Let $X, Y$ be left invariant vector fields on a Lie group $G$, equipped with come arbitrary connection $\nabla$. Also, let $L_g : G \to G$ be left-multiplication by $g \in G$, so that for example $X \circ L_g (.) = X_{L_g(.)} : G \to TG$. I would assume that the following identity holds; $$\nabla_{(X \circ L_g)} (Y\circ L_g) = \nabla_X Y \circ L_g\hspace{2mm}.$$
It seems really simple to prove, but I'm stuck. Could someone give a hint?
I know that the value of $\nabla_X Y |_p$ is determined by the value of $X$ only at p and on Y on a neighbourhood of p. However I can't figure out how to handle the "$\circ L_g$" in $\nabla_X Y \circ L_g$.
Edit: the vector fields X and Y are assumed to be left-invariant.
You are trying to prove the impossible.
Counterexample: let $G=\mathbb{R}$, with global coordinate $t$. What you are asserting is that every connection has to be $$ \nabla_{\frac{\mathrm{d}}{\mathrm{d}t}}\frac{\mathrm{d}}{\mathrm{d}t}=c\frac{\mathrm{d}}{\mathrm{d}t} $$ where $c$ is constant. But this is not true, e.g., $\nabla_{\mathrm{d}/\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}t}=\sin(t)\frac{\mathrm{d}}{\mathrm{d}t}$ is a connection (and descends to a connection on $S^1$ if you want a compact example).