit under $G$ of $p$. It is for this reason that I said this set is certainly discrete.
I am working on a more elaborate duality between points in hyperbolic space and isometries of that space using quaternion algebras. I am holding back some info because this is part of my thesis which is not published, and my school would not be too happy if I posted certain ideas online at this point. So suffice it to say I have interesting applications using the group $H$ generated by that set.