*Edit: In my original post I failed to require the group to be a manifold group. The answer below from @BenLinowitz works in that case. I am really interested though in when the group is torsion-free, so have edited this to reflect that.*

For a non-elementary finite-covolume Kleinian group $\Gamma$, let $\mathbb{Q}(\mathrm{tr}\Gamma)$ and $k\Gamma$ be the trace field and the invariant trace field of $\Gamma$, in the sense of Maclachlan-Reid and Neumann-Reid.

If $\Gamma$
is derived from a quaternion algebra,
then $\mathbb{Q}(\mathrm{tr}\Gamma)=k\Gamma$.
What I want to know is,
what is an example of a case where $\Gamma$
is arithmetic *and torsion-free* (but not derived from a quaternion algebra, of course),
and $\mathbb{Q}(\mathrm{tr}\Gamma)\neq k\Gamma$?

**Definitions:**

The *trace field* of $\Gamma$
is $\mathbb{Q}(\mathrm{tr}\Gamma):=\mathbb{Q}\big(\{\mathrm{tr}\gamma\mid\gamma\in\Gamma\}\big)$.

The *invariant trace field* of $\Gamma$
is $k\Gamma:=\mathbb{Q}(\mathrm{tr}\Gamma^{(2)})$,
where $\Gamma^{(2)}:=\langle\gamma^2\mid\gamma\in\Gamma\rangle$ (the group generated by squares in $\Gamma$).

$\Gamma$ is *derived from a quaternion algebra*
if there exists a quaternion algebra $B$
that has a unique complex place,
and is ramified at all real places,
and an order $\mathcal{O}\subset B$
so that $\Gamma$
is a finite-index subgroup of $\mathrm{P}\mathcal{O}^1:=\{x\in\mathcal{O}\mid\mathrm{nrd}(x)=1\}/\{\pm1\}$.

$\Gamma$ is *arithmetic* if it is commensurable to a group derived from a quaternion algebra.

**Properties:**

$\mathbb{Q}(\mathrm{tr}\Gamma)$ and $k\Gamma$ are number fields.

If $\Gamma$ and $\Gamma'$ are commensurable (up to conjugation), then $k\Gamma=k\Gamma'$.

The definition given of arithmeticity is equivalent to $\Gamma$ being Kleinian and and an arithmetic group in the sense of Borel (the ramification conditions guarantee discreteness).