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).

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