# Wreath product

1. Jun 6, 2007

### happyg1

1. The problem statement, all variables and given/known data
Hello,
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. Jun 6, 2007

### Chris Hillman

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.

HTH

3. Jun 6, 2007

### happyg1

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.

Then:
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.
CC

Last edited: Jun 6, 2007
4. Jun 6, 2007

### NateTG

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

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

And this readily generalizes to any finite power of $$H$$.

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

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

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

So, if we consider the elements of the cartesian product $G \times H^n$, we can define a group operation on them:
$$(g_1,\vec{h}_1) * (g_2,\vec{h}_2) = (g_1 g_2, g_1 \cdot (\vec{h}_1 \vec{h}_2))$$

Last edited: Jun 6, 2007
5. Jun 6, 2007

### happyg1

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.
CC