TITLE: Kernel of an action
QUESTION [5 upvotes]: $$G = \left\{ 
 \begin{pmatrix}
  a & b \\
  0 & 1  \\
  \end{pmatrix}
\text{with $a$ in $\{1, -1\}$ and $b$ in } \mathbb{Z}\right\}$$
G is a subgroup of the matrix group $GL_2(\mathbb{Q})$. 
$\phi : G \rightarrow \{1,-1\} \times \mathbb{Z}/2\mathbb{Z}$ is given by $ (
 \begin{smallmatrix}
  a & b \\
  0 & 1  \\
  \end{smallmatrix})
\rightarrow (a,\overline{b})$
Is the kernel of $\phi$ equal to $ \left\{ 
 \begin{pmatrix}
  1 & 2 \\
  0 & 1  \\
  \end{pmatrix}^n
\text{with $n$ in $\mathbb{Z}$ } \right\}$? Give a proof or a counterexample.
I'm totally stuck on this question. How would one go about answering this? I know the kernel of an action $G$ on $S$ is defined as $\{g \in G \vert g \cdot s = s \text{ for all $s$ in $S$}\}$, but I don't know how to go from there. For instance, what is $S$ in this case? Or is that even the way to go?

REPLY [2 votes]: The answer is yes. The group $\{1,-1\}\times \mathbb Z/2\mathbb Z$ has $(1, \bar 0)$ as its identity.
The kernel of $\phi$ is therefore all matrices of the form $ \begin{pmatrix}
        1 & 2n \\
        0 & 1 \\
        \end{pmatrix}$
But ${\begin{pmatrix}
       1 & 2 \\
       0 & 1
     \end{pmatrix}}^n=\begin{pmatrix}
        1 & 2n \\
        0 & 1 \\
        \end{pmatrix}$ as can be seen by induction.