The problem is the following which was left by my lecturer:
Let $n \geq 3$ be an integer. Assume that $G$ is a subgroup of $S_n$ which is transitive on $B=\{1,2, \ldots, n\}$. That is, for each pair $i, j$ of elements of $B$ there exists an element $g \in G$ such that $g(i)=j$. Let $K_i$ be the subgroup of $G$ that leaves each of the elements $1,2, \ldots, i$ fixed: $$ K_i=\{g \in G: g(k)=k \quad \text { for } k=1,2, \ldots, i\} $$ for $i=1,2, \ldots, n$. Prove that if $K_i \neq K_j$ for all pairs $i, j$ such that $i \neq j$ and $i<n-1$, then $G=S_n$.
Firstly, it is noted that $K_{i+1}$ is a subgroup of $K_i$, and there is an element $g_i$ that belongs to $K_i$ but does not belong to $K_{i+1}$ for $i=1,2, \ldots, n-2$. At this point, it seems possible to use $g_i$ to construct a sequence of left cosets of $K_{i+1}$ in $K_i$, but this fails because the transitivity of G does not work on the subgroup $K_i$ of $G$ since $K_i$ may not be transitive. So I can't continue to do it anymore.
Any help is appreciated!
$K_{n-2} \ne K_{n-1}$ implies that $K_{n-2} = \langle (n-1,n) \rangle$ has order $2$ and acts as $S_2$ on the set $\{n-1,n\}$.
Now (if $n>2$) $K_{n-3} \ne K_{n-2}$ implies that $K_{n-3}$ acts as $S_3$ on $\{n-2,n-1,n\}$, etc.
Formally, using $i=2$ as the base case, prove by induction on $i$ that $K_{n-i}$ acts as $S_i$ on $\{n-i+1,\ldots,n\}$ for $i=2,3,\ldots,n$, where $K_0 = G$.
Note that for the base case we are using $K_i \ne K_j$ with $i = n-2$ and $j=n-1$, which the wording of the question allows. Also the assumption that $G$ is transitive is used to show that $G=K_0 \ne K_1$.