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Math Help - F20's Conjugacy Classes

  1. #1
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    F20's Conjugacy Classes

    The Frobenius Group F20 has 5 conjugacy classes.F = {a,b|a^5 - b^4 = 1, bab^-1 = a^2}.I think the conjugacy classes are [1],[a],[b],[b^2], and [b^3], but I'm having a hard time figuring out why [b^3] is a conjugacy class. I know, if a and b are cojugate, then |a| = |b|.|1| = 1, |a| = 5, |b| = 4, |b^2| = 2. So there are at least 4 conjugacy classes.But I'm not sure what to do now?
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  2. #2
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    Re: F20's Conjugacy Classes

    since b3 has order 4, the only conjugacy class out of [1],[a],[b],[b2] b3 could possibly belong to is [b].

    so assume that b3 = xbx-1. we can automatically rule out all powers of b, since b commutes with all powers of b.

    note we can write any element of F20 as akbm, for k = 0,1,2,3,4 and m = 0,1,2,3 because of the relation bab-1 = a2,

    so that ba = a2b.

    and (akbm)b(akbm)-1= akba5-k,

    so it suffices to prove that no conjugate of b by a power of a gives b3.

    now aba-1 = aba4 = a(ba)a3 = a(a2b)a3 = a3(ba)a2 =

    a3(a2b)a2 = ba2 = (ba)a = (a2b)a = a2(ba) = a4b.

    continuing: a2ba-2 = a2ba3 = a(aba-1)a4 = a(a4b)a4

    = ba4 = (a2b)a3 = a4ba2 = a6ba = aba = a3b.

    and next: a3ba-3 = a2(aba-1)a3 = a2(a4b)a3

    = aba3 = a(ba)a2 = a(a2b)a2 = a3(ba)a = a5ba = ba = a2b.

    and: a4ba-4 = a4(ba) = a4(a2b) = ab.

    so [b] = {b,ab,a2b,a3b, a4b}.

    since b3 is not in this conjugacy class, [b3] must be a 5th conjugacy class.
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  3. #3
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    Re: F20's Conjugacy Classes

    Okay great, thanks! And then to show that there's only 5, and not more than 5 conjugacy classes, do we just go through each of the 20 elements and show that they belong to one of those 5?
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