## In a symmetric group, the normalizer of a subgroup stabilizes is associated sets of fixed and moved points

Let $A$ be a set, let $H \leq S_A$ be a subgroup and let $F(X)$ and $M(X)$ be the sets of fixed and moved points as defined in this previous exercise. Prove that if $\tau \in N_{S_A}(H)$ then $\tau$ stabilizes $F(H)$ and $M(H)$.

We begin with some lemmas.

Lemma 1: If $\varphi : A \rightarrow B$ is injective and $\{X_i\}_I$ is a collection of subsets of $A$, then $\varphi[\bigcap_I X_i] = \bigcap_i \varphi[X_i]$. Proof: $(\subseteq)$ If $y \in \varphi[\bigcap_I X_i]$, then $y = \varphi(x)$ for some $x \in \bigcap_I X_i$. Thus $y \in \varphi[X_i]$ for all $i \in I$, hence $y \in \bigcap_I \varphi[X_i]$. $(\supseteq)$ If $y \in \bigcap_I \varphi[X_i]$, then for every $i \in I$ there exists $x_i \in X_i$ such that $y = \varphi(x_i)$. Since $\varphi$ is injective, we in fact have $x_i = x_j = x$ for all $i,j \in I$. Hence $x \in \bigcap_I X_i$, so that $y \in \varphi[\bigcap_I X_i]$. $\square$

Lemma 2: If $\varphi : A \rightarrow B$ is injective and $X,Y \subseteq A$, then $\varphi[X \setminus Y] = \varphi[X] \setminus \varphi[Y]$. Proof: Let $b \in \varphi[X \setminus Y]$. Then $b = \varphi(a)$ for some $a \in X$ with $a \notin Y$. If $b \in \varphi[Y]$, then since $\varphi$ is injective, we have $a \in Y$, a contradiction. We thus have $b \in \varphi[X] \setminus \varphi[Y]$. $(\supseteq)$ Let $b \in \varphi[X] \setminus \varphi[Y]$. Now $b = \varphi(a)$ for some $a \in X$. If $a \in Y$, then $b = \varphi(a) \in \varphi[Y]$, a contradiction. Thus $a \notin Y$, so that $b \in \varphi[X \setminus Y]$. $\square$

If $\tau H \tau^{-1} = H$, then (using the lemmas to the previous exercise) we have

1. $\tau[F(H)] = \tau[\bigcap_{\sigma \in H} F(\sigma)]$ $= \bigcap_{\sigma \in H} \tau[F(\sigma)]$ $\bigcap_{\sigma \in H} F(\tau \sigma \tau^{-1})$ $= F(\tau H \tau^{-1}) = F(H)$ and
2. $\tau[M(H)] = \tau[A \setminus F(H)] = \tau[A] \setminus \tau[F(H)]$ $= A \setminus F(H) = M(H)$.

Thus $\tau$ fixes both $F(H)$ and $M(H)$.