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$\begingroup$

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 $S_n$?2012-07-19
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    Stabilizers in primitive (hence 2-transitive) groups are maximal.2012-07-19

1 Answers 1

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None.

Let $S_n 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 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.2012-07-19
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    If by none you mean two... (the inclusions are not necessarily proper on either end!)2012-07-19
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    I often put in ~'s for cycle notation. e.g. `$(1~2~3)$`=$(1~2~3)$2012-07-19
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    @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.2012-07-19
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    @anon: thanks for the tip.2012-07-19