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Let $A$ be a non empty set and $k \in \mathbb{N}, k \leq |A|$. The symmetric group $S(A)$ acts on set $B$ containing all the subsets of $A$ of cardinality $k$ by $$f(\{a_1, a_2, ..., a_k\}) = \{f(a_1), f(a_2), ...,f(a_k)\}$$ For what values of $k$ is the action faithful? I have tried the problem and got that this action is faithful for all $k < |A|$. But I'm not able to write a proof.

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  • $\begingroup$ Canyou explain what the group action does when $k = 1$ and when $k = \mathrm{card}(A)$? $\endgroup$ – Eric Towers Sep 8 '17 at 13:35
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If $k = |A|$, then the only subset of $A$ with $k$ elements is $A$ itself. Now let an arbitrary $f \in S(A), f\neq Id$ act on $A$:

$f(\{a_1, ..., a_n\}) = \{f(a_1), ..., f(a_n)\} = \{a_1, ..., a_n\}$

since we only permute the elements in a non-ordered set, which yields the same set again. So this means that in the case $k = |A|$ that the action is not faithful (Please note that there are several equivalent definitions of faithfulness).

Now prove that in the case $k \neq |A|$ for every $f \in S(A), F\neq Id$ you can always find a subset $M \subset A, |M| = k$ with $f(M) \neq M$, which means that the action is faithful.

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