Centralizers of connected linear group and its Lie algebra If we have that $G$ is a connected linear group and $H<G$, where $H$ is also connected, with $\mathfrak{h}$ the lie algebra of $H$ and we define the centralizers of the elements in the following way:
$Z(H):=\{a\in G| aha^{-1}=h\forall h\in H\}$ and $Z(\mathfrak{h})=\{a\in G|ad(a)Y=Y\forall Y\in \mathfrak{h}\}$
I have a question which is asking me to show that these two are the same but I am unsure how to go about doing this, is there some property of the adjoint that I am not seeing here that may be useful?
Thanks for any help

For these sorts of questions, you want to transition between Lie algebras and Lie groups using the exponential map.  The useful facts for this problem are:


*

*$\exp(Ad_a Y) = a (\exp Y) a^{-1}$. This follows from $Ad_a$ being the derivative of conjugation and the naturality of $\exp$.

*If $X \in \mathfrak g$ then $X \in \mathfrak h \subset\mathfrak g$ if and only if $\exp X \in H$.

*A connected Lie group is generated by the image of the exponential map.


So for this problem, suppose first $a \in Z(H)$ and let $Y \in \mathfrak h$.  Then by 1.,
$$
\exp(Ad_a Y) = a \exp(Y) a^{-1} \in H
$$
so that $Ad_a Y \in \mathfrak h$ by 2.  Thus $a \in Z(\mathfrak h)$.
Conversely, suppose $a \in Z(\mathfrak h)$ and let $h \in H$.  Then since $H$ is connected, by 3. we can write $h = \exp(Y_1) \cdots \exp(Y_n)$ for $Y_j \in \mathfrak h$, giving us
$$
a h a^{-1} = a\exp(Y_1) \cdots \exp(Y_n)a^{-1} = (a \exp(Y_1) a^{-1})(a \exp(Y_2) a^{-1}) \cdots (a \exp(Y_n) a^{-1})
$$
which lies in $H$ since each $a \exp(Y_j) a^{-1} \in H$.  Thus $a \in Z(H)$.
