Centralizer of element in $SO(3)$ I'm asked to find an element $g$ in $SO(3)$ such that the centralizer of $g$, $Z(g)$, is a disconnected subgroup of $SO(3)$, whose identity component is isomorphic to $SO(2)$ (maximal torus). I attempt to transfer this question to a geometry question. The fact that $g$ commutes with $SO(2)$ means $g$ must be a 2D rotation around certain axis, but then the centralizer will just be $SO(2)$, which is connected. So I actually think such element does not exist. Am I missing something?
Define:
$$g := \begin{pmatrix}
1 &amp; 0 &amp; 0\\
0 &amp; -1 &amp; 0\\
0 &amp; 0 &amp; -1
\end{pmatrix}\in SO(3)$$
The eigenspace of $1$ is one-dimensional and generated by $e_1$. If $h\in SO(3)$ commutes with $g$, then:
$$ghe_1=hge_1=he_1$$
Since $\|he_1\|=\|e_1\|=1$, this implies $he_1=\pm e_1$. Therefore, $h$ is of one of the following forms for an appropriate $\phi\in\mathbb R$:
$$h = \begin{pmatrix}
1 &amp; 0 &amp; 0\\
0 &amp; \cos\phi &amp; -\sin\phi\\
0 &amp; \sin\phi &amp; \cos\phi
\end{pmatrix}\text{ or }
h = \begin{pmatrix}
-1 &amp; 0 &amp; 0\\
0 &amp; \cos\phi &amp; \sin\phi\\
0 &amp; \sin\phi &amp; -\cos\phi
\end{pmatrix}$$
Clearly every matrix of one of those forms is an element of $SO(3)$ and commutes with $g$. The connected components of the centraliser of $g$ correspond to cases $he_1=+e_1$ and $he_1=-e_1$ respectively.
You are right. Moreover, in any connected compact Lie group $G$ the following statement is true. Let $S$ be a connected Abelian subgroup of $G$, and let $g\in G$ commute with all elements of $S$. Then there is a torus $T$ containing $g$ and $S$. This is Proposition 4.25 from Adams' book 'Lectures on Lie groups'.
This is in case you are familiar with Lie group theory.