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## Compute the action of two group elements on a set

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.9

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 $A^k$ 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 16 elements of $\{ 1, 2, 3, 4 \}^2$.

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 $$(1\ 2) \cdot (1,1) = (2,2)$$ $$(1\ 2) \cdot (1,2) = (2,1)$$ $$(1\ 2) \cdot (1,3) = (2,3)$$ $$(1\ 2) \cdot (1,4) = (2,4)$$ $$(1\ 2) \cdot (2,1) = (1,2)$$ $$(1\ 2) \cdot (2,2) = (1,1)$$ $$(1\ 2) \cdot (2,3) = (1,3)$$ $$(1\ 2) \cdot (2,4) = (1,4)$$ $$(1\ 2) \cdot (3,1) = (3,2)$$ $$(1\ 2) \cdot (3,2) = (3,1)$$ $$(1\ 2) \cdot (3,3) = (3,3)$$ $$(1\ 2) \cdot (3,4) = (3,4)$$ $$(1\ 2) \cdot (4,1) = (4,2)$$ $$(1\ 2) \cdot (4,2) = (4,1)$$ $$(1\ 2) \cdot (4,3) = (4,3)$$ $$(1\ 2) \cdot (4,4) = (4,4)$$ Or, in cycle notation, $$((1,1)\ (2,2))((1,2)\ (2,1))((1,3)\ (2,3))((1,4)\ (2,4)) ((3,1)\ (3,2))((4,1)\ (4,2)).$$and $$(1\ 2\ 3) \cdot (1,1) = (2,2)$$ $$(1\ 2\ 3) \cdot (1,2) = (2,3)$$ $$(1\ 2\ 3) \cdot (1,3) = (2,1)$$ $$(1\ 2\ 3) \cdot (1,4) = (2,4)$$ $$(1\ 2\ 3) \cdot (2,1) = (3,2)$$ $$(1\ 2\ 3) \cdot (2,2) = (3,3)$$ $$(1\ 2\ 3) \cdot (2,3) = (3,1)$$ $$(1\ 2\ 3) \cdot (2,4) = (3,4)$$ $$(1\ 2\ 3) \cdot (3,1) = (1,2)$$ $$(1\ 2\ 3) \cdot (3,2) = (1,3)$$ $$(1\ 2\ 3) \cdot (3,3) = (1,1)$$ $$(1\ 2\ 3) \cdot (3,4) = (1,4)$$ $$(1\ 2\ 3) \cdot (4,1) = (4,2)$$ $$(1\ 2\ 3) \cdot (4,2) = (4,3)$$ $$(1\ 2\ 3) \cdot (4,3) = (4,1)$$ $$(1\ 2\ 3) \cdot (4,4) = (4,4).$$Or, in cycle notation,$$((1,1)\ (2,2)\ (3,3))((1,2)\ (2,3)\ (3,1))((1,3)\ (2,1)\ (3,2)) ((1,4)\ (2,4)\ (3,4))((4,1)\ (4,2)\ (4,3)).$$