Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. A linear representation $\Pi : G \to GL_n(\mathbb{C})$ of $G$ induces a linear representation of $\mathfrak{g}$, namely the differential $d \Pi$ at the identity element.
My question: Assuming that the representation $\Pi$ of $G$ is faithful, is the representation $d\Pi$ of $\mathfrak{g}$ faithful as well?
Yes, it is. Suppose otherwise. Then there is some $X\in\mathfrak{g}\setminus\{0\}$ such that $\mathrm d\Pi(X)=0$. But then$$(\forall t\in\mathbb R)(\forall v\in\mathbb R^n):\Pi\bigl(\exp(tX)\bigr).v=\exp\bigl(t\mathrm d\Pi(X)\bigr).v=v$$and so $\exp(tX)\in\ker\Pi$. But then, since $\Pi$ is faithful, $(\forall t\in\mathbb R):\exp(tX)=e$ and so$$X=\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}\exp(tX)=0.$$