Prove that for $\pi_g(\alpha)=\alpha g$, $\pi$ is a group action

Question

Let $R$ a principal ideal domain and let $G$ the group of all the invertible members of $M_2(R)$. Let $\Omega :=R^2$. For all $g\in G$ let $\pi_g:\Omega\to\Omega$ defined by $\alpha \pi_g=\alpha g$ for all $\alpha=(x,y)\in\Omega$. Prove that the map $\pi:G\to Sym(\Omega)$ defined by $g\mapsto \pi_g$ is an action of $G$ on $\Omega$.

In the solution I've managed to show that $\pi(I,\alpha)=\alpha$, (for $I=\bigl(\begin{smallmatrix} 1 &0 \\ 0 & 1 \end{smallmatrix}\bigr)$), but I cant prove $$\ \alpha g h\overset{def}{=}\pi(gh,\alpha)=\pi(g,\pi(h,\alpha))\overset{def}{=}\alpha hg \ $$

Answer 1

I think your notation is giving you some roadblocks here.

A (right) group action of a group $G$ on a set $\Omega$ is a map $G\times \Omega\to \Omega$ satisfying:

• $x^e = x$ for any $x\in \Omega$
• $(x^g)^h = x^{gh}$

where the image of $(g,x)$ is denoted as $x^g$.

In this context, our action is defined as

$$\alpha^g := \pi(g,\alpha) := \alpha g.$$

I’m going to abandon the notation relying on $\pi$. If we want to prove that this is truly an action, then

$$(\alpha^g)^h = (\alpha g)^h = (\alpha g)h = \alpha (gh) = \alpha^{gh}.$$

Your mistake is that you stated that $\pi(gh,\alpha) = \pi(g, \pi(h,\alpha))$, or that $\alpha^{gh} = (\alpha^h)^g$, or that $\alpha(gh) = (\alpha h)g$. This doesn’t follow immediately from definitions — if it were true it would require some additional convincing.