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Thread: Frobeniusgroup

  1. #1
    Junior Member
    Mar 2009


    Let F_{20} be the Frobeniusgroup of order 20.

    1) Show that F_{20} has the trivial Center.
    2) Determine for every divisor d of 20 the number of elements in F_{20} of order d.
    3)Determine the number of 5-Sylowgroups in F_{20} and teh number of 2-Sylowgroups in F_{20} .

    4) Determine the normal subgroups of F_{20} .
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  2. #2
    Apr 2009
    The Frobeniusgroup of order 20
    is the set of maps from F_5 to F_5 of the form
    x \mapsto ax+b where a,b \in F_5, b \neq 0.
    I'll answer your questions in
    slightly different order:

    d=1: 1 element, the identity x
    d=2: 5 elements, -x+b, b \in F_5
    d=4: 10 elements, ax+b, \; a\in \{2,3\}, b \in F_5
    d=5: 4 elements, x+b, \; b \in F_5, b \neq 0

    A presentation for F_{20} is

    <br />
\alpha^5 = \beta^2 = \gamma^4 = 1, \gamma^2 = \beta, \alpha^\beta=\alpha^{-1}, \alpha^\gamma=\alpha^2<br />

    We take
    <br />
\alpha = x+1, \beta = -x, \gamma = 2x<br />
    The elements of order 5 and two together lie in the
    Dihedral group of generated by \alpha and \beta (of order 10).
    This group has trivial center. Hence F_{20} has trivial center.

    3) There is only one Sylow 5 subgroup, the group
    generated by \alpha = x+1 (of order 5).
    There are 5 Sylow 2 subgroups (of order 4).
    They are the groups generated by \gamma = 2x and its conjugates
    2x+b \; (b \neq 0).

    4) There are two normal subgroups, both mentioned already:
    The Sylow 5 subgroup generated by \alpha,
    and the Dihedral group of order 10 generated by \alpha and \beta.


    Last edited by ZeroDivisor; Apr 26th 2009 at 07:17 AM. Reason: Latex formatting
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