Prove that $\displaystyle A_6$ has no subgroup of prime index.
you can prove a much stronger result: if $\displaystyle n \geq 5$ and $\displaystyle H \lneq A_n,$ then $\displaystyle [A_n:H] \geq n.$ this is a quick result of this well-known fact that for any group G and any subgroup H of G of finite index, there
exists a normal subgroup N of G contained in H such that $\displaystyle [G:N] \mid [G:H]!.$