If $H$ is a closed subgroup of $G$, is the Lie algebra of $H$ contained in the Lie algebra of $G$?

Question

Let $H \subseteq G$ be connected linear algebraic groups with $H$ closed in $G$, and let $e$ be the identity of $G$. The Lie algebra of $G$ is the tangent space $T_eG$ of $G$ at $e$, which we can identify with the vector space of $k$-derivations of $k[G]$ into $k_e$ (here $k_e = k$ with the $k[G]$-module structure $f \cdot a = f(e)a$). The inclusion morphism $\phi: H \rightarrow G$ induces a $k$-linear transformation $$T_e H \rightarrow T_e G$$ given by $\delta \mapsto \delta \circ \phi^{\ast}$. My question is, is this transformation always injective? In the examples that I know, it is. For example, if $G = \textrm{GL}_n, H = \textrm{SL}_n$, $T_e G$ can be identified with $n$-by-$n$ matrices, and $T_e H$ can be identified with trace zero the $k$-vector space of $n$ by $n$ matrices. But it does not seem immediate from the definition that $\delta \mapsto \delta \circ \phi^{\ast}$ is injective.

Answer 1

Note that $H\subseteq G$ is a closed immersion of smooth varieties. In particular, it is unramified so for example this Lemma in the Stacks project states that the corresponding map on tangent spaces is injective.

Answer 2

Answer: Let $\phi: H:=Spec(B) \rightarrow G:=Spec(A)$ be a closed subgroup scheme with corresponding surjective map $f: A \rightarrow B$ of rings. Assume $G,H$ are of finite type over a field $k$. Let $\mathfrak{m}_e \subseteq B$ be the ideal of the identity element in $H$ and let $\mathfrak{n}_e:=\phi^{-1}(\mathfrak{m}_e)$ be the ideal of the identity in $G$. There is a canonical surjective map

$$\mathfrak{n}_e/\mathfrak{n}_e^2 \rightarrow \mathfrak{m}_e/\mathfrak{m}_e^2.$$

When you dualize you get an injection

$$ Lie(H):=T_e(H) \rightarrow T_e(G):=Lie(G).$$