Equivalent Categories kG-Mod & Rep G

Let $G$ be a group and $k$ a field. \medskip Step 1: From representations to $kG$-modules. Let $\rho : G \to \operatorname{GL}(V)$ be a representation of $G$ over $k$. We define a $kG$-module structure on $V$ as follows. Recall that $kG$ consists of finite sums $\sum_{g \in G} a_g g$ with $a_g \in k$. For $v \in V$, define $$ \left( \sum_{g \in G} a_g g \right)\cdot v \;:=\; \sum_{g \in G} a_g \, \rho(g)(v). $$ We verify this defines a $kG$-module structure. Linearity in $v$ and in the coefficients $a_g$ is immediate from linearity of each $\rho(g)$. For multiplicativity, let $$ x = \sum_g a_g g, \qquad y = \sum_h b_h h. $$ Then $$ xy = \sum_{g,h} a_g b_h \, gh. $$ Hence $$ (xy) \cdot v = \sum_{g,h} a_g b_h \, \rho(gh)(v) $$ $$ = \sum_{g,h} a_g b_h \, \rho(g)\rho(h)(v) $$ $$ = \sum_g a_g \rho(g)\!\left( \sum_h b_h \rho(h)(v) \right) \\ = x \cdot (y \cdot v). $$ Thus $V$ becomes a left $kG$-module. Step 2: From $kG$-modules to representations. Conversely, let $V$ be a $kG$-module. For $g \in G$, define $$ \rho(g) : V \to V \qquad by \qquad \rho(g)(v) := g \cdot v. $$ Since the $kG$-action is $k$-linear, each $\rho(g)$ is a $k$-linear map. We check that $\rho$ is a group homomorphism. For $g,h \in G$, $$ \rho(g)\rho(h)(v) = g \cdot (h \cdot v) = (gh) \cdot v = \rho(gh)(v), $$ where the middle equality uses associativity of the module action. Also, $$ \rho(1)(v) = 1 \cdot v = v. $$ Thus $\rho : G \to \operatorname{GL}(V)$ is a representation. Step 3: The constructions are inverse to each other. Starting with a representation $\rho$, constructing the associated $kG$-module, and then restricting the action back to $G$, clearly recovers the original $\rho$. Conversely, starting with a $kG$-module $V$, defining $\rho(g)=g\cdot$, and then extending linearly to $kG$, recovers the original module structure. Thus the two constructions are mutually inverse. Therefore there is a one-to-one correspondence between representations of $G$ over $k$ and left $kG$-modules.