Let $G$ be a group and $\Omega$ be a nonempty set.   

Determine whether the following definition of $\omega \cdot g$ (for $\omega \in \Omega$ and $g \in G$) defines a group action.

$\Omega = \{H \mid H \leq G\}$, and $\omega \cdot g = C\_G(\omega)$

I have the following solution and am curious if anyone could explain it in further detail. Are we choosing an $\omega = H$ since $\Omega$ is the set of $H$ elements? Why does this break one of the conditions of $G$ acting on $\Omega$?

$\textbf{Definitions:}$ $G$ $\textbf{act}$ on $\Omega$ means $\Omega \neq \emptyset$. $G$ $\textbf{acts}$ on $\Omega$ means the following conditions hold:
(i) $\omega \cdot 1 = \omega$ $\forall \omega \in \Omega$
(ii) $(\omega \cdot g) \cdot h = \omega \cdot (gh)$ $\forall g,h \in G$ and $\omega \in \Omega$
where $\cdot$ is the binary operation on $\Omega$ and $G$ uses right multiplication.

If $G$ is non abelian, then for some $g \in G$, the set $C\_G(g)=\{x \in G \mid xg=gx\}$ of all elements that commute with g. We call $C\_G(g)$ the $\textbf{centralizer}$ of $g$ in $G$. Note: $C\_G(g) \leq G$.

The $\textbf{centralizer of a set}$ $X \leq G$, $C\_G(X) =\{y \in G \mid xy=yx \,\, \forall x \in X\}$. Thus, $C\_G(X)=\bigcap\limits\_{x \in X} C\_G(x)$. So $X=G \implies C\_G(G)$ is the $\textbf{center}$ of $G$ and denoted $Z(G)$.

$\textbf{Solution:}$ $H \cdot g = C\_G(\omega)$. So, $H=H \cdot 1 = C\_G(H)$.   

$\therefore$ This definition does not define a group action.