# Problem related to group action of $GL_2(\mathbb Z)$ on $\mathbb Z^2$

Let $$G = GL_2(\mathbb Z)$$ and $$X = \mathbb Z^2$$ with elements of $$X$$ written as column vectors.

(a) Show that $$\phi: G\to \text{Perm}(X)$$ given by $$\phi(g)\left(\begin{matrix} m \\ n\end{matrix}\right) = g\left(\begin{matrix} m \\ n\end{matrix}\right)$$ is a group action.

(b) Show that the orbits of this action are the same as the equivalence classes under the relation $$(m,n)\sim (a,b)$$ if $$\gcd(m,n) = \gcd(a,b)$$.

(c) Let $$(m,n)\in\mathbb Z^2$$. Show that there exists $$g\in G$$ such that $$g\ \text{stab}_G\left(\begin{matrix} m \\ n\end{matrix}\right)g^{-1} = \left\{\left(\begin{matrix} 1 & a \\ 0 & b\end{matrix}\right) : a\in\mathbb Z, b=\pm 1\right\}$$

Before anything, it's important to note that for all $$g\in G$$, $$\det g = \pm 1$$.

I was able to do (a). $$\left(\begin{matrix} 1 & 0 \\ 0& 1\end{matrix}\right) \in G$$ is the identity element $$e$$ for this group action, and $$g\cdot (h\cdot x) = (gh)\cdot x$$ for all $$x\in X$$ follows from associativity of matrix multiplication.

I tried (b). Firstly, $$\sim$$ is indeed an equivalence relation - this is easy to check. Now, I considered arbitrary $$x\in X$$ and looked at $$G\cdot x = \{g\cdot x: g\in G\}$$, the orbit of $$x\in X$$. I presume if I start working with arbitrary matrix entries at this point, things will get messy, so there must hopefully be a nicer way. Also, I noted that if $$(m,n)\sim (a,b)$$ then there exist integers $$a_1,b_1,m_1,n_1$$ such that $$mm_1 + nn_1 = aa_1 + bb_1$$. What do I do next? Don't have much clue about (c) either.

I'd appreciate any hints or solutions that would help since I have been working on this for over a week now without much progress.

P.S. $$\text{Perm}(X)$$ denotes the set (actually a group under composition itself) of all bijections $$X\to X$$.

Hint for part (b): let $$d$$ be the greatest common divisor of $$m$$ and $$n$$, which is also the greatest common divisor of $$a$$ and $$b$$. Try showing there is a $$g \in \operatorname{GL}_2(\mathbb Z)$$ such that

$$g \begin{pmatrix} m \\ n \end{pmatrix} = \begin{pmatrix} d \\ 0 \end{pmatrix}. \tag{\ast}$$

By the same reasoning, there will exist another matrix $$h \in \operatorname{GL}_2(\mathbb Z)$$ such that

$$h \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} d \\ 0 \end{pmatrix}.$$

It follows that $$h^{-1}g$$ will send $$\begin{pmatrix} m \\ n \end{pmatrix}$$ to $$\begin{pmatrix} a \\ b \end{pmatrix}$$, so they will be in the same orbit.

How do you know that such a matrix $$g$$ exists? You know that $$d$$ is a divisor of $$m$$ and $$n$$, so $$\frac{m}{d}$$ and $$\frac{n}{d}$$ are integers. You also know that there exist integers $$x$$ and $$y$$ such that $$xm + ny = d$$. Now you can take

$$g = \begin{pmatrix} x & y \\ ? & ? \end{pmatrix}$$

so that the top row of $$\ast$$ comes out correctly. What should go in the blanks '?' so that the determinant of $$g$$ is $$1$$ and the bottom row of $$\ast$$ comes out correctly?

The above reasoning shows that if $$(a,b)$$ and $$(m,n)$$ have the same GCD $$d$$, then they are in the same orbit, since they are both in the orbit of $$(d,0)$$. You still need to show conversely that if $$g \in \operatorname{GL}_2(\mathbb Z)$$, and $$g.(m,n) = (a,b)$$, then $$(m,n)$$ and $$(a,b)$$ have the same GCD.

Hint for part (c): in general, if $$G$$ is a group acting on a set $$X$$, then for $$g \in G$$ and $$x \in X$$, we can consider the stabilizers $$H_x$$ and $$H_{g.x}$$ of the elements $$x$$ and $$g.x$$ of $$X$$, respectively. It's quite easy to see that these subgroups of $$G$$ are related by conjugation: $$gH_x g^{-1} = H_{g.x}$$.

So the question is, what element of $$\mathbb Z^2$$ is

$$\{ \begin{pmatrix}1 & a \\ 0 & b \end{pmatrix} a \in \mathbb Z, b = \pm 1 \}$$ the stabilizer of?

• Thanks a lot for the hints, I shall retry this problem and let you know! Mar 8, 2021 at 3:08