While giving my answer here, I thought to the following generalization of the claim therein stated:
Claim. Let $p>3$ be a prime. There doesn't exist any $H\lhd S_p$, $|H|=p$, such that $S_p/H\cong S_{p-1}$.
Proof sketch. By contradiction, suppose that such a $H$ does exist. Then, there is a surjective homomorphism $\varphi$ from $S_p$ to $S_{p-1}$ with kernel $H$. Such a $\varphi$ sends conjugacy classes to conjugacy classes. $H$ is made up of $p$ $p$-cycles; the number of $p$-cycles in $S_p$ is $(p-1)!>p$ (for $p>3$, as assumed), and thence $H\setminus\{Id\}\subsetneq \operatorname{Cl}((1…p))$. Therefore, any element of $H\setminus \{Id\}$ is sent into $\varphi(\operatorname{Cl}((1…p)))$, which does not contain the identity of $S_{p-1}$; but any such element is sent to $Id$ by definition of kernel. Contradiction.
(As "minimal corollary", take $p=5$ to get the case addressed in the opening link.)
Is this all correct?
Best Answer
The generalization is good, but you're proving by contradiction, not by contraposition.
If such a normal $H$ subgroup exists, it must have $p$ elements by the homomorphism theorems. Since $p$ is prime, the group is cyclic and generated by a $p$-cycle that, upon relabeling of the elements we permute on, it can be assumed to be $(123\dots p)$.
Since $(12)(1234\dots p)(12)=(2134\dots p)$ is not a power of $(1234\dots p)$, $H$ is not normal. Contradiction.
The assumption $p>3$ is necessary: indeed, the same argument applied to $(123)$ yields $(213)=(123)^2$ and, indeed, $S_3/A_3\cong S_2$.
The statement is however also true for every $n>4$ and follows from the simplicity of $A_n$.