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Math Help - A4 has no subgroup of order 6

  1. #1
    Super Member Bernhard's Avatar
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    A4 has no subgroup of order 6

    In Chapter 11 of Armstrong: Groups and Symmetry, Armstrong gives an argument (see bottom of page 59 and top of page 60 of Armstrong - attached)

    The argument runs as follows: (see attached)

    A_4 does not contain a subgroup of order 6. For suppose that H is a subgroup of A_4 which has six elements. If a 3-cycle belongs to H, its inverse must also belong to H, so the number of 3-cycles in H is even. There cannot be six as we need room for the identity element. Suppose there are four, say \alpha, \alpha^{-1}, \beta, \beta^{-1}. Then \epsilon, \alpha, \alpha^{-1}, \beta, \beta^{-1}, \alpha \beta, \alpha \beta^{-1} are all distinct and belong to H, contradicting our assumption that |H| = 6.

    Finally, if only two 3-cycles lie in H, then H must contain the subgoup { \epsilon, (12)(34), (13)(24), (14)(23)}. But 4 is not a factor of 6 so Lagrange's Theorem rules out this case too. We conclude that no subgroup of order 6 exists.

    ================================================== ===

    Firstly, why does Armstrong focus on 3-cycles in his reasoning regarding a subgroup of order 6?

    Secondly, I cannot follow the statement:

    "Finally, if only two 3-cycles lie in H, then H must contain the subgoup { \epsilon, (12)(34), (13)(24), (14)(23)}. "

    Why does this if-then statement follow?

    Can someone please help?

    Peter
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Re: A4 has no subgroup of order 6

    Quote Originally Posted by Bernhard View Post
    I cannot follow the statement:

    "Finally, if only two 3-cycles lie in H, then H must contain the subgoup { \color{red}\epsilon, (12)(34), (13)(24), (14)(23)}. "

    Why does this if-then statement follow?
    A_4 consists of 12 elements, 8 of which are 3-cycles. If a subgroup contains 6 elements, only 2 of which are 3-cycles, then it must contain all the elements that are not 3-cycles.
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  3. #3
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    Re: A4 has no subgroup of order 6

    one reason to focus on 3-cycles, is that for n ≥ 3, the 3-cycles generate An.

    there are a number of ways to show that H must contain {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} if it only contains 2 3-cycles. the most elementary way, is to show that if we only have 2 3-cycles, then those 4 elements of A4 is all that is left (because the other 8 elements of A4 are all 3-cycles:

    (1 2 3)
    (1 3 2)
    (1 2 4)
    (1 4 2)
    (1 3 4)
    (1 4 3)
    (2 3 4)
    (2 4 3).

    but even if we had not bothered to count the 3-cycles in A4, the only possible elements in A4 are the identity, 3-cycles, and 2,2-cycles (disjoint).

    if we only have 2 3-cycles in H, the remaining elements must be 2,2-cycles. but any pair of 2,2-cycles when multiplied, yields a 3rd. that is:

    (a b)(c d)(a c)(b d) =

    a-->c-->d
    b-->d-->c
    c-->a-->b
    d-->b-->a = (a d)(b c), so any 2 2,2-cycles generate a group of order 4 (which is necessarily abelian, it is easy to check that (a c)(b d)(a b)(c d) = (a d)(b c) as well).

    the case where A4 has just 2 3-cycles needs to be ruled out, because any group of order 6 is either cyclic, or isomorphic to S3. clearly we can rule out the cyclic case, becase S4 has no elements of order 6 at all. but S3 has no elements of order 6, either, but has just 2 elements of order 3. if it were not the case that the 2,2-cycles commuted, then we might have a subgroup of A4 isomorphic to S3 (compare this to when we have two transpositions, which do NOT commute, and the product of two transpositions can lead to a 3-cycle).
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