Action of a group G on $\mathbb{C}$[G] makes $\mathbb{C}$[G] a G-module I am trying to show that the action of the group G on the vector space of functions $\mathbb{C}$[G] defined by [g.$\psi$] (h)=$\psi(g^{-1}h$) makes a G-module. My main problem is that I can't get any intuition behind this action and the space $\mathbb{C}[G]$. And in general when I show that something, say M, is a G-module there is a defined rule for multiplication of the elements of G with the elements of M. What is confusing me is whether the the group operation . that rule and whether the h is there just because we want to evaluate the function somewhere. 
Do I have to check that:
(1) [$g_1.g_2$ . $\psi$] (h)=$[g_1.(g_2.\psi)]$(h) where $g_1, g_2 \in G$
(2) $1_G . \psi (h) = \psi (h)$ 
(3) $ [g. (\alpha\phi +\beta\psi)](h)= \alpha[(g.\phi)](h) + \beta[(g.\psi)](h)$ for $\alpha, \beta \in \mathbb{C}, \psi, \phi \in \mathbb{C}[G], g \in G$
Thanks!
 A: Let's unpack what the formulation $[g\cdot \psi] (h)=\psi(g^{−1}h)$ really defines, and then go over what would establish this defines a G-module.
First, we presume $\mathbb{C}[G]$ to be the complex vector space of functions $\psi:G \to \mathbb{C}$, a vector space defined by the usual obvious rules for adding two such functions and for multiplying one of these by a complex scalar:
$$ (\alpha \phi + \beta \psi)(h) = \alpha \phi(h) + \beta \psi(h) $$
where $\alpha,\beta \in \mathbb{C}$ and $\phi,\psi \in \mathbb{C}[G]$.  Thus $\phi$ and $\psi$ may be applied to an element $h \in G$, producing a complex number as their result.  The formula above therefore describes a function in $\mathbb{C}[G]$ by defining its value as applied to $h$, a typical element.  This definition does not use any group properties of $G$, which might as well be simply a set as far as the vector space operations are concerned.
The group action of $G$ on $\mathbb{C}[G]$ does exploit the group structure of $G$.  That is, we want to tell what the image $[g\cdot \psi]$ should be in $\mathbb{C}[G]$, given $g \in G$ and $\psi \in \mathbb{C}[G]$.  Explaining this requires us to say what $g\cdot \psi$ does with an element $h \in G$, and that's the point of the equation:
$$ [g\cdot \psi] (h)=\psi(g^{−1}h) $$
Here the group element $g$ that is "acting" on $\psi$ does so by affecting the argument $h \in G$ being mapped.  In your words, "the $h$ is there just because we want to evaluate the function somewhere." It should now be clear that the recipe for this group action depends on the group operation of $G$ through the formation of term $g^{-1}h$.
To prove this is a (left) group action we need these two compatibility conditions, which do not involve the additive structure on $\mathbb{C}[G]$:
$$ (g_1 g_2)\cdot \psi = g_1 \cdot (g_2 \cdot \psi) $$
$$ 1_G \cdot \psi = \psi $$
A (left) G-module is constituted iff this (left) group action of $G$ on $\mathbb{C}[G]$ satisfies a compatibility condition with the additive structure of $\mathbb{C}[G]$:
$$ g\cdot(\phi + \psi) = (g\cdot \phi) + (g\cdot \psi) $$
for all $g \in G$ and all $\phi,\psi \in \mathbb{C}[G]$.  While there is an additional compatibility with the scalar multiplication on $\mathbb{C}[G]$, such that $g\cdot (\alpha \psi) = \alpha (g\cdot \psi)$, this is not required to demonstrate a G-module.
The Answer by @DonAntonio gives the essential details of verifying these properties.
A: $$\begin{align*}\bullet&\;\;(g_1g_2)\psi(h):=\psi\left((g_1g_2)^{-1}h\right)=\psi\left(g_2^{-1}g_1^{-1}h\right)=g_2\psi(g_1^{-1}h)=g_1(g_2\psi)(h)\\
\bullet&\;\;1_G\psi(h):=\psi(1_G^{-1}h)=\psi(h)\\
\bullet&\;\;g(\alpha\phi+\beta\psi)(h)=(\alpha\phi+\beta\psi)(g^{-1}h)=\alpha\phi(g^{-1}h)+\beta\phi(g^{-1}h) =\ldots etc\end{align*}$$
