Exercise 4.6.1($A_n$ does not have a proper subgroup of index $<n$ for all $n\geq 5$)

Proof. Put $G=A_n$, where $n\ge 5$. We know that $G$ is a simple group.

Suppose that $H$ is a subgroup of index $|G:H|<n$. We show that $H$ is not a proper subgroup, i.e., $H=G$.

Consider the set $X$ of left cosets of $H$:

of cardinality $k=|G:H|$.

Then $G$ acts on $X$ by left translation. This action affords a homomorphism

Since $n\ge 5>2$,

The inequality $|A_n|>|S_X|$ shows that $\phi$ cannot be injective, thus

Since $G=A_n$ is simple, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=A_n$, so the action of $G$ on $X$ is trivial. In particular, for all $g\in G$, $\mathit{𝑔𝐻}=H$, thus $g\in H$. This shows that $H=G$ is not a proper subgroup.

In conclusion, $A_n$ does not have a proper subgroup of index $|A_n:H|<n$ for all $n\ge 5$. □