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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 $$\ \alpha g h\overset{def}{=}\pi(gh,\alpha)=\pi(g,\pi(h,\alpha))\overset{def}{=}\alpha hg \ $$

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  • $\begingroup$ 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. $\endgroup$ – Santana Afton Jan 17 at 14:33
  • $\begingroup$ @SantanaAfton I don't understand. Don't I need to show $\alpha gh= \alpha hg$? Because $\pi(gh,\alpha)=\alpha(gh)=\alpha gh$ and $\pi(g,\pi(h,\alpha))=\pi(g,\alpha h)=(\alpha h)g=\alpha hg$. $\endgroup$ – J. Doe Jan 17 at 15:46
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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.

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  • $\begingroup$ 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 $\endgroup$ – J. Doe Jan 24 at 10:09
  • $\begingroup$ @J.Doe Why do we need to show $(\alpha)^{gh} = (\alpha^h)^g$? $\endgroup$ – Santana Afton Jan 24 at 13:55

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