Let be $A_n$ the alternating group $n\geq 5$. How to prove that $\operatorname{Stab}_{A_n}(x) \cong A_{n-1}$, for all $x \in \{1, ..., n\}$?

Question

MY ATTEMPT:

I have 3 properties in hand:

1. $H\trianglelefteq G \iff Hg=gH, \; \forall g \in G \iff gHg^{-1}=H, \; \forall g \in G$;
2. When $H=gHg^{-1}, \; \forall g \in G$ then we have $H\trianglelefteq G \iff G=N_{G}(H)$;
3. When we have an action over conjugative of a subgroup than $\operatorname{Stab}^{*}(H)=N_{G}(H)$

So my try is: By the properties above we have $H \trianglelefteq A_n \iff A_n=N_{A_n}(H)=\operatorname{Stab}_{A_n}(x)$.

So, How can I find that $\operatorname{Stab}_{A_n}(x) \cong A_{n-1}$? For me doesn't make sense.

$*$ Where Stab means stabilizer.

Answer 1

I'm not sure why you involve normal subgroups and the sort. What I believe you wish to show is the following:
Consider the set $[n] = \{1, \ldots, n\}$ and the action of $A_n$ given on it by $\sigma\cdot x = \sigma(x)$.

Now, the stabiliser of an element $x\in [n]$ will be the set (subgroup) of all those $\sigma \in A_n$ such that $\sigma(x) = x$.
In other words, $\sigma$ restricted to $X = \{1, \ldots, n\}\setminus\{x\}$ is a bijection from $X$ onto itself.

Thus, the stabiliser is isomorphic to the symmetric group on $X$. (This is because composition of restrictions behaves well and thus, the group operation is respected.)

As $|X| = n-1$, the result follows.