TITLE: Semigroup implies exponential
QUESTION [5 upvotes]: Let $S :[0,1]\to \mathbb R^{n\times n}$ be a continuous function which satisfies
$$
S(0)=I \quad\text{and}\quad S(s+t)=S(s)S(t),
$$
for all $s,t\in[0,1]$ with $s+t\in[0,1]$. (Here $I$ is the identity matrix in $\mathbb R^{n\times n}$).
Show that there exists a matrix $A\in\mathbb R^{n\times n}$, such that $S(t)=\mathrm{e}^{tA}$.
Comment. Once we show that $S$ is differentiable at $0$, then setting $A=S'(0)$, it is not hard to prove the above, as we would then have $S'(t)=AS(t),\,\,S(0)=I$, which is a system of ODEs with unique solution the exponential.

REPLY [4 votes]: For $0<t<1$, define the integrated matrix   $H(t)x= \int_0^t S(w)x\,dw$.
Note that ${1\over t}H(t) x\to x$ as $t\to 0$, so for $t$ small enough, $H(t)$ is invertible. 
Now fix such a $t$.
Your semigroup is differentiable from the right on $H(t)(\mathbb{R}^n)=\mathbb{R}^n$ 
since direct calculations give
$${S(s)(H(t)x)-H(t)x\over s}\to (S(t)-I)x,$$ as $s\downarrow 0$. The required matrix is 
$A=(S(t)-I)H(t)^{-1}$.