Centralizer of element in $SO(3)$

Question

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?

Answer 1

Define: $$g := \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{pmatrix}\in SO(3)$$ 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 $\phi\in\mathbb R$: $$h = \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\phi & -\sin\phi\\ 0 & \sin\phi & \cos\phi \end{pmatrix}\text{ or } h = \begin{pmatrix} -1 & 0 & 0\\ 0 & \cos\phi & \sin\phi\\ 0 & \sin\phi & -\cos\phi \end{pmatrix}$$ 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.