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Let $S_{2n}$ be the symmetric group of degree $2n$. Let $H=\{s\in S_{2n}:i\leq n\rightarrow s(i)\leq n \}$ and let $A$ be the subset of $S_{2n}$ which contains all left and right shifts and the identity. ($(3\ 4\ 5)$ is a right shift but $(3\ 5\ 6)$ isn't, $(5 4 3)$ is a left shift.)

Question: Show that there is a set $I\subset S_{2n}$ s.t. $|I|=\frac{{2n \choose n}}{n+1}$ and $S_{2n}=HIA$, where $HIA=\{hia:h\in H, i\in I, a\in A\}$.

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    What is then "left/right" shift"? A permutation that maps **some** elements of $\;\{1,2,...,2n\}\;$ to the immediate element to its right? But then how is $\;(3 4 5)\;$ a right shift? Here $\;5\;$ is mapped to $\;3\;$ ...Is it perhaps something else...? In group theory, left shift *by an element* $\;g\;$ is a map $\;G\to G\;$ s.t. $\;x\to gx\;$ , and likewise right shift. – 2017-01-21
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    Sorry I should be more clear, a right/left shift here is any cycle which can be written as $(x_1 ... x_k)$ where $x_i$ are ascending/descending integers. – 2017-01-21
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    You mean that the $x_i$'s must be *consecutive* ascending/descending integers, I guess? Otherwise I don't see why $(3\ 5\ 6)$ is not a right shift. – 2017-01-21
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    Yes, I've ment consecutive! – 2017-01-21
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    I don't know if it helps, but the numbers $\frac{(2n)!}{n!(n+1)!}=\frac{{2n}\choose n}{n+1}$ are known as Catalan numbers, and have several combinatorial interpretations, some of which are related to permutations of $\{1,\dots,2n\}$. This could give some clues to solve your problem. See https://en.wikipedia.org/wiki/Catalan_number#Applications_in_combinatorics – 2017-01-21
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    Your question is an exact (to the letter) duplicate of this (unanswered) one : http://math.stackexchange.com/questions/1970895/question-on-a-subgroup-of-s-2n?rq=1 Is it from some textbook, or homework? – 2017-01-21
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    Thank you for the reference! I already read through the whole article a while ago when I first tried to solve it. No it's not from a textbook, it's the exact duplicate because I posted that question a while ago (and lost my account), so I couldn't close it to repost. It even seems to be the case that there exists $I_1,...,I_{n+1}\subset S_{2n}/H$ s.t. the $I_j's$ are pairwise disjoint, $I_1\cup ...\cup I_{n+1}=S_{2n}/H$, they all have the same size and $AI_{j}=S_{2n}/H$, where $x\in AI_j$ iff $x=(ai)H$ from some $a\in A$ and $iH\in I_j$. – 2017-01-21
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    @cousle: You may ask for your old account to be merged into the new one (thus recovering all the privileges that you used to have) by [contacting](http://math.stackexchange.com/contact) the SE staff (choose "I need to merge user profiles" as the subject). Of course, given that you have used an email address to create the other acount, you may also use it to reset your password. – 2017-01-21

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