Differentiability of Matrix function satisfying exponential functional equation Let $\{U(t)\;|\;t \in \mathbb{R}\}$ be a subgroup of $\text{GL}_n(\mathbb{R})$ such that $t \mapsto U(t)$ is continuous and satisfies $U(0) = E_n$ and $$U(t+s) = U(s)U(t)$$

Is it true then that $t \mapsto U(t)$ is differentiable at $0$ (and
  hence for all $t \in\mathbb{R}$)?

If yes, I would appreciate a hint on how to prove this since it is clear by definition that $$U(h) - U(0) = o(1) \text{ for } h \to 0$$ but not $$U(h) - U(0) = o(||h||)\text{ for } h \to 0$$
 A: Yes. Since $I$ is non-singular, we may define in its small neighbourhood a matrix logarithm. It follows that $f(t) = \log U(t)$ is well defined in some neighbourhood of zero because $U$ is continuous and $U(0)=I$.
By definition, the subgroup $\{U(t): t\in\mathbb R\}$ is Abelian. Hence $\log\left(U(s)U(t)\right)=\log U(s)+\log U(t)$. Consequently, if we denote by $f_{ij}$ the $(i,j)$-th component of $f$, then we can reduce the original functional equation to
$$
f(s+t)=f(s)+f(t)
$$
or $f_{ij}(s+t)=f_{ij}(s)+f_{ij}(t)$ for each $(i,j)$. This is basically Cauchy's functional equation, with the exception that the domain of $f_{ij}$ is a neighbourhood of zero instead of the whole real line. However, the same solution method applies: using the additive property, one can show that there exists a real number $c_{ij}$ such that $f_{ij}(q) = c_{ij}q$ for every rational number $q$ inside the domain of $f_{ij}$. Then, using the continuity of $f_{ij}$ and the denseness of rational numbers in $\mathbb R$, we conclude that $f_{ij}(t)=c_{ij}t$.
Hence $f(t)=tC$ for some constant matrix $C$. That is, $U(t)=e^{tC}$ on some neighbourhood of zero. Thus $U$ is differentiable at zero.
A: Let $s(x) =\sum_k \frac{(-1)^k}{k} x^k $ then it is easy to see that $s\circ U $ is differentiable and thus so is $U.$
