Conjugate transpose and discreteness, for Kleinian groups 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.
 A: Here is a thought about your question. Suppose that $G$ contains a unipotent element $u$. Then we can conjugate $G$ to $G'=hGh^{-1}$ such that $u$ conjugates to $u'=huh^{-1}$ which is close to $1\in SL(2,C)$. Note that if $u_1, u_2$ are two such commuting unipotent elements then $v_i=(u_i')^{\dagger} u_i'$, $i=1, 2$, will not (generically) commute. Therefore, by Jorgensen's inequality (or Zassenhaus Lemma if you prefer), the subgroup generated by $v_1, v_2$ will be nondiscrete. I suspect, one can play a similar game in greater generality and prove that for any infinite discrete subgroup, its conjugate $hG h^{-1}$ leads to  nondiscrete subgroup. This definitely applies to "generic" geometrically infinite finitely generated Kleinian groups. But for convex-cocompact groups cannot use Jorgensen inequality argument and one needs a different approach. 
A: The map that you have written down is equivariantly identified with the map
\begin{align}
\mathbb{F}(\mathbb{H}^3)\rightarrow \mathbb{H}^3
\end{align}
where $\mathbb{F}(\mathbb{H}^3)$ is the bundle of oriented, orthonormal frames over hyperbolic 3-space $\mathbb{H}^3.$  Being a fiber bundle with compact fibers, this maps discrete sets to discrete sets.  In particular, this does not require that the discrete set of points in $\text{PSL}(2,\mathbb{C})$ forms a group.
