Let $K$ be a field and $G$ be a group. Consider the group algebra $KG = \{\sum_{g}'\alpha_gu_g: \alpha _g \in K\}$ where $\{u_g: g \in G\}$ are a $K$-basis. Consider the following operations:
$$f: \{K-\mathrm{representations \ of \ G}\}\longrightarrow\{\mathrm{left\ KG-modules}\}$$ $$(V, \rho) \mapsto V$$ where $V$ has the 'scalar' multiplication $$(\sum_g \alpha_g u_g).v = \sum_g \alpha_g (\rho(g)(v))$$
$$g: \{\mathrm{left\ KG-modules}\}\longrightarrow \{K-\mathrm{representations \ of \ G}\}$$ $$M \mapsto (M, \rho)$$ where $\rho(g)v = u_g.v$ where the $.$ denotes multiplication in the $KG$-module $M$
In the book I'm reading, it is claimed that $f$ and $g$ are mutual inverses. However, I struggle to show this. In particular, I'm stuck at showing that $f \circ g$ is the identity.
Attempt: Start with a left $KG$-module $M$ with scalar multiplication $\cdot$
Then we can associate to this multiplication $\rho(g)(v) = u_g.v$
We want to show that the associated module structure with $(M, \rho)$ coincides with the scalar multiplication $.$, i.e. the we started with. That is, we have to show
$$(\sum_{g} \alpha_g u_g).v = \sum_{g}\alpha_g (\rho(g)v)$$
Expanding the right side, $$\sum_{g}\alpha_g (\rho(g)v)= \sum_g\alpha_g(u_g.v)$$ Expanding the left side, $$(\sum_{g} \alpha_g u_g).v= \sum_g (\alpha_g u_g).v$$ so it suffices to show $$\sum_{g} \alpha_g (u_g.v) = \sum_{g}(\alpha_g u_g).v \quad \quad (*)$$
How can I show $(*)$? Or what am I missing? Maybe I defined the scalar multiplication wrong?
This is just the associativity of scalar multiplication. More precisely, the scalar multiplication by $a_g$ in $V$ is, by definition, the scalar multiplication by $a_gu_e$ (where $e$ is the identity element of $G$) in $M$, so, by associativity in $M$, $a_g(u_g.v)=(a_gu_e).(u_g.v)=(a_gu_e\cdot u_g).v=(a_gu_g).v$, as desired.