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Homework Help: Wreath product

  1. Jun 6, 2007 #1
    1. The problem statement, all variables and given/known data
    My book has all of 1 page about wreath products and standard wreath products. I'm really lost on this topic and I have 4 homework problems using it. Does anyone know where I could find a little more info on the topic?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 6, 2007 #2

    Chris Hillman

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    Intuition for wreath products

    What book is that? Cameron, Permutation Groups is short and very readable and quite fascinating book, which might help. I don't seem to see the "relevant equations" (am I the only one?), but the basic idea is pretty simple: given two groups G,H acting on X,Y, to form the wreath product action you simply make |X| copies of the H-set Y and then let G act on that by permutating the copies as per the G-set X.

  4. Jun 6, 2007 #3
    This book is by Robinson "A Course in the Theory of Groups"
    The "Relevant Equations" look like this

    Let H and K be permutation groups acting on the sets X and Y respectively. Then H~K=<H(y),K^*|\y in Y>is the wreath product.

    If H and K are arbitrary groups, we can think of them as permutation groups an their underlying sets via the right regular representation and form of their wreath product W=H~K: this is called the standard wreath product.

    On of my HW problems is:
    Prove that the standard wreath product Z~Z is finitely generated but that it has a non-finitely generated subgroup.

    I'm just lost on this concept. I'll have to come back to it later. This book doesn't say anything about |X| copies. I'll try to find your book at the library.
    Last edited: Jun 6, 2007
  5. Jun 6, 2007 #4


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    This only deals with the finite case.
    You're familiar with the direct product, right?

    So, for example if we have a group [itex]H[/itex], then elements of
    [itex]H \times H[/itex] are ordered pairs [itex](h_1,h_2)[/itex] and the product is defined in the obvious way:
    [tex](h_1,h_2) \times (g_1,g_2)=(h_1 g_1, h_2 g_2)[/tex]

    And this readily generalizes to any finite power of [tex]H[/tex].

    Now, let's say we have some finite group [itex]G[/itex] with [itex]n=|G|[/itex]. Then, we know that [itex]G < S_n[/itex] from Caley's theorem - that is, we can consider the elements of [itex]G[/itex] to be permutations of [itex]n[/itex] elements.

    Now, consider [itex]H^n[/itex] (which is the direct product of [itex]n[/itex] copies of [itex]H[/itex]). Each element of this group is an ordered [itex]n[/itex]-tuple of elements of [itex]H[/itex].

    If we consider the elements of [itex]G[/itex] as permutations, we can allow them to act on the elements of [itex]H^n[/itex] by permuting them:
    [tex]\vec{h}=\left( h_1,h_2...h_n \right)[/tex]
    [tex]g \cdot \vec{h} = \left(h_{g(1)},h_{g(2)}...h_{g(n)}\right)[/tex]

    So, if we consider the elements of the cartesian product [itex]G \times H^n[/itex], we can define a group operation on them:
    [tex](g_1,\vec{h}_1) * (g_2,\vec{h}_2) = (g_1 g_2, g_1 \cdot (\vec{h}_1 \vec{h}_2))[/tex]
    Last edited: Jun 6, 2007
  6. Jun 6, 2007 #5
    Wow! Thanks! That really helps! I am MUCH more clear on the concept now. I'll give my HW problem another look with your info at hand.
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