Exercise 4.3.18 (If $ au \in N_{S_A}(H)$, then $ au$ stabilizes the sets $F(H)$ and $M(H) = A - F(H)$)

Question

Let $A$ be a set, let $H$ be a subgroup of $S_A$ and let $F(H)$ be the fixed points of $H$ on $A$ as defined in the preceding exercise. Prove that if $	au \in N_{S_A}(H)$ then $	au$ stabilizes the set $F(H)$ and its complement $A - F(H)$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $b \in 	au(F(H))$. We want to prove $b \in F(H)$.
Consider any $\lambda \in H$. Since $	au \in N_{S_A}(H)$, then $	au^{-1} \in N_{S_A}(H)$, thus $\sigma = 	au^{-1} \lambda 	au \in H$, so $\lambda = 	au \sigma 	au^{-1}$ for some $\sigma \in H$.

Since $b \in 	au(F(H))$, $b = 	au(a)$ for some $a \in F(H)$. By definition of $F(H)$, $\sigma(a) = a$ (because $\sigma \in H$), thus $\sigma 	au^{-1}(b) = 	au^{-1}(b)$, so $(	au \sigma 	au^{-1})(b) = b$, which gives $\lambda(b) = b$. Since this is true for all $\lambda \in H$, we obtain $b \in F(H)$. We have proved
$$	au(F(H)) \subseteq F(H).$$

Conversely, suppose that $a \in F(H)$. Put $c = 	au^{-1}(a) \in A$, so that $a = 	au(c)$.

Let $\sigma$ be any element in $H$. Since $	au \in N_{S_A}(H)$, then $	au \sigma 	au^{-1} \in H$, where $a \in F(F)$, thus $(	au \sigma 	au^{-1})(a) =a$, therefore $\sigma(c) = c$. Since this is true for every $\sigma \in H$, $c \in F(H)$, where $a 	au(c)$, so $a \in 	au(H)$. This proves $F(H) \subseteq 	au(F(H))$, and so
\begin{align}
	au(F(H)) = F(H).
\end{align}
Since $	au$ is bijective, the complement $A - 	au(F(H))$ of $	au(F(H))$ in $A$ is $	au(A - F(H))$. So, if we take the complementary sets in (1), we obtain
\begin{align}
	au(A -F(H))= A -F(H),
\end{align}
or equivalently,
\begin{align*}
	au(M(H)) = M(H).
\end{align*}
So, if $	au \in N_{S_A}(H)$, then $	au$ stabilizes the sets $F(H)$ and $M(H) = A - F(H)$.

\end{proof}

Exercise 4.3.18 (If $\tau \in N_{S_A}(H)$, then $\tau$ stabilizes the sets $F(H)$ and $M(H) = A - F(H)$)

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Exercise 4.3.18 (If $\tau \in N_{S_A}(H)$, then $\tau$ stabilizes the sets $F(H)$ and $M(H) = A - F(H)$)

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