First, we shall recall the definition of an irreducible lattice.
Let $G$ be a Lie group which admits a direct product decomposition into simple non-compact factors $G_1\times\dots\times G_k$. A lattice $\Gamma$ in $G$ is irreducible if the intersection of $\Gamma$ with any factor $G_{i_1}\times\dots\times G_{i_j}$ is not a lattice [Wikipedia].
Now, let $G$ be a Lie group which admits a direct product decomposition into simple factors $G_1\times G_2$. Let $\Gamma$ be a lattice in $G$ and assume that $\Gamma\cap G_1$ is not a lattice in $G_1$. What conditions are sufficient to say $\Gamma$ is irreducible?
Obviously we can simply satisfy the above definition, but in the case of two factors of $G$ are there other equivalent conditions which could hold that would not in the case of $3$ or more factors?
Let $G_1,G_2$ be simple, noncompact connected Lie groups with finite center.
For a lattice $\Gamma$ in $G=G_1\times G_2$, the following are equivalent:
(The negation of either of these conditions is called being "irreducible".)
With more factors, the definition is a little more tricky. Let $I$ be a finite set, $G_i$ be simple noncompact connected Lie groups with finite center. For $J\subset I$, write $G_J=\prod_{j\in J}G_j$, and $G=G_I$. Let $\Gamma$ be a lattice in $G$. The following are equivalent
Note that the condition "for every $i$, the projection of $\Gamma$ on $G_i$ is dense", sometimes abusively called "irreducible" by confusion with the case $|I|=2$, is not equivalent: it is false for $|I|=1$, and strictly weaker if $|I|\ge 4$.
I think all this material is more or less covered in Raghunathan's book.