If an element $a$ in a Lie group commutes with another one, does this element also commutes with a one-parameter subgroup?

Question

Let $G$ be a Lie group and $b=exp(X) \in G $, where $X \in Lie(G)$. Suppose that there exists an element $a \in G$ such that $a$ and $b$ commutes.

Now I'm wondering if it is true that $a$ commutes with every element of the one-parameter subgroup $ \{exp(tX) \vert t \in \mathbb{R} \}$. If so, any hints for the proof would be much appreciated !

Answer 1

No, take $Gl(n,R)$ and $X$ such that there exists $t$ with $exp(tX)=Id$, and $X$ is not zero, ($X$ generates a subgroup isomorphic to $S^1$) for example $n=2, X=\pmatrix{0&-1\cr 1&0}$.

Answer 2

No, it is not true. Take $G=GL(2,\mathbb R)$ and $X=\left[\begin{smallmatrix}0&-2\pi\\2\pi&0\end{smallmatrix}\right]$. Then $\exp(X)=\operatorname{Id}$ and so every element of $GL(2,\mathbb R)$ commutes with $\exp(X)$. But not all elements of $GL(2,\mathbb R)$ commute with every element of $G$ of the form $\exp(tX)$, of course.

Answer 3

No.

For example, take the unit quarternions $S^3$, and let $X=\pi i\in T_1S^3=\operatorname{Im}\mathbb{H}$. Then the element $j\in S^3$ commutes with $\exp(X)=\exp(\pi i)=-1$ obviously, but it doesn't commute with $\exp(tX)$ if $t\notin\mathbb{Z}$.