## The set of permutations which move a finite number of elements is normal

Let $A$ be a nonempty set and let $X \subseteq S_A$. Let $F(X) = \{ a \in A \ | \sigma(a) = a\ \mathrm{for\ all}\ \sigma \in X \}$ be the fixed set of $X$. Let $M(X) = A \setminus F(X)$ be the elements which are moved by some element of $X$.

Define $D = \{ \sigma \in S_A \ |\ |M(\sigma)| < \infty \}$. Prove that $D$ is a normal subgroup of $S_A$.

The identity permutation moves zero objects, and so $1 \in D$. Now suppose $\sigma,\tau \in D$. Now $M(\tau^{-1}) = M(\tau)$, and $M(\sigma \circ \tau) \subseteq M(\sigma) \cup M(\tau)$, so that $D$ is closed under identity, inversion, and multiplication, and hence is a subgroup.

Next we prove some lemmas.

Lemma 1: For all $\sigma \in S_A$ and $B \subseteq A$, $\sigma[A \setminus B] = A \setminus \sigma[B]$. Proof: $(\supseteq)$ Let $x \in A \setminus \sigma[B]$. Then $x \notin \sigma[B]$, hence $\sigma^{-1}(x) \notin B$. Thus $\sigma^{-1}(x) \in A \setminus B$, and so $x \in \sigma[A \setminus B]$. $(\subseteq)$ Let $x \in \sigma[A \setminus B]$. Then $x = \sigma(y)$ for some $y \in A \setminus B$. So $\sigma^{-1}(x) \notin B$, hence $x \notin \sigma[B]$, and thus $x \in A \setminus \sigma[B]$. $\square$

Lemma 2: If $F(X)$ is defined as above, then for all $\tau, \sigma \in S_A$, $\tau[F(\sigma)] = F(\tau \sigma \tau^{-1})$. Proof: $(\subseteq)$ If $x \in \tau[F(\sigma)]$, then $x = \tau(y)$ for some $y \in F(\sigma)$. Thus $\tau^{-1}(x) \in F(\sigma)$. Now $(\tau\sigma\tau^{-1})(x) = \tau(\sigma(\tau^{-1}(x))) = (\tau\tau^{-1})(x) = x$, so that $x \in F(\tau\sigma\tau^{-1})$. $(\supseteq)$ If $x \in F(\tau\sigma\tau^{-1})$, then $(\tau\sigma\tau^{-1})(x) = x$. Thus $\sigma(\tau^{-1}(x)) = \tau^{-1}(x)$, so that $\tau^{-1}(x) \in F(\sigma)$. Hence $x \in \tau[F(\sigma)]$. $\square$

Lemma 3: If $M(X)$ is defined as above, then for all $\tau, \sigma \in S_A$, $M(\tau \sigma \tau^{-1}) = \tau[M(\sigma)]$. Proof: We have $M(\tau \sigma \tau^{-1}) = A \setminus F(\tau \sigma \tau^{-1})$ $= A \setminus \tau[F(\sigma)]$ $= \tau[A \setminus F(\sigma)]$ $= \tau[M(\sigma)]$. $\square$

Now to the main result.

Let $\sigma \in D$ and $\tau \in S_A$. Then $M(\tau \sigma \tau^{-1}) = \tau[M(\sigma)]$ by Lemma 3. Since $M(\sigma)$ is finite and $\tau$ is a bijection, $M(\tau \sigma \tau^{-1})$ is also finite. Hence $\tau\sigma\tau^{-1} \in D$. Thus $D \leq S_A$ is normal.