# Subgroups between $S_n$ and $S_{n+1}$

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 ?

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What can you say about an element of $H$ which is not in $S_n$? –  Qiaochu Yuan Jul 19 '12 at 15:33
Stabilizers in primitive (hence 2-transitive) groups are maximal. –  user641 Jul 19 '12 at 18:25

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 $S_{n+1}$.

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Sorry I used a non-standard notation for cycles (the commas aren't usually there) but I couldn't find a way to manage spaces in this TeX environment. –  Andrea Mori Jul 19 '12 at 15:39
If by none you mean two... (the inclusions are not necessarily proper on either end!) –  Qiaochu Yuan Jul 19 '12 at 15:40
I often put in ~'s for cycle notation. e.g. $(1~2~3)$=$(1~2~3)$ –  anon Jul 19 '12 at 15:40
@QiaochuYuan: of course you are right. The "none" incipit was chosen for dramatic effects :) I guess that I should say that I was impliciting assuming that $H$ is not one of the trivial possibilities. –  Andrea Mori Jul 19 '12 at 15:45
@anon: thanks for the tip. –  Andrea Mori Jul 19 '12 at 15:46