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Let the group $SL(2,\mathbb{Z}) = \left\{ \begin{pmatrix}a & b \\ c & d \end{pmatrix}\vert ad-bc = 1\right\}$ act on $\mathbb{Z}^{2}$ by matrix multiplication in the following way: for $M \in SL(2,\mathbb{Z})$ and $x \in \mathbb{Z}^{2}$, $M.x$ is the matrix product $Mx$ of the $2\times 2$ matrix $M$ and the vector $x$ of dimension $2$ written in the column form.

First, I had to find the stabilizer $H = Stab(x_{0})$ of $x_{0} = \begin{pmatrix} 1 \\ 0\end{pmatrix}$, and I found it to be $\left\{ \begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix} \vert b \in \mathbb{Z}\right\}$.

Now, I need to show that $H = \left\{ \begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix} \vert b \in \mathbb{Z}\right\}$ is a cyclic group.

I know that any infinite order cyclic group is isomorphic to $\mathbb{Z}$, so I attempted to set up an isomorphism between $H$ and $\mathbb{Z}$. I chose the mapping $f:\begin{pmatrix}1 & b \\ 0 & 1 \end{pmatrix} \mapsto b$; however, when I tried to show the mapping was a group homomorphism, I failed:

Suppose $x = \begin{pmatrix} 1&b \\ 0 & 1 \end{pmatrix}$, $y = \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}$ $\in H$.

The problem is is that

$f(xy) = f\left(\begin{pmatrix} 1&b \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}\right) \\ = f\left(\begin{pmatrix} 1 & b+c \\ 0 & 1\end{pmatrix}\right) = b+c \neq b\cdot c = f\left( \begin{pmatrix} 1&b \\ 0 & 1 \end{pmatrix}\right) \cdot f\left( \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}\right) = f(x) f(y)$,

so $f$ is NOT even a homomorphism, let alone an isomorphism.

Pouring over various theorems in my notes, I have no found another way to approach this problem that looks promising, given that $H$ is not of finite order. Is there an isomorphism between $H$ and $\mathbb{Z}$ that gets the job done that I'm overlooking? If so, what is it?

Thank you.

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    The proof is alright. The point where you have mistaken is that $\mathbb{Z}$ is a group with the law of composition being addition and not multiplication. So, $$f(xy) = f\left(\begin{pmatrix} 1&b \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}\right) \\ = f\left(\begin{pmatrix} 1 & b+c \\ 0 & 1\end{pmatrix}\right) = b+c = f\left( \begin{pmatrix} 1&b \\ 0 & 1 \end{pmatrix}\right) + f\left( \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}\right) = f(x) + f(y)$$ – 2017-01-17
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    $\mathbb{Z}$ under multiplication is not even a group ($\mathbb{Z}\setminus \{0\}$ is a commutative monoid). "cyclic group with infinite order" is really the definition of $\mathbb{Z},+$ – 2017-01-17

2 Answers 2

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The map $f:H = \left\{ \begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix} \vert b \in \mathbb{Z}\right\}\rightarrow \mathbb{Z}$, given by $f(z)=\begin{pmatrix}1 & z \\ 0 & 1\end{pmatrix}$ is indeed an homomorphism because $f(xy)=f(x)+f(y)$. Here $\mathbb{Z}$ is a group with respect to addition, so $f(x)\circ f(y)=f(x)+f(y)$. Clearly $f$ is surjective and injective.

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Remember that the group operation on $\mathbb{Z}$ is addition, not multiplication. So your $f$ actually is a homomorphism: it turns the group operation on $H$ into the group operation on $\mathbb{Z}$.