Let $X$ be a left-invariant vector field. If $\exp(X)\in Z(G)$ then $X$ is right-invariant.

Question

Let $G$ be a Lie Group and $X$ a left-invariant vector field. Show that if $\exp(X)$ is in the center of $G$, i.e $\exp(X)\in Z(G)$, then $X$ is right-invariant.

I'm trying to use the fact that $X$ is a left-invariant and right-invariant vector field iff $\mathrm{Ad}(g)(X)=X$ for all $g\in G$. Using this, and the relation $C_g(\exp(X))=\exp(\mathrm{Ad}(g)(X))$, where $C_g(h)=ghg^{-1}$, we have: $$ \exp(X)=\exp(\mathrm{Ad}(g)(X)), \quad \forall g \in G. $$

But I don't know how to conclude that $\mathrm{Ad}(g)(X)=X$ for all $g\in G$.

Answer 1

Since $\mbox{exp}(X)$ is in the center, $C_h(\mbox{exp}(X))=\mbox{exp}(X)$ for all $h$ in $G$. Take the curve $\mbox{exp}(tX)$ (starting at the identity with velocity X) to compute the differetial $(dC_g)_1 X$.