Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.8
Let $A$ be a nonempty set and let $k \in \mathbb{Z}^+$ such that $k \leq |A|$. The symmetric group $S_A$ acts on the set $B$ of all subsets of $A$ of cardinality $k$ by $\sigma \cdot \{ a_i \}_{i = 1}^k = \{ \sigma(a_i) \}_{i=1}^k$.
(1) Prove that this is a group action.
(2) Describe explicitly how the elements $(1\ 2)$ and $(1\ 2\ 3)$ act on the six 2-element subsets of $\{ 1, 2, 3, 4 \}$.
Solution:
(1) We have $$\mathsf{id}_A \cdot \{ a_i \}_{i = 1}^k = \{ \mathsf{id}_A(a_i) \}_{i = 1}^k = \{ a_i \}_{i = 1}^k.$$ Moreover, if $\sigma, \tau \in S_A$, we have $$\sigma \cdot (\tau \cdot \{ a_i \}_{i = 1}^k) = \sigma \cdot \{ \tau(a_i) \}_{i = 1}^k = \{ \sigma(\tau(a_i)) \}_{i = 1}^k = \{ (\sigma \circ \tau)(a_i) \}_{i = 1}^k = (\sigma \circ \tau) \cdot \{ a_i \}_{i = 1}^k.$$ Thus the mapping is a group action.
(2)We have $$(1\ 2) \cdot \{1,2\} = \{1,2\}$$ $$(1\ 2) \cdot \{1,3\} = \{2,3\}$$ $$(1\ 2) \cdot \{1,4\} = \{2,4\}$$ $$(1\ 2) \cdot \{2,3\} = \{1,3\}$$ $$(1\ 2) \cdot \{2,4\} = \{1,4\}$$ $$ (1\ 2) \cdot \{3,4\} = \{3,4\}$$ and $$(1\ 2\ 3) \cdot \{1,2\} = \{2,3\}$$ $$ (1\ 2\ 3) \cdot \{1,3\} = \{1,2\}$$ $$(1\ 2\ 3) \cdot \{1,4\} = \{2,4\}$$ $$(1\ 2\ 3) \cdot \{2,3\} = \{1,3\}$$ $$(1\ 2\ 3) \cdot \{2,4\} = \{3,4\}$$ $$(1\ 2\ 3) \cdot \{3,4\} = \{1,4\}.$$
