Let G be a Lie group and $\mathfrak{g}=Lie(G)$ and $\mathfrak{h}\subset \mathfrak{g}$ a subalgebra.
I saw in some proof that if $t\in \Bbb R$ and $X\in \mathfrak{h}$ and $ad(tX)\subset \mathfrak {h}$ then $\exp(ad(tX))\subset \mathfrak {h}$
I don't see why it is the case, to my knowledge $exp\not \in \operatorname{End}(\mathfrak {h})$.
The proof was meant to show that $Ad(\exp(tX))\subset \mathfrak {h}$ using the identity $Ad\circ \exp = \exp \circ \,ad$.
Many thanks for any help/hint in the right direction.
Pardon the simplistic language, as I am a physicist, so I'm not sure the street version of the structure would satisfy you. The fundamental identity invoked is what many call Hadamard's lemma in deriving the CBH formula.
That is, for $t\in \Bbb R$ and for any $X,Y\in \mathfrak{h}$, and $ad(tX)\subset \mathfrak {h}$ you have
$$\exp(ad(tX)) Y =Y+ ad(tX)Y+ \frac{1}{2!} ad(tX)ad(tX) Y+...= Y + [X,Y]+ \frac{1}{2!}[X,[X,Y]]+ ... \in \mathfrak {h}$$ since all commutators of the Lie sub algebra elements are in it.
You apply the identity $Ad\circ \exp = \exp \circ \,ad$ on the left-hand side.
I am more used to starting with the Ad(exp(tX)) being in the subagebra and differentiating w.r.t. t to go backwards.