Exercise
I've been given the task to show, given a flat $G$-structure, we have that $\text{Hol}=\text{Id}$ (here "Hol" is the holonomy group; furthermore a flat $G$-structure is defined be one such that at every $x\in M$ there exist local coordinates $x^1,\ldots ,x^n$ in a neighborhood of $x$ such that $(dx^1,\ldots ,dx^n)$ is local section of $\mathcal{F}_G$). It is presented as a Corollary to the following
Let $G\subseteq\text{GL}(V)$, let $\mathcal{F}_{G}\to M$ be a $G$-structure, and let $W\subseteq V^{\otimes k}\otimes V^{*\otimes l}$ be a $G$-submodule. Let $E_\rho\to M$ be the vector bundle induced by $\rho : G\to\text{GL}(W)$. If there exists $\phi\in\Gamma(E_\rho)$ such that $\nabla\phi=0$, then $$ \text{Hol}^{\theta}_{x}\subseteq G_{\phi,x}:=\{g\in\text{GL}(T_xM)\,|\,\rho(g)\phi_x=\phi_x\} $$
Thoughts
My first attempt would be to produce some $\phi\in\Gamma(E_\rho)$ such that $\nabla\phi=0$ and $\rho(g)\phi_x\neq\phi_x$ for all $\text{Id}\neq g\in G$, so I proceed naively:
If $\mathcal{F}_G\to M$ is flat, then there exists local coordinates $(x^1,\ldots ,x^n)$ about $x$ such that $(dx^1,\ldots ,dx^n)$ is a local section of $\mathcal{F}_G$. Define $$ \phi:=dx^1\otimes\cdots\otimes dx^n. $$ It follows that $\nabla\phi=0$ and so the quoted proposition applies. Furthermore, it doesn't seem hard to verify that if $\rho(g)\phi_x=\phi_x$, then $g=\text{Id}$ (through homomorphism of representation).
Other Thoughts
I feel like this is correct, but at the same time I fear that I've fallen into the fallacy of forcing my argument to meet the conclusion instead of the other way around. Any feedback is greatly appreciated.
I might be using non-standard terminology, so if anything needs clarification, please let me know and I will do my best to resolve any issues. Thanks again.
Edit: thinking further about the problem I believe we are supposed to assume $M$ simply-connected for this to work.