Let $G$ be $l$-dimensional compact Lie group and consider any smooth $F : \mathbb{R}^n \to G$.

Then, the first-order derivative of $F$ at each $x \in \mathbb{R}^n$ can be regarded as a linear mapping $D_x F : \mathbb{R}^n \to \mathbb{R}^l$. Let us fix a matrix norm $\lVert \cdot \rVert$ and assume that \begin{equation} x \mapsto \lVert D_x F \rVert \text{ is polynomially bounded} \end{equation} We can extend the above notion of polynomial boundedness to higher order derivatives as well.

Now, let us define a set $\mathfrak{G}$ as the set of $F : \mathbb{R}^n \to G$ such that $F$ is smooth and all its derivatives are polynomially bounded in the above sense.

Now, my question is that

Is this $\mathfrak{G}$ a group under pointwise multiplication and inverse?

That is,

For any $F_1, F_2 \in \mathfrak{G}$ and $x \in \mathbb{R}^n$, let $(F_1 F_2)(x) := F_1(x) F_2(x)$. Then, do we have $F_1 F_2 \in \mathfrak{G}$?

For any $F \in \mathfrak{G}$ and $x \in \mathbb{R}^n$, let $F^{-1}(x) := [F(x)]^{-1}$. Then, do we have $F^{-1} \in \mathfrak{G}$.

I think the item 1 must be true, but not sure about item 2. For example, can a derivative of $F$ decay exponentially at infinity, making a derivative of $F^{-1}$ "grow exponentially" at infinity?

Could anyone please clarify for me? If I am wrong about $\mathfrak{G}$ being a group, what additional conditions should I impose to make it into a group?

Add) I think this is indeed the case for $\operatorname{SU}(N)$ with $N \geq 2$ or any direct product of them. Still, I haven't figured out for general $G$.

Add 2) This ME post shows that my guess is also true for $\operatorname{U}(1)$.