Proving a group, $G$, is a group action onto some set, $X$

Question

I want to prove that a function defines a group action:

We have group $G$ of diagonal $2\times 2$ matrices under matrix multiplication, and the set $X$ of points of the Cartesian plane, eg:

$G = \left\{ \begin{bmatrix} a &0\\0&b \end{bmatrix} : a,b\in \mathbb{R} - \{0\} \right\}$, $X=\{(x,y): x,y \in \mathbb{R}\}$

For each $g =\begin{bmatrix} a &0\\0&b \end{bmatrix}\in G$ and $(x,y)\in X$ where I use minus to denote that $0\not\in$ this set, define the function

$g((x,y)) = (ax,by)$

How to prove closure, identity and composition? Refer to edits for effort shown. Question de-cluttered so people won't deem it too much effort.

Answer 1

In fact, if we work with column vectors, the group action you described is just the multiplication of matrices.

$$g(x,y)= \begin{pmatrix}a&0\\0&b\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax \\ by \end{pmatrix}$$

Now the fact that this is indeed a group action follows from the well-known properties of matrix multiplication:

• Closure: If we multiply $2\times 2$ and $2\times 1$-matrix, we get again $2\times 1$-matrix.
• Identity: Multiplication by the identity matrix does not change anything.
• Compatibility is a consequence of associativity of matrix multiplication.

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Another possibility how to look at this problem is to view it as the coordinatewise multiplication of 2-dimensional vectors. (If you identify the matrix $\begin{pmatrix}a&0\\0&b\end{pmatrix}$ with the vector $(a,b)$.)

Answer 2

You might have seen that a group action $G \times X \to X$ is actually the same things as a group morphism $G \to \operatorname{Bij}(X)$. Namely, for a group action $\varphi \colon G \times X \to X$, the group morphism is $\psi \colon g \mapsto \varphi(g,\cdot)$ ; conversely, any group morphism $\psi \colon G \to \operatorname{Bij}(X)$ gives rise to a group action $\varphi \colon (g,x) \mapsto \psi(g)(x)$.

Here, viewing $2\times 2$-matrices as linear endomorphisms of the plane, the inclusion $i \colon G \hookrightarrow \operatorname{Bij}(\mathbb R^2)$ is a group morphism giving rise to the group action $(g,(x,y)) \mapsto i(g)(x,y) = g(x,y)$ which is precisely the function of the exercise.

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Remark that here the image of the inclusion $i$ is actually included in the group $\operatorname{Aut}(\mathbb R^2)$ of linear automorphisms of the plane, not only in the group of set-theoretic bijections. It is what we call a (faithful here) linear representation of the group $G$. Representation theory is a beautiful theory, you can look it up if you are curious.