Can the derived subgroup be realized as an intersection of stabilizers?

For any group $G$, we have that $Z(G)=\bigcap_{x\in G}C_G(x)$ and that each $C_G(x)$ is the stabilizer of $x$ when $G$ acts on itself by conjugation. Is there a similar representation for $G'$? That is

Can the derived subgroup be realized as an intersection of stabilizers for some action of $G$?

I know that $G'$ is the intersection of all normal subgroups $A_\alpha$ of $G$ such that $G/A_\alpha$ is abelian. Could we realize each of these $A_\alpha$ as the stabilizer of some element of some action?

• Every subgroup $H$ of $G$ is a stabilizer in the action of $G$ by multiplication on the cosets of $H$ in $G$. (Take right or left cosets according to whether you want a right or left action.) – Derek Holt Jul 21 '14 at 10:47
• @DerekHolt Well that does answer my question, but that is a kind of dull action. I know this is getting specific, but can we think of an action of $G$ such that the $A_\alpha$ are the stabilizers? – Robert Wolfe Jul 21 '14 at 10:56
• Also, every action of $G$ in which the subgroup $H$ is a stabilizer is equivalent to the coset action, so you are not going to find a different action with that property. – Derek Holt Jul 21 '14 at 11:37
• @DerekHolt Now I'm confused. A centralizer for some element of $G$ arises as a stabilizer when $G$ acts on itself by conjugation. And it arises as a stabilizer when $G$ acts on the cosets of the centralizer by left multiplication. These actions are not equivalent, are they? – Robert Wolfe Jul 21 '14 at 11:52
• Sorry, I did not state that result correctly. Two transitive actions are equivalent when they have the same stabilizers. So the action by left multiplication on the left ocsets of $C_G(x)$ in $G$ is equivalent to the action of $G$ by conjugation on the conjugacy class of $x$. – Derek Holt Jul 21 '14 at 12:39