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

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 $\{ 1, 2, 3, 4 \}^2$.

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, $((1,1)\ (2,2)\ (3,3))((1,2)\ (2,3)\ (3,1))((1,3)\ (2,1)\ (3,2))$ $((1,4)\ (2,4)\ (3,4))((4,1)\ (4,2)\ (4,3))$.