$\tau\in N_{S_A}(H)$ then $\tau$ stabilizes both $F(H)$ and $A-F(H)$

Question

(I have seen this question:Prove that if $\tau \in N_{S_A}(H)$ then $\tau$ stabilizes the sets $F(H)$ and $A \backslash F(H)$, but the notation in both the question and the comment was unfamiliar to me, so I got nowhere.)

This is Exercise 18 from Dummit&Foote Chapter 4.3, and is connected to the previous exercise.

Let $A$ be a set, let $H$ be a subgroup of $S_A$ and let $F(H)$ be the fixed points of $H$ on $A$, i.e $$F(H)=\{a\in A \mid \sigma(a)=a \,\text{for all}\, \sigma\in H\}$$ Prove that if $\tau\in N_{S_A}(H)$, then $\tau$ stabilizes the set $F(H)$, and $M(H):=A-F(H)$.

What I know: I have shown in the previous exercise that

• $F(\{\phi, \psi\})=F(\phi) \cap F(\psi)$ $\cdots$ (1),
• $M(\{\phi, \psi\})=M(\phi) \cup M(\psi)$ $\cdots$ (2),
• and for any $\tau\in S_A$ and $\sigma \in D:=\{\sigma \in S_A \mid |M(\sigma)|<\infty\}$, $F(\tau\sigma\tau^{-1})=\tau F(\sigma)$ and $M(\tau\sigma\tau^{-1})=\tau M(\sigma)$ $\cdots$ (3)
• $D$ is normal in $S_A$ $\cdots$ (4)

What I (think I have) proved: My "proof" goes as follows. Let $\tau\in N_{S_A}(H)$, i.e $\tau H \tau^{-1}=H$ where $\tau\in S_A$. Then, $$\tau(F(H))\overset{(1)}{=}\bigcap_{\sigma \in H}\tau F(\sigma)\overset{(3)}{=}\bigcap_{\sigma \in H}F(\tau\sigma \tau^{-1})\overset{(4)}{=}\bigcap_{\sigma' \in H}F(\sigma')\overset{(1)}{=}F(H)$$ and similarly for $M(H)$, $$\tau(M(H))\overset{(2)}{=}\bigcup_{\sigma \in H}\tau M(\sigma)\overset{(3)}{=}\bigcup_{\sigma \in H}M(\tau\sigma \tau^{-1})\overset{(4)}{=}\bigcup_{\sigma' \in H}M(\sigma')\overset{(2)}{=}M(H)$$ so we see that $\sigma$ stabilizes both $F(H)$ and $M(H)$.

I want to know if my proof is valid. Also, I still really couldn’t understand the concepts of fixed points and the meaning of the group $D$. I just used the definitions of each objects so I’m not convinced to my proof, and I’m wondering if there is a better proof using some insight.

Answer 1

Your proof seems unnecessarily complicated. Let $a \in F(H)$. We have to show that $\tau(a) \in F(H)$. So, for $\sigma \in H$, we have to show that $\sigma(\tau(a)) = \tau(a)$.

But $\tau \in N_{S_A}(H)$, so $\sigma\tau =\tau \sigma'$ for some $\sigma' \in H$.

Then, since $\sigma'(a)=a$, we have $\sigma(\tau(a)) = \tau(\sigma'(a)) = \tau(a)$, as required.

To show that $\tau$ fixes $A \setminus F(H)$, note that, if $b \in A \setminus F(H)$ and $\tau(b) = a \in F(H)$, then $\tau^{-1}(a) \not\in F(H)$. But $\tau^{-1} \in N_{S_A}(H)$, so this contradicts the first part.