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Math Help - Problem: Abelian/non-abelian subgroups

  1. #1
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    Problem: Abelian/non-abelian subgroups

    Hi,
    Can anyone help with the following please?

    A non-Abelian group can have both Abelian and non-Abelian subgroups. The symmetric groups Sn are non-Abelian for n ≥ 3.
    a. Prove that the alternating group, An, which is a subgroup of Sn, is non-Abelian for n ≥ 3.
    b. Construct an Abelian and a non-Abelian subgroup of S4 (excluding the trivial cases G = {e} and G = S4).

    For a) i was thinking i need to find two subgroups say (1 2 3...) and (1 3 2...) (in cycle notation) and show that they are not commutative but (1 2 3)(1 3 2) = (1)(2)(3)... and (1 3 2) (1 2 3) = (1)(2)(3)...so this example is commutative but yields a result which is not an element of A4 (the set of even permutations). Am I on the right lines with this?

    b) I'm unsure of but solving a) would help I assume.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by jackiemoon View Post
    Hi,
    a. Prove that the alternating group, An, which is a subgroup of Sn, is non-Abelian for n ≥ 3.
    Try (123)(132) and (132)(123).

    b. Construct an Abelian and a non-Abelian subgroup of S4 (excluding the trivial cases G = {e} and G = S4).
    The subgroup \left< (1234) \right> is cyclic.
    While, S(1) the permutations that fix 1 is non-abelian.
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