Is the stabilizer of an element $\delta$ in the stabilizer of $\omega$ in G equal to the pointwise stabilizer of $\{ \delta, \omega \}$

Question

i.e., is $(G_{\delta})_{\omega} = G_{( \{\delta, \omega\} )}$?

I know that \begin{eqnarray*} (G_{\delta})_{\omega} &=& \{ \forall g \in G_{\delta} \,|\, \omega^g = \omega \} \\ &=& \{ \forall g \in G \,|\, \delta^g = \delta, \, \omega^g = \omega \} \,\text{(is this correct?)} \\ &=& G_{(\{ \delta, \omega \})} \end{eqnarray*}

Answer 1

I suppose that

• $G$ is acting on a set $\Omega$, to which $\delta$ and $\omega$ belong,
• $\delta \ne \omega$, and
• $G_{(\{\delta, \omega\})}$ denotes the stabilizer of $\{\delta, \omega\}$ in the action of $G$ on the subsets of $\Omega$.

Then, no, it is not true in general.

• Consider $G = S_{n}$ acting on $\Omega = \{1, 2, \dots, n \}$.
• Then $(G_{n-1})_{n}$ is $S_{n-2} \le S_{n}$.
• But $G_{(\{n-1, n \})}$ also contains the $2$-cycle $(n-1, n)$, and it is actually the direct product $S_{n-2} \times \langle (n-1, n) \rangle$.

For instance, when $n = 3$ you have $(G_{2})_{3} = \{ I \}$, where $I$ is the identity map, while $G_{(\{2, 3 \})} = \{ I, (23) \}$.