Do $Ad_t(X)$ and $X$ commute?

Question

Let $G$ be a compact lie group. Let $t \in G$. Then is it true that $Ad_t(X)$ and $X$ commute?

In case the group $G$ is abelian $Ad_t(X)=X$. But how can we prove this if the group is non abelian? Any hints will be appreciated. Thank you.

Answer 1

I think this is simply wrong and you can already get a counter-example in $SU(2)$ and taking $X\in\mathfrak{su}(2)$ to be diagonal with entries $i$ and $-i$. Then any matrix commuting with $X$ has to preserve the eigenspaces of $X$ and thus be diagonal. But of course that are matrices unitarily conjugate to $X$ which are not diagonal, for example using $t=\begin{pmatrix} a & a\\ -a & a\end{pmatrix}$ where $a=1/\sqrt{2}$.