Exponentiation of Center of Lie Algebra Let $G$ be a Lie Group, and $g$ its Lie Algebra. Show that the subgroup generated by exponentiating the center of $g$ generates the connected component of $Z(G)$, the center of $G$. 
Source: Fulton-Harris, Exercise 9.1
The difficulty lies in showing that exponentiating the center of $g$ lands in the center of $G$. Since the image of $exp(Z(g))$ is the union of one parameter subgroups that are disjoint, we know it connected. Also I can show this for the case when $G = Aut(V) $ and $g = End(V)$ since we have $G \subset g $.
EDIT: G not connected

$G$ is connected and so is generated by elements of the form $\exp Y$ for $Y \in g$.  Therefore it is sufficient to show that for $X \in Z(g)$ $\exp X$ and $\exp Y$ commute.  Now define $\gamma: \mathbb R \to G$ by $\gamma(t) = \exp(X)\exp(tY)\exp(-X)$.  Then 
$$
\gamma'(0) = Ad_{\exp(X)} Y = e^{ad_X} Y = (1 + ad_X + \frac{1}{2}ad_X \circ ad_X + \cdots)(Y) = Y.
$$
But it is also easily seen that $\gamma$ is a homomorphism, i.e. $\gamma(t+s) = \gamma(t)\gamma(s)$.  This characterizes the exponential map so that $\gamma(t) = \exp(tY)$.  Taking $t=1$ shows that $\exp X$ and $\exp Y$ commute.
