If $\Phi(0)=I$ (identity) and that $\Phi(t+s)=\Phi(t)\Phi(s)$ for all $t,s\in\mathbb{R}$ then $\Phi(t)=e^{tA}$ I'm trying to solve the next problem on linear differential equations. 

Let $\Phi(t)$ be a $n\times n$ matrix of functions of class $C^1$. Suppose that $\Phi(0)=I$ (identity) and that $\Phi(t+s)=\Phi(t)\Phi(s)$ for all $t,s\in\mathbb{R}$. Prove that there exists only one matrix $A$ such that $\Phi(t)=e^{tA}$.

I'm not sure where to start from, so i'd appreciate some hints. Thanks in advance.
 A: Sorry this isn't much in the way of a hint (I'm not too good at hints), but:
We first note that
$\Phi(s) \Phi(t) = \Phi(t) \Phi(s) \tag{1}$
for all $s, t \in \Bbb R$, the real numbers.  This binds since
$\Phi(s) \Phi(t) = \Phi(s + t)$
$= \Phi(t + s) = \Phi(t) \Phi(s). \tag{2}$
We next compute $\Phi'(t)$ for any $t \in \Bbb R$:
$$\begin{align}\Phi'(t) &= \lim_{h \to 0} \dfrac{\Phi(t + h) - \Phi(t)}{h}\\
&=\lim_{h \to 0} \dfrac{\Phi(t) \Phi(h) - \Phi(t)}{h}\\
&= \lim_{h \to 0} \dfrac{\Phi(h) \Phi(t) - \Phi(t)}{h}\\ &= \lim_{h \to 0} \dfrac{\Phi(h) - I}{h} \Phi(t)\\
&=  \lim_{h \to 0} \dfrac{\Phi(h) - \Phi(0)}{h} \Phi(t), \tag{3}\end{align}$$
since $\Phi(0) = I$; we have used (1) in deriving (3).  We observe that, since $\Phi(t)$ is of class $C^1$, the rightmost limit occurring in (3) exists and is $\Phi'(0)$; we set
$A = \Phi'(0); \tag{4}$
$A$ is an $n \times n$ matrix; then we have established that
$\Phi'(t) = A \Phi(t) \tag{5}$
for all $t \in \Bbb R$.  Consider the $C^1$ matrix function
$B(t) = e^{-At} \Phi(t); \tag{6}$
we have
$B(0) = e^0 \Phi(0) = I \tag{7}$
and
$B'(t) = (e^{-At} \Phi(t))' = (e^{-At})' \Phi(t) + e^{-At} \Phi'(t)$
$= -Ae^{-At} \Phi(t) + e^{-At} A \Phi(t) = A(-e^{-At} \Phi(t) + e^{-At} \Phi(t)) = 0, \tag{8}$
whence $B(t)$ is a constant matrix; by (7), we have
$B(t) = I; \tag{9}$
thus
$e^{-At} \Phi(t) = I, \tag{10}$
implying that
$\Phi(t) = e^{At} \tag{11}$
for all $t$.  The uniqueness of such $A$ follows from (3); or, asuuming that
$\Phi(t) = e^{A_1 t} \tag{12}$
for some other matrix $A_1$, taking derivatives shows that
$A_1 e^{A_1 t} = \Phi'(t) = A e^{At}; \tag{13}$
setting $t = 0$ yields
$A_1 = A, \tag{14}$
and the uniqueness of $A$ is thus established.  QED!
