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Suppose $S_n$ acts on a set of $k$ element subsets of $\{1,\ldots,n\}$, by $\sigma\cdot\{a_1,\ldots,a_k\}=\{\sigma(a_1),\ldots,\sigma(a_k)\}$ where $k<n$ and $\sigma\in S_n$. Show that action is faithful.

Proof: Suppose action not faithful, then there exist $\sigma, \tau\in S_n$ such that $\sigma\cdot\{a_1,\ldots,a_k\}=\tau\cdot\{a_1,\ldots,a_k\}$ for all $k$ element subsets $\{a_1,\ldots,a_k\}$, but $\sigma\ne \tau$. Then there exists an $x\in\{1,\ldots,n\}$ such that $\sigma(x)\ne\tau(x)$. But I can't see how to show $\sigma\cdot\{x,\ldots,a_k\}\ne\tau\cdot\{x,\ldots,a_k\}$.

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You need to assume that $1 \le k$ as well as $k<n$.

Suppose that $1 \ne \sigma \in S_n$. So there exists $x$ with $\sigma(x) \ne x$. Then $\sigma^{-1}(x) \ne x$, so there exists a $k$-subset $S$ with $x \in S$ but $\sigma^{-1}(x) \not\in S$. Then $x \not\in \sigma(S)$, so $\sigma(S) \ne S$. So the action of $S_n$ on the set of $k$-subsets is faithful.

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  • $\begingroup$ Why does $\sigma(S) \ne S$ imply the action is faithful? $\endgroup$ – Silent Mar 15 '18 at 14:18
  • $\begingroup$ That follows immediately from the definition of a faithful action. $\endgroup$ – Derek Holt Mar 15 '18 at 14:30

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