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.

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 $a$.
