A permutation group $P \leq S_n$ can be represented as a subgroup $H \leq GL(n,2)$ if a permutation $\sigma\in P$ acts in the indices of the vectors of $e_n$. Then every permutation of $P$ has a unique invertible matrix on $H$. We can represent it as a function $\varphi$ that relates both structures:
$$\varphi : P \to H \quad \varphi(\sigma) \mapsto [e_{\sigma(1)}, \cdots, e_{\sigma(n)}]$$
In the other hand, we take our group as the multiplicative group of integers modulo a prime $p$, $G=\mathbb{Z}_p^*$. We are interested on represent it as a permutation group, and later as a matrix group $H\leq GL(p-1,2)$. If we take the group action of $G$ on $X=\{1,\cdots,p-1\}$ to be:
$$\phi: G \times X \to X$$
$$\phi(g,x) = g\cdot x \mapsto gx \pmod p$$
Then a permutation $\sigma_g$ is composed of all the bijections given by $\phi(g,x)$ for each $x\in X$. We can obtain it's cycle decomposition as follows:
$$\sigma_g = (\phi(g,1), \cdots, \phi^{p-1}(g,1))$$
As we are interested on representing $G$ in $GL(p-1,2)$, the final step is to represent $\sigma_g$ as an invertible matrix:
$$\varphi(\sigma_g) \mapsto [e_{\sigma_g(1)}, \cdots, e_{\sigma_g(p-1)}]$$
As I noticed, we just have obtained the left representation of $G$ this is because the action of mapping every element of $g\in G$ as a matrix in $GL(V)$ is the same as mapping each element of $G$ in $S_6$, then map each resulting element in $GL(V)$.
$$L_p: G \to GL(V) \quad L_p(g) \mapsto [e_{g\cdot 1}, \cdots, e_{g\cdot (p-1)}]$$
So the relationship can be stated as
$$L_p(g) = \varphi(\sigma_g)$$
Would be my proof alright if we assume that every permutation group is isomorphic to a subgroup of $GL(V)$?
Am I right if I state that the representation matrix of $g$ is the same as the representation matrix of $\sigma_g$ (the permutation representation of $g$)?