Here is the problem:
Show by example that the subgroup of an algebraic group generated by two non-irreducible closed subsets need not be closed.
and a hint is given:
Use the cyclic subgroups of $GL(2, \mathbb{C})$ generated by $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$.
Now, let $G_1 = \{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \} $ and $G_2 = \{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} \}$ (the cyclic subgroups generated by $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$ ). Both are closed and non-irreducible subsets (subgroups). The subgroup generated by $G_1$ and $G_2$ is $G = \{ \begin{pmatrix} 1 & n_1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & n_2 \\ 0 & -1 \end{pmatrix} : n_1, n_2 \in \mathbb{Z} \} $. As $G$ is a set of discrete points in $GL(2, \mathbb{C} )$, can it be not closed?
In another way, set $G_1 = \{ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} b & 0 \\ 0 & -b \end{pmatrix} : a,b \in \mathbb{C}^* \} $ and $G_2 = \{ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} b & b \\ 0 & -b \end{pmatrix} : a,b \in \mathbb{C}^* \}$. They are closed and non-irreducible. And the subgroup of $GL(2, \mathbb{C})$ generated by $G_1$ and $G_2$ is $G= \{ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} b & 0 \\ 0 & -b \end{pmatrix}, \begin{pmatrix} c & c \\ 0 & -c \end{pmatrix}, \begin{pmatrix} d & d \\ 0 & d \end{pmatrix} : a,b,c,d \in \mathbb{C}^* \} $. Isn't it closed? And why?
I don't know if I am wrong somewhere.
To prove a variety is closed, we often make the target variety into the inverse image of a closed variety.
But how to prove a variety is not closed (in a certain larger one)?
Many thanks.