I know that for a compact Lie group, its Lie algebra is perpendicular (with respect to the killing form) to the center of said Lie algebra, so that $T_{e}(G)^{\perp} = \ker \text{ad}$. Also, since G is a complex Lie group, I know $T_{e}(G)$ is a $\mathbb{C}$-Lie subalgebra of $\mathfrak{gl}(n, \mathbb{C})$.
My guess is that I need to prove that for $X,Y \in T_{e}(G)$, if $\text{ad}(X) = 0$, and $B(X,Y) = 0$, then I must have that $Y=0$. I can then conclude that $T_{e}(G) \subset T_{e}(G)^{\perp \perp} = \{0\}$ and so $G$ must be finite, but I'm not sure how I get this from the fact that $T_{e}(G)$ is a $\mathbb{C}$-Lie subalgebra.
Am I missing something simple?