Let $n \geq 5$ and let $X$ be a set on which $S_n$ acts. I'd like to show that every $S_n$-orbit has size $1$, $2$ or $\geq n$.
This isn't hard if I know that $A_n$ is simple: For $x \in X$, let $\text{Stab}(x)$ be the stabilizer of $x$. Then $|S_n x| = [S_n : \text{Stab}(x)]$. See the answer here.
However, proofs that $A_n$ are simple are all a bit messy. I'd like a really quick, slick way to do this, for a student who has seen groups and group actions (and hence subgroups and orbit stabilizer), but doesn't know $A_n$ is simple. Any ideas?
A note as to motivation: The way that I understand the cubic and biquadratic formulas is using resolvents. We start with a degree $n$ polynomial, for $n=3$ or $4$, with roots $\alpha_1$, $\alpha_2$, ..., $\alpha_n$. We find a multivariate polynomial $f(x_1, x_2, \ldots, x_n)$ such that the orbit of $f$ under the $S_n$ action on $\mathbb{Q}[x_1, x_2, \ldots, x_n]$ has size $d<n$. Then $f(\alpha_1, \alpha_2, \ldots, \alpha_n)$ satisfies a polynomial of degree $d$. For example, with $n=4$, we can find a cubic equation satisfied by $\alpha_1 \alpha_2 + \alpha_3 \alpha_4$. The reason that this method is not useful for $n \geq 5$ is that, for any $f(x_1, \ldots, x_n) \in \mathbb{Q}[x_1, x_2, \ldots, x_n]$, either $f$ is symmetric, or $f$ is anti-symmetric, or else $|S_n f| \geq n$.
I think this is a very nice application of group actions. I'd like to be able to discuss it as soon as I introduce group actions, which means before we know that $A_n$ is simple.