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.
Note that for a group $G$ and a set $A$, a group action is a function $\cdot: G \times A \to A$ (or $A \times G \to A$ such that it satisfies
1) $e \cdot a = a, \forall a \in A$
2) $g_1 \cdot (g_2 \cdot a) = (g_1g_2)\cdot a, \forall a \in A, g_1,g_2 \in G$
Now, in your example, we have a group $G$ and set $\Omega$, which is the set of all subgroups of $G$. And notice that 1) must hold for all elements in the set, on which the group acts. So, can we find a group where centralizer of every subgroup $H \le G$ is $H$, that is, $C_G(H) = H, \forall H \le G$?
If $G$ is trivial group, then there won't be any problem and we will indeed have a group action. But for the non-trivial case, one can easily show that the centralizer of trivial subgroup $\{e\}$ will be $G$ itself so we have $\{e\} \cdot e = C_G(\{e\}) = G \ne \{e\}$ since $G$ is non-trivial.