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)).

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Comments

  • Brian  On November 28, 2011 at 1:18 am

    Wouldn’t be nicer to write this rather concise way:
    (1 2) induces the permutation ({1 3}{2 3}) ({1 4}{2 4})
    (1 2 3) induces the permutation ({1 2}{2 3}{1 3})({1 4}{2 4}{3 4})

    • nbloomf  On November 28, 2011 at 10:28 am

      I’m not sure how to interpret that notation, but I agree that cycles would be helpful here.

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