# Proving action on group and set

Assume $G$ is a group and $X$ is a set, and $\phi$ : G × X → X is an action.

Let $R$ = {f : X → $\mathbb R$} be the set of all functions from X to $\mathbb R$.

Define $$m : G × R → R$$ be the function $$m(g, f) = f ◦ \phi_{g^{−1}}$$

How do I even go about proving m is an action on G and R?

I know to have an action, you need a group and a set. It also needs to satisfy two axioms:

$$e.x = x$$
$$(gh).x = g.(h.x)$$

I guess my difficulties lie in different functions and how they help me prove m as an action.

In order to describe the action of $G$ on $R$, you have to specify what the function $(g.f):X\to\mathbb{R}$ is. To do this, you would evaluate at an element $x\in X$ and according to what was written above $$(g.f)(x)=f(g^{-1}.x).$$ Okay, so now you have to check that this definition satisfies the axioms:
1. $(1.f)=f$? Sure, $(1.f)(x)=f(1^{-1}.x)=f(1.x)=f(x)$.
2. $(g.(h.f))=((gh).f)$? Again, yes. You just need to check that $((g.(h.f))(x)=((gh).f)(x)$ for any $x\in X$ (spoiler below):
$$(g.(h.f))(x)=(h.f)(g^{-1}.x)=f(h^{-1}.(g^{-1}.x)=f((h^{-1}g^{-1}).x)=f((gh)^{-1}.x)=((gh).f)(x).$$
$m(e,f)=f \circ \phi_{e^{-1}} = f \circ \phi_{e}= f \circ 1_X=f$ where $\phi_e = 1_X$ because $\phi$ is action, thus $m(e,f)=f$. $m(gh,f)=f \circ \phi_{(gh)^{-1}} = f \circ \phi_{h^{-1}g^{-1}} = f \circ \phi_{h^{-1}}\circ \phi_{g^{-1}} = m(g,f \circ \phi_{h^{-1}}) = m(g,m(h,f))$ where $\phi_{h^{-1}g^{-1}} = \phi_{h^{-1}}\circ \phi_{g^{-1}}$ because $\phi$ is action, thus $m(gh,f)=m(g,m(h,f))$ and $m$ is action.
• Just check the domains of every function and evaluate them on elements of $X$ and use the fact that $\phi$ is action. It's just a rewriting exercise. – Gilberto López Apr 12 '17 at 18:20