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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.