Set of permutations that move a finite number of elements is normal Let $A$ be a nonempty set and let $X$ be a subset of $S_A$. Let $D$ be the set of permutations $\sigma$ that move a finite number of elements of $A$. To prove: $D$ is normal in $S_A$.
Attempt:
I can show that $D$ is a subgroup, but I can't seem to show that $D$ is normal. What I need to do is take an arbitrary permutation $\tau \in S_A$ and $\sigma \in D$ and show that $\tau \sigma \tau^{-1} \in D$, i.e. that $\tau \sigma \tau^{-1}$ also moves a finite number of elements.
I proceed by contradiction. Suppose that $\tau \sigma \tau^{-1}$ moves an infinite number of elements of $A$. First of all, this means that $\tau$ also moves an infinite number of elements, because if $\tau$ moved only a finite number of elements then so would $\tau^{-1}$, and therefore $\tau \sigma \tau^{-1}$ would also move only a finite number of elements, contrary to my assumption. Thus there exist some distinct elements $a_1, a_2 \in A$ such that $\sigma$ fixes both $a_1$ and $a_2$, and that $\tau(a_1) = a_2$. Then $\tau \sigma \tau^{-1}(a_2) = \tau \sigma(a_1) = \tau(a_1) = a_2$. And here is where I get stuck, because I don't see any contradictions to be reached.
 A: Suppose that $a\in A$ is moved by $\tau\sigma\tau^{-1}$. Then $\tau^{-1}(a)$ is moved by $\sigma$. Since $\sigma$ moves only a finite number of elements of $A$ and $\tau$ is injective, there can only be finitely many such $a$.
A: The subset of $A$ of the elements moved by $\tau\in\operatorname{Sym}(A)$ is named support of $\tau$:
$$\operatorname{supp}(\tau):=\{a\in A\mid\tau(a)\ne a\}\tag 1$$
Therefore (for convenience, let me use $\tau^{-1}\sigma\tau$ in place of $\tau\sigma\tau^{-1}$, as it's the same for our aim):
\begin{alignat}{1}
\operatorname{supp}(\tau^{-1}\sigma\tau) &=\{a\in A\mid (\tau^{-1}\sigma\tau)(a)\ne a\} \\
&=\{a\in A\mid \tau^{-1}(\sigma(\tau(a)))\ne a\} \\
&\stackrel{(*)}{\subseteq}\{a\in A\mid \sigma(\tau(a)))\ne \tau(a)\} \\
&=\{a\in A\mid \tau(a)\in \operatorname{supp}(\sigma)\} \\
&=\tau^{\leftarrow}(\operatorname{supp}(\sigma)) \\
\tag 2
\end{alignat}
Being $\tau$ injective, if $\operatorname{supp}(\sigma)$ is finite (i.e. if $\sigma\in D$), then $|\tau^{\leftarrow}(\operatorname{supp}(\sigma))|=|\operatorname{supp}(\sigma)|$, and thence (by $(2)$) $|\operatorname{supp}(\tau^{-1}\sigma\tau)|\le|\operatorname{supp}(\sigma)|$; so, $\tau^{-1}\sigma\tau \in D$, and $D\unlhd S_A$.

$^{(*)}$This inclusion holds because, by the injectivity of $\tau$, we have that:
$$P(a)\colon\space\tau^{-1}(\sigma(\tau(a)))\ne a \space\Longrightarrow\space Q(a)\colon\space\sigma(\tau(a))\ne \tau(a)$$
and thence $\tilde a \in \{a\mid P(a)\} \Rightarrow \tilde a \in \{a\mid Q(a)\}$, that's to say $\{a\mid P(a)\}\subseteq \{a\mid Q(a)\}$.
