The subgroup of an algbraic group generated by two non-irreductible closed subsets need not to be closed

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Actually I have difficulties with the Section 7 Exercise 10 of the book J. E. Humphrey "Linear algebraic groups".

This exercises is about what is mentionned in the title. We have to make a demonstration by example and the book alrady gave one: Consider in $GL_2(\mathbb{C})$ the subgroup $H$ generated by the elements $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)$ and $\left( \begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array} \right)$

In addition, I would like to find its Zariski closure $\overline{H}$ and I don't know how to.

Thank you in advance for any hint to solve this. K. Y.

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First hint: Let $G$ be a subgroup of $GL_n(\mathbb{C})$, then if $G$ is both countable and Zariski closed, it is finite. Use this to deal with the exercise.

Second hint: Show that the kernel of the determinant map restricted to $H$ is of index $2$ in $H$ and explicitly compute its kernel. Deduce from this the connected component of the Zariski closure of $H$ and then $\overline{H}$.