Alternating groups of order n, group action on set of m elements

Question

Let $A_n$, $n \geq 5$ act transitively on a set with $m>1$ elements, then I need to show that $m \geq n$. I have been thinking of $A_n$ being simple, and somehow figure out something out of this. Perhaps, I should choose a group action $A_n$ x $\{1,...,m\}$ $\to$ $\{1,...,m\}$ given by $\sigma . i$:= $\sigma^{-1} (i)$. Answer to this will be a great help. Thanks in advance.

Answer 1

Recall a group $G$ acting on $X$ is a homomorphism $$ \rho\colon G\to\operatorname{Sym}(X). $$ So if $A_n$ ($n\geq 5$) acts on an $m$-element set, we have a homomorphism $A_n\to S_m$. The kernel is a normal subgroup so is either $1$ or $A_n$ by simplicity of $A_n$. But $m>1$ and the action is transitive, so the kernel must be $1$. So $\lvert S_m\rvert\geq\lvert A_n\rvert$, yielding $m\geq n$.