Let $G$ be a lie group. Let $\pi$ be a continuous representation of $G$ in $V$ a finite dimensional complex linear space. Let $W$ be a linear subspace of $V$. The following is lie subgroup $H:=\{x\in G: \pi(x)W \subset W\}$. I want to show that it's lie algebra $\mathfrak{h}$ is given by $\{X\in G: \pi_*(X)W \subset W\}$, where $\pi_*$ is induced representation of the lie algebra $\mathfrak{g}$.
By the definition of a lie algebra of a lie subgroup, $\mathfrak{h}=\left\{X\in\mathfrak{g}\,\middle|\,(\forall t\in\mathbb{R}):\exp(tX)\in H\right\}$. Now if $\exp(tX)\in H$ then $\pi(\exp(tX))W\subset W$. So we need to show that $\pi_*(X)W\subset W$ if and only if $\pi(\exp(tX))W\subset W$. This almost seems like it should follow from the definition of $\pi_*$ but I cannot see why this is the case?
Given $w\in W$, $c(t):=\pi(\exp(tX))(w)$ is a smooth curve in $V$ that by assumption has values in $W$. Its derivative at $t=0$, which then also lies in $W$, by definition is $\pi_*(X)(w)$. Since this holds for any $w\in W$, you see that $\pi_*(X)W\subset W$. For the converse, you need the fact that $\pi(\exp(tX))=e^{t\pi_*(X)}=\sum_{k\in\mathbb N}t^k\pi_*(X)^k$ (choosing as basis, you can interpret the right hand side as a matrix exponential). Assuming that $\pi_*(X)W\subset W$, you see that for any $N\in\Bbb N$ and $w\in W$, the sum $\sum_{k=0}^Nt^k\pi_*(X)^k(w)$ lies in $W$. Since linear subspaces in finite dimensional vector spaces are automatically closed, the limit for $N\to\infty$ lies in $W$, too.