Lie algebra associated to Lie group of algebra automorphisms I'm working through Fulton and Harris's Representation Theory, and I'm stuck on Exercise 8.27.
I'm trying to show that if $A$ is an algebra and $G$ is the Lie group of algebra automorphisms (interpreted as a subgroup of $GL(A)$) then the Lie algebra associated to $G$ is the algebra of derivations on $A$ ($Der(A)$).
I've shown that if $\gamma : [0,1] \to G$ is some path (writing $\gamma(t)$ as $\gamma_t$) with $\gamma_0 = id$ and $\gamma'_0 = X$ then just by differentiating with respect to $t$
$$
\gamma_t(ab) = \gamma_t(a)\gamma_t(b)
$$
we get
$$
\gamma'_t(ab) = \gamma'_t(a)\gamma_t(b) + \gamma_t(a)\gamma'_t(b),
$$
which at $t=0$ gives
$$
X(ab) = X(a)b + aX(b),
$$
so $X$ is a derivation. But I can't show that any derivation must be a tangent to the identity, so that the associated Lie algebra to $G$ is actually $Der(A)$. 
Any help with this would be great, thanks!
 A: I guess this should work. Take $X \in \text{Der}(A)$ and consider $\gamma_t = \exp(tX)$. $\gamma$ is a curve on $GL(A)$ such that $\gamma_0=\text{Id}$ and $\gamma'_0=X$. We want to show that $\gamma$ is actually $G$-valued. Remark that $\gamma'_t=X \circ \gamma_t=\gamma_t \circ X$.
Let's fix $a,b \in A$ and consider $\phi : t \mapsto \gamma_t(ab)$ and $\psi : t \mapsto \gamma_t(a)\gamma_t(b)$.
Deriving $\phi$, we get :
$$\begin{array}{rcl}
\phi'(t) & = & \gamma'_t(ab)\\
         & = & X\circ\gamma_t(ab) \\
         & = & X \left[ \phi(t) \right]  
\end{array}$$
In the same way :
$$\begin{array}{rcl}
\psi'(t) & = & \gamma'_t(a)\gamma_t(b)+\gamma_t(a)\gamma'_t(b)\\
         & = & X\left[\gamma_t(a) \right] \gamma_t(b) + \gamma_t(a) X \left[ \gamma_t(b) \right] \\
         & = & X \left[ \gamma_t(a)\gamma_t(b) \right] \; \; \; \; \; \text{as} \; X \; \text{is a derivation}\\
         & = & X \left[ \psi(t) \right] 
\end{array}$$
Then, $\phi$ and $\psi$ satisfy the same ODE. Yet, $\phi(0)=\psi(0)=ab$, hence $\phi=\psi$.
Thus, for all $a,b \in A$, $\gamma_t(ab)=\gamma_t(a)\gamma_t(b)$, i.e. $\gamma_t$ is in $G$.
