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

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 \$$

• What makes you think that $\alpha gh = \alpha hg$? For a group action, you want to prove that $\alpha(gh) = (\alpha g)h$, along with what you’ve proven already. – Santana Afton Jan 17 at 14:33
• @SantanaAfton I don't understand. Don't I need to show $\alpha gh= \alpha hg$? Because $\pi(gh,\alpha)=\alpha(gh)=\alpha gh$ and $\pi(g,\pi(h,\alpha))=\pi(g,\alpha h)=(\alpha h)g=\alpha hg$. – J. Doe Jan 17 at 15:46

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.

• You showed $(\alpha^g)^h=\alpha^{gh}$ i.e. $\pi(\textbf{h},\pi(\textbf{g},\alpha))=\pi(gh,\alpha)$ but we need to show $\pi(\textbf{g},\pi(\textbf{h},\alpha))=\pi(gh,alpha)$ @santana – J. Doe Jan 24 at 10:09
• @J.Doe Why do we need to show $(\alpha)^{gh} = (\alpha^h)^g$? – Santana Afton Jan 24 at 13:55