Finding the action to Dummit and Foote 4.3.17

Question

Let $A$ be a nonempty set, $X$ any subset of $S_A$, $F(X)=\{a\in A:\sigma(a)=a\;\forall \sigma\in X\}$, and $M(X)=A-F(X)$ be the elements that are moved by some element of $X$. I need to show that $D=\{\sigma \in S_A: |M(\sigma)|<\infty\}$ is normal in $S_A$.

I know that I need to show that $D$ is the kernel of some action, but I do not know how define that action. (Sorry for the bad notation.)

Answer 1

Let $\sigma \in D$, and let $\tau \in S_{A}$. Our goal is to show that $\tau \sigma \tau^{-1} \in D$, i.e. that $\tau \sigma \tau^{-1}$ stabilizes all but finitely many elements of $A$.

Indeed, let $\Omega = \tau(M(\sigma))$. Since $M(\sigma)$ is finite and $\tau$ is a bijection, $|\Omega| = |M(\sigma)| < \infty$. Further, if $a \in A \setminus \Omega$, then $\tau^{-1}(a) \notin M(\sigma)$, i.e. $\sigma(\tau^{-1}(a)) = \tau^{-1}(a)$, so

$$(\tau\sigma\tau^{-1})(a) = \tau(\sigma(\tau^{-1}(a))) = \tau(\tau^{-1}(a)) = a$$

Hence, $M(\tau\sigma\tau^{-1}) \subset \Omega$, whence $|M(\tau\sigma\tau^{-1})| < \infty$, as desired.

Answer 2

$\def\inv{^{-1}}$ $\def\Fix{\operatorname{Fix}}$ This is just the same solution as given by Alex, but potentially with less extra notation.

Let's search for fixed points of $\tau \sigma \tau\inv$: $$ \tau \sigma \tau\inv(a) = a \Leftrightarrow \sigma (\tau\inv(a)) = \tau\inv(a). $$ Show that this implies $$ F(\tau \sigma \tau\inv) = \tau(F(\sigma)), $$ and hence also

$$ M(\tau \sigma \tau\inv) = \tau(M(\sigma)). $$ Conclude $M(\tau \sigma \tau\inv)$ is finite iff $M(\sigma)$ is finite.

Answer 3

I woud prefer to make this a comment, but it's too long.

I was curious about the possibility suggested by the OP of defining an action with kernel $D$. So first let $S_A$ act on the power set of $A$. Then define an equivalence relation on $\mathcal P(A)$ by declaring two sets to be equivalent if their symmetric difference is finite. The action of $S_A$ respects equivalence, so induces an action on $X_A = \mathcal P(A)/\sim$. It's pretty clear that $D$ is contained in the kernel of this action. We would still have to show that if $\tau \not\in D$ then $\tau$ acts non-trivially on $X_A$. In fact, I claim that one can find an infinite set $B$ such that $B \cap \tau(B) = \emptyset$. Unfortunately, this seems to require an actual argument, depending on the orbit structure of $\tau$, but it seems to work. (Split into cases that $\tau$ has an infinite orbit or not.) Clearly this argument is not to be prefered, as it is more complicated, but nevertheless I thought it worth mentioning.