## For n at least 5, the index of a subgroup of Alt(n) is at least n

Prove that $A_n$ does not have a proper subgroup of index less than $n$ for all $n \geq 5$.

We begin with a lemma.

Lemma: If $n \geq 3$, then $n!/2 > (n-1)!$. Proof: For the base case, note that $3!/2 = 3 > 2 = 2!$. For the inductive step, suppose $n!/2 > (n-1)!$. Now $n+1 > n$, so that $(n+1)!/2 > n!$. $\square$

Let $n \geq 5$, and let $H < A_n$ be a proper subgroup. Note that $A_n$ acts on the left cosets of $H$ by left multiplication; let $\varphi : G \rightarrow S_{[A_n : H]}$ be the permutation representation induced by this action, and let $K$ be the kernel of this action. Note that $K \leq \mathsf{stab}(H) = H < A_n$. Since $A_n$ is simple and $K$ is normal, we have $K = 1$. Thus $\varphi$ is injective.

If $[A_n : H] < n$, we have $A_n \leq S_{n-1}$. By the lemma this is a contradiction. Hence no proper subgroup has order less than $n$.

It will be useful to note Poincare’s theorem(http://groupprops.subwiki.org/wiki/Poincare’s_theorem): If $G$ has a subgroup $H$ of index $n$, then $H$ contains a normal subgroup $N$ of $G$ such that $n | [G:N] | n!$.