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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.

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  • $\begingroup$ "Make a given group act by fixing..." is not a valid definition. $\endgroup$ – YCor Nov 10 '18 at 9:27
  • $\begingroup$ 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. $\endgroup$ – user131200 Nov 10 '18 at 9:37
  • $\begingroup$ Thanks, this makes sense. Would you edit the post accordingly? $\endgroup$ – YCor Nov 10 '18 at 10:27
  • $\begingroup$ 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. $\endgroup$ – Ben McKay Nov 10 '18 at 14:29
  • $\begingroup$ 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 ? $\endgroup$ – J. Darné Nov 10 '18 at 15:34

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