This weird question comes from Abstract Algebra: Dummit and Foote.
Let $A$ be a non-empty set. $X\subseteq Sym(A)$.
Define $F(X)=\{a\in A$ s.t. $\sigma(a)=a\space \forall\sigma\in X\}$ - fixed set of $X$, and $M(X)=A\setminus F(X)$ - moved set of $X$. Let $D=\{\sigma\in Sym(A)$ s.t. $|M(\sigma)|<\infty\}$. Prove $D\unlhd Sym(A)$.
I don't quite get the meaning of $D$. So I came up with a few examples:
Let $A=\{1,2,3,4,5\}$, $X=\{1,(1\space 2\space 3)\}$.
Then $F(X)=\{4,5\}$ and $M(X)=\{1,2,3\}$.
Is $D=Sym(5)$, the symmetric group $Sym(A)$ itself? So I should not bother with finite sets $A$?
Another example is shown by letting $A=\{1,2,3,\dots\}$, $X=\{1,(1\space 2\space 3)\}$.
Then $F(X)=\{4,5,\dots\}$ and $M(X)=\{1,2,3\}$.
Now I suspect that actually means $F(\sigma)$ is the set of objects that are 'fixed' by $\sigma$ and so $M(\sigma)$ is the set of objects that are 'moved' by $\sigma$, so is $D$ 'the set of permutations that only 'moves' finitely many objects'? So that for all $\sigma\in D$, the cycle decomposition of $\sigma$ consists of finitely many cycles, each of which has finite length? So that for all $\tau\in Sym(A)$, $\tau\sigma\tau^{-1}$, with the same cycle type as $\sigma$, also consists of finitely many cycles, each of which has finite length? Is this proof correct?
Let $\sigma \in D$ and $g \in Sym(A)$ be arbitrary. Consider the permutation $g^{-1}\sigma g$ and assume that $a \in A$ is a fixed point of the latter permuation, i.e. $g^{-1}\sigma g \cdot a = a$. Thus you have $\sigma g \cdot a = g \cdot a$, in other words $g \cdot a $ is a fixed point of $\sigma$. What can you conclude if you assume that there would exist infinitely many fixed points of $g^{-1}\sigma g$? ($g$ is a permutation, i.e. a bijective map!)