Show that a group homomorphism $\phi : G\to GL_n(F),$ where $G$ is a group and $GL_n(F)$ is the set of $n\times n$ invertible matrices with entries in $F$, extends to a ring homomorphism $\psi : F[G]\to M_n(F)$. Here, $F[G]$ is the group algebra consisting of elements of the form $\sum_{g\in G}\beta_g g , \beta_g \in F, \beta_g \neq 0$ for only finitely many $g$, equipped with the natural addition and multiplication operations and with identity element $1_F\cdot 1_G.$ Also, $M_n(F)$ is the set of $n\times n$ matrices with entries in $F$.
Show that if $V$ is the set of $n\times 1$ column vectors with entries in $F,V$ is a left $F[G]$-module.
I think I can extend $\phi$ by setting $\psi (\sum_{g\in G} \beta_g g)=\sum_{g\in G}\beta_g \phi(g).$ It's clearly well-defined and maps $1_F\cdot 1_G$ to the identity matrix as $\phi$ is a homomorphism. Also, $\psi ((\sum_{g\in G} \beta_g g) \cdot (\sum_{h\in G} \alpha_h h))=\psi(\sum_{g,h\in G} \beta_g\alpha_h (gh)) = \sum_{g,h\in G}\beta_g \alpha_h\phi(g)\phi(h) = \psi(\sum_{g\in G} \beta_g g)\psi (\sum_{h\in G} \alpha_h h))$ and $\psi ((\sum_{g\in G} \beta_g g) + (\sum_{h\in G} \alpha_h h))=\psi(\sum_{g\in G} (\beta_g+\alpha_g) g) = \sum_{g\in G}(\beta_g +\alpha_g)\phi(g) = \psi(\sum_{g\in G} \beta_g g)+\psi (\sum_{h\in G} \alpha_h h))$.
I know that a left $F[G]$-module $M$ is an abelian group with a map $\cdot : F[G]\times M\to M$ so that $(rs)\cdot m = r\cdot (s\cdot m), r\cdot (m+n) = r\cdot m + r\cdot n, (r+s)\cdot m = r\cdot m+s\cdot m$ and $1\cdot m=m$ for $r,s \in F[G]$ and $m,n \in M.$ $V$ is an abelian group under $+$ and I think the map $\cdot : F[G]\times M\to M, \cdot(f, m) = \psi(f)m$ makes $V$ into an $F[G]$-module, and this is verified using the fact that $\psi$ is a ring homomorphism.