This is one of our homework problem.
Let $G$ be a Lie group be defined as a manifold with group structure such that the map $F:G\times G \mapsto G, F(a, b)=ab^{-1}$ is smooth. Show that $$dF_e(a,b)=a-b,$$ where $e$ is the identity.
I assume the general form of $dF_e(a,b)=Aa+bB$ where $A$ and $B$ are matrices. Then use composition of $F_e$ with other functions I show that $dF_e$ can only be $a-b$. But it doesn't seem to be the rigorous solution so I'm wondering what should be the most normal way of proving it.
One way to go about proving this identity is to prove the following : $$d_e\iota=-\mathrm{id}_{\mathfrak g}$$ where $\iota:G\to G,g\mapsto g^{-1}$ is the inverse, $\mathfrak g=T_eG$ is the tangent space at the identity of $G$, and $$d_{(e,e)}\mu(X,Y)=X+Y$$ for all $X, Y\in\mathfrak{g}$, where $\mu:G\times G\to G, (g,h)\mapsto gh$ is the product map, and compose the two.
In order to prove these two identities, just use the fact that the differential of a map $f:M\to N$ evaluated on a tangent vector $X\in T_mM$ equals the derivative of $f\circ c$ at $0$ for any smooth path $c$ with $c'(0)=X$. There are on a Lie group natural paths for the question at hand given by the exponential of said Lie group, or by integrating left (or right) invariant vector fields. These paths are one-parameter subgroups of $G$ (at least those that start at the identity of $G$) and thus behave well with respect to passing to inverses.
For instance with the inverse, observe that for all $X\in\mathfrak g$, $\iota(\exp_G(X))=\exp_G(-X)$. Now since $d_0\exp_G=\mathrm{id}_{\mathfrak g}$ it follows that $$d_e\iota=-\mathrm{id}_{\mathfrak g}$$
For the product map, use the natural (inverse) isomorphisms $$(d_{(m,n)}\pi_M,d_{(m,n)}\pi_N):T_{(m,n)} (M\times N)\stackrel{\sim}{\leftrightarrow} T_mM\times T_nN:(d_mi_n,d_ni_m)$$ where $\pi_M,\pi_N$ are the natural projections and $i_n: M\to M\times N$ and $i_m: N\to M\times N$ are the natural injections.
Since $i_e:G\to G\times G, g\mapsto (g,e)$ composed with $\mu$ gives $\mathrm{id}_G$ (and similarly for $j_e:G\to G\times G, g\mapsto (e,g)$) it follows that $$\mathfrak g\times \mathfrak g\xrightarrow{(d_ei_e,d_ej_e)}T_e(G\times G)\xrightarrow{d_e\mu}\mathfrak g$$ sends $(X,Y)$ to $X+Y$.
Now combine the two and you'll find (using the same isomorphism to identify $T_e(G\times G)$ with $\mathfrak g\times \mathfrak g$ as earlier) that the differential of $(g,h)\mapsto gh^{-1}$ at $(e,e)$ sends the pair $(X,Y)$ to $X-Y$.