proof $S_n$ has no subgroup of order $ \frac {n!}{4}$for $n>5$.
We know only subgroup of index $2$ in $S_n$ is $A_n$ and $A_n$ is simple group And we know every non trivial subgroup of $S_n$ has atleast one even permutation now let $K$ be subgroup of $S_n$ with order $ \frac {n!}{4}$ then $K \cap A_n$ is not trivial and $K \cap A_n$ is subgroup of $A_n$ but how we can show $K \cap A_n$ is not normal in $A_n$ ?
This is a standard trick I learned from my supervisor Jef Laga.
$n\geq 5$. Suppose $K\leq S_n$ with index $4$. From $A_n$ simple some easy manipulations would tell you the only normal subgroups of $S_n$ are $1$, $S_n$ or $A_n$.
Consider the action of $S_n$ on the cosets of $K$ by left multiplication. This gives a homomorphism $\varphi: S_n\rightarrow S_4$, and $\ker \varphi \trianglelefteq S_n$. One could easily see $\ker \varphi \neq S_n$ as the action is transitive by definition; and $\ker \varphi \neq 1$ as $\varphi$ cannot be injective ($n>4$). Therefore $\ker \varphi =A_n$. This would clearly cause a contradiction, as $\ker \varphi \leq K$.
Similarly, one could show that the index $t$ of a subgroup in $S_n$ cannot have $2<t<n$.