This is an exercise in Rotman's Advanced Modern Algebra. Keep in mind no Sylow subgroups theory is developed. Only until group actions.
So $p=2$ is fine, since $A_{6}$ is simple. Suppose that $H$ is a subgroup of index $3$. Consider the action on the set of subgroups of $A_{6}$. Since the order of the orbit of $H$ is equal to $[A_{6}:N_{A_{6}}(H)]$ and $3=[A_{6}:H]=[A_{6}:N_{A_{6}}(H)][N_{A_{6}}(H):H]$, and the action on the set of subgroups of $A_{6}$ cannot be transitive, otherwise $H$ is normal, I get that $H$ is equal to its normalizer. But I don't know much about the normalizer, and fiddling around I don't seem to get a contradiction.
I don't know whether I should be using the order of $H$, maybe?
Any hint or help is appreciated. Thanks in advance!
Best Answer
In general, if $G$ is a simple group and $H$ is a subgroup of $G$ of index $m > 1$, then the homomorphism from $G$ to $S_m$ coming from $G$ acting on the cosets of $H$ will be injective (since it is not trivial and $G$ is simple).
Thus, we get that $n$ divides $|G|$ by Lagrange, and that $|G|$ divides $m!$ (also by Lagrange).
This means that if $A_n$ has a subgroup of index $p$ for some prime $p$ (and for $n\geq 5$), then $\frac{n!}{2}$ divides $p!$, which can only happen if $n = p$. On the other hand, as pointed out by Jack Schmidt in the comments, $A_{p-1}$ does indeed have index $p$ in $A_p$.