I am reading Dave Witte-Morris' book on arithmetic groups and having trouble getting a handle on his definition of an irreducible lattice.
A lattice $\Gamma\subset G$ in a semisimple Lie group $G$ is a discrete subgroup such that the quotient $G/\Gamma$ has finite volume (with respect to Haar measure).
In chapter 4.3 an irreducible lattice is defined to be a lattice $\Gamma$ such that $\Gamma N$ is dense in $G$ for every noncompact, closed, normal subgroup $N$ of $G^{\circ}$ (where $G^{\circ}$ is the identity component of $G$).
How does this definition work intuitively and how can I work with it?
My intuition behind a reducible lattice $\Gamma\subset G$ is that it is 'made up' of a product of lattices so that there exists some subgroup $H\le G$ so that $\Gamma\cap H\subset H$ is also a lattice.
I kind of understand that for example $\mathrm{SL}_{2}(\mathbb{Z})\times\mathrm{SL}_{2}(\mathbb{Z})\subset \mathrm{SL}_{2}(\mathbb{R})\times \mathrm{SL}_{2}(\mathbb{R})$ is reducible. In this case if we consider $\mathrm{SL}_{2}(\mathbb{R})\times \{I_{n}\}$ then $$\left(\mathrm{SL}_{2}(\mathbb{Z})\times\mathrm{SL}_{2}(\mathbb{Z})\right)\cap \left(\mathrm{SL}_{2}(\mathbb{R})\times \{I_{n}\}\right)= \mathrm{SL}_{2}(\mathbb{Z})\times \{I_{n}\}\subset \mathrm{SL}_{2}(\mathbb{R})\times \{I_{n}\}$$ and this is clearly a lattice so it is reducible.
But then why is $\mathrm{SL}_{n}(\mathbb{Z})\subset \mathrm{SL}_{n}({\mathbb{R}})$ irreducible?
