Show that the additive group $\mathbb R$ acts on the $x, y$ plane $\mathbb R \times \mathbb R$ by $r\cdot (x, y)=(x+ry,y)$.

Question

Show that the additive group $\mathbb R$ acts on the $x, y$ plane $\mathbb R \times \mathbb R$ by $r\cdot (x, y)=(x+ry,y)$.

I am completely lost with this one partly because I do not understand group action very well. I can understand the concept of group action through the example $D_{2n} \times \{1,2,...,n\} \to \{1,2,...,n\}$ because the group action of $D_{2n}$ permutates the numbers $1,2,...,n$, so its a homomorphism from $G \times A \to A$. Beyond the example of dihedral groups, I am unable to generalise this idea of $G \times A \to A$ to any group.

Must the set $A$ be part of the group? i.e. $G=(A,\cdot)$? Or $G$ and $A$ can be ANY arbitrary group and set?

Additionally, shouldn't a group be acting on its own set via its binary operator? A group itself already encompasses a set. Why would a group extend its operator to act on other sets?

Back to the main question, how do you solve the above problem? Thank you!

Answer 1

A group action on a set $A$ is a map $\cdot\,\colon G\times A\to A$ such that

$$(gh)\cdot a=g\cdot(h\cdot a),\ \forall g,h\in G, a\in A,$$

$$e\cdot a = a,\ \forall a\in A.$$

In your case, you have additive group, so you must check that

$$(r+s)\cdot(x,y) = r\cdot(s\cdot(x,y)),$$

$$0\cdot(x,y) = (x,y).$$

This shouldn't be hard. Good luck.

Answer 2

If a group $G$ acts on a set $X$ then for every $g \in G$ we get a function $\alpha_g : X \to X$.

We have a family of functions (parametrised by $G$) from $X$ to $X$.

For this to make sense, we want the group operation to agree with function composition. If $e \in G$ is the identity element and $g,h \in G$ are arbitrary elements of the group, then the axioms are

1. $\alpha_e \equiv \mathrm{id}$
2. $\alpha_g \circ \alpha_h \equiv \alpha_{gh}$

You have the group $(\mathbb R,+)$ acting on $\mathbb R^2$ with action $$\alpha_r(x,y)=(x+ry,y)$$

The identity in $(\mathbb R,+)$ is $0$, and so to meet the first axiom we need $\alpha_0(x,y)=(x,y)$. Well, by definition $\alpha_0(x,y)=(x+0\cdot y,y)=(x,y)$. So the first axiom is met.

Consider $s \in (\mathbb R,+)$ and its corresponding action $\alpha_s(x,y) = (x+sy,y)$. To satisfy the second axiom, we need $\alpha_r \circ \alpha_s \equiv \alpha_{rs}$. In other words $\alpha_r(\alpha_s(x,y)) = \alpha_{r+s}(x,y)$. Remember that addition is the group operation in $(\mathbb R,+)$ and so the general juxtaposition $rs$ represents $r+s$.

$\alpha_s(x,y)=(\color{blue}{x+sy},y)$ and so $$\alpha_r(\alpha_s(x,y)) = (\color{blue}{x+sy}+ry,y)=(x+(s+r)y,y)=\alpha_{r+s}(x,y)$$