# Is there a good way to show that a subgroup is Zariski-dense?

Let $G$ be a semisimple algebraic group, and let $\Gamma$ be a finitely generated subgroup. Given the generators of $\Gamma$, is there a good way to determine if $\Gamma$ is Zariski-dense in $G$?

For example, let $\Gamma \subseteq \mathrm{PSL}_2$ where $\displaystyle \Gamma = \left\langle \gamma_1 , \gamma_2 \right\rangle$ and $\gamma_1 = \begin{pmatrix} 1 & i \newline 0 & 1 \end{pmatrix}$ and $\gamma_2 = \begin{pmatrix} 1 & 0 \newline 1 & 1 \end{pmatrix}$.

2. In $PSL_2,$ Zariski-dense is equivalent to non-elementary, so your group is Zariski-dense.
• By "good," I meant that I was looking for some criterion for Zariski-density based on the generators of the group. I was looking for a way to create examples of Zariski-dense subgroups that are a little more interesting than $\mathrm{SL}_2(\mathbb{Z})$ in $\mathrm{SL}_2$ and most importantly, be able to prove that these examples are Zariski-dense. Thank you very much for your answer; it sounds like it's what I was looking for. – JS_UNCUVa Feb 23 '12 at 20:56