TITLE: Subgroups between $S_n$ and $S_{n+1}$
QUESTION [14 upvotes]: Lets look at $S_n$ as subgroup of $S_{n+1}$. 
How many subgroups $H$, $S_{n} \subseteq H \subseteq S_{n+1}$ there are ?

REPLY [16 votes]: None.
Let $S_n<H<S_{n+1}$ and suppose that $H$ contains a cycle $c$ involving $n+1$, say
$$
c=(a,\cdots,b,n+1).
$$
Then by composing to the left with a suitable permutation $\sigma\in S_n<H$ such that $\sigma(a)=b$ we have
$$
\sigma c=\sigma^\prime (b,n+1)
$$
where $\sigma^\prime(n+1)=(n+1)$, i.e. $\sigma^\prime\in H$. Thus the transposition $(b,n+1)$ is in $H$. Of course we may assume that $b=1$ and so all transpositions $(1,2)$, $(1,3)$, ..., $(1,n+1)$ are in $H$. These transpositions are known to generate $S_{n+1}$.