Action of alternating group

Question

Group $A_n$ ($n>4$) acts on a set of cardinality $m$, and not all elements of this set are fixed points. Prove, that $m \geq n$.

I tried to apply Burnside's lemma but got nothing good.

Answer 1

As indicated in the comments, I will assume Abel's theorem on the simplicity of $A_n$ for $n > 4$.

Now, we have some set $X$ of cardinality $m$ with a nontrivial (let's say left) action of $A_n$. By definition, that means we have a map $A_n \times X \longrightarrow X$, which we write as $(\sigma, x) \mapsto \sigma x$. Now, we can modify this to a map $\lambda: A_n \longrightarrow \Sigma(X)$, where $\Sigma(X)$ denotes the group of permutations on $X$. This is defined by taking $\sigma \in A_n$ to the permutation $\lambda_\sigma(x) = \sigma x$. You can check that the inverse of $\lambda_\sigma$ is $\lambda_{\sigma^{-1}}$. In fact, the axioms of a group action will translate precisely to the fact that $\sigma \mapsto \lambda_\sigma$ is a group homomorphism.

Now, as $A_n$ is simple for $n > 4$, the map $\lambda: A_n \longrightarrow \Sigma(X)$ is either trivial or injective, as the kernel is either the trivial subgroup or all of $A_n$. If $ker(\lambda) = A_n$ then $\lambda_\sigma = id$ for all $\sigma \in A_n$. Hence, for all $x \in X$ we have $\sigma x = \lambda_\sigma(x) = x$. Thus, every element of $X$ is a fixed point of the action of $A_n$, but you explicitly assumed that this was not the case. Hence, we must have $ker(\lambda) = \{e\}$ so $\lambda$ must be an injection.

Now, $|A_n| = \frac{n!}{2}$ and $|\Sigma(X)| = m!$ (in fact, $\Sigma(X) \cong S_m$). So the injectivity of $\lambda$ tells us that $\frac{n!}{2} \leq m!$. We rewrite this as $\frac{n!}{m!} \leq 2$. Now, if $m < n$ then $\frac{n!}{m!} = n(n-1) \cdots (m + 1)$, which is in particular at least $n$. But $n > 4$ so we have just shown $4 < \frac{n!}{m!} \leq 2$ - a contradiction. Hence, we must have $m \geq n$.

This technique of turning a group action $G \times X \longrightarrow X$ into a group homomorphism $G \longrightarrow \Sigma(X)$ is very fruitful. For instance, if $X$ has additional structure, say it's a vector space, then we can consider groups "acting linearly" on $X$ by considering actions which coincide with morphisms $G \longrightarrow GL(X)$. This particular example leads to representation theory, but you could replace $GL(X)$ with the homeomorphisms, diffeomorphisms, group isomorphisms, or any other sort of group of automorphisms $X \longrightarrow X$.

Answer 2

No need for Burnside:

An action gives a homomorphism $h:A_n\to S_m $. The assumption that not everything is fixed implies the kernel can't be $A_n $. Thus since $A_n $ is simple, the homomorphism is injective. Thus $m!\ge n!/2\implies m\ge n $.