# A permutation group acting on subsets

Consider the the set $$X = \prod_{1 \leq k \leq n-2} \binom{ \bf{n}}{k}$$ where $$\binom{ \bf{n}}{k}$$ denotes the set of subsets with $$k$$ elements of the set $${\bf n} = \{1, \cdots , n\}$$.

For each $$i \in {\bf n}$$ and each permutation $$g \in S_{{\bf n} \backslash \{i\}}$$, we define a permutation $$\tilde{g_k}$$ of $$\binom{\bf n}{k}$$ as follows: if the $$k$$-subset $$Y$$ of $${\bf n}$$ contains $$i$$ then $$\tilde{g_k}$$ fixes $$Y$$, otherwise $$\tilde{g_k}$$ sends $$Y \subset {\bf n} \setminus \{i\}$$ to $$g(Y)$$. The cartesian product of the $$\tilde{g_k}$$ taken together define a permutation $$\tilde{g}$$ of $$X$$. For each $$i$$, define a subgroup $$G_i$$ of the permutation group $$S_X$$ as follows:

$$G_i = \{ \tilde{g} : g \in S_{{\bf n} \backslash \{i\}} \};$$

note that $$G_i \cong S_{n-1}$$. Let $$G = \langle G_i : i \in {\bf n}\rangle$$ be the subgroup of $$S_X$$ generated by all these subgroups.

Can you describe $$G$$ and its action on $$X$$? Thank you in advance.

• "Make a given group act by fixing..." is not a valid definition. – YCor Nov 10 '18 at 9:27
• Sorry. I mean that, given $i$, a permutation in $S_{n-1}$ acts on a subset which does not contain $i$ in the natural way (i.e., by taking the image of that subset once one identifies $S_{n-1}$ with the permutations in $S_n$ fixing $i$) and it acts on a subset containing $i$ by fixing it. The action is then extended on the product in the obvious way. – user131200 Nov 10 '18 at 9:37
• Thanks, this makes sense. Would you edit the post accordingly? – YCor Nov 10 '18 at 10:27
• Do you want generators and relations for $G$, or just the cardinality of $G$, or the character table, or something else? "Can you describe" is a touch vague. – Ben McKay Nov 10 '18 at 14:29
• If I understand correctly, a more concise definition would be : $G$ is the subgroup of permutations of $\mathcal P(\textbf n)$ generated by the $\varphi_{i, \sigma}$ ($i \in \textbf n$, $\sigma \in \Sigma_{\textbf n - \{i\}}$) sending $Y$ to $Y$ if $i \in Y$, and $Y$ to $\sigma(Y)$ else ? – J. Darné Nov 10 '18 at 15:34