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This is a part of Exercise 1.6.14 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE. The first part of the exercise is here, a question of mine.

The Details:

[The details are the same as in the first part.]

The Question:

Paraphrased:

Show that $Z(H\wr K)=1$, for abelian $H\neq 1$ and arbitrary, infinite $K$.

Thoughts:

I found Arturo Magidin's answer to the first part helpful in understanding standard wreath products. However, I'm not sure how $K$ being infinite would imply $Z(H\wr K)=1$, not $\lvert Z(H\wr K)\rvert$ being infinite (or at least being $\lvert H\rvert$), since ($H$ is assumed to be abelian and), according to this perspective, the elements of $H\wr K$ are the elements of $H$ indexed by $K$.

A Potential Example:

I don't know much about the lamplighter group, but apparently it's the (restricted) wreath product $\Bbb Z_2\wr \Bbb Z$. I haven't found anything about its centre being trivial - which is what this result would imply - and yet I would imagine it'd be stated clearly somewhere.

Please help :)

Shaun
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    In this setting, the base group should be $H^{\oplus K}$ instead of $H^K$ in order to obtain a true statement, at least according to the results in the first part. – Berci Mar 30 '21 at 11:49
  • Okay, @Berci; thank you. What do you mean by $H^{\oplus K}$? – Shaun Mar 30 '21 at 11:51
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    It's the subgroup of $H^K$ consisting those sequences $(h_k)_{k\in K}$ for which $h_k=0$ for all but finitely many elements $k$. – Berci Mar 30 '21 at 11:54
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    Informally the argument goes like this: the elements in the centre of the unrestricted wreath product are those elements in the base group that are constant on all of their components - these are the only elements in the base group centralized by the subgroup $K$. In the restricted wreath product, in which all but finitely many components of base group elements must be trivial, the only element in which these components are constant is the identity element. – Derek Holt Mar 30 '21 at 12:43
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    On browsing Robinson's book, I think he is less than clear on whether he is looking at the unrestricted wreath product (where the "base" group is the unrestricted direct product of copies of $H$), or the restricted wreath product (where the "base" is the restricted direct product of copies of $H$; i.e., the subgroup of the cartesian consisting of the tuples of finite support). I can't find anywhere in his definition that would suggest the latter, but this statement clearly implies he is thinking of the latter, as previous comments suggest. No wonder you are confused. – Arturo Magidin Mar 30 '21 at 15:36
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    But here's the key: when he defines "direct products", he uses $\mathrm{Cr};G_{\lambda}$ for what he calls the "cartesian (or unrestricted direct) product"; and he uses $\mathrm{Dr};G_{\lambda}$ for what he calls the "external direct product", consisting the $(g_{\lambda})$ with $g_{\lambda}=1$ for almost all $\lambda$. (Pages 20 and 21 in my copy; Section 1.4). When he defines the "standard wreath product" in Section 1.6, he says "its base group is $\mathrm{Dr};H_k$, where $H_k\cong H$..." that is, he is looking at the restricted wreath product. – Arturo Magidin Mar 30 '21 at 15:40

1 Answers1

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Looks like my previous answer was incomplete, or perhaps not entirely concordant with Robinson's nomenclature.

Robinson uses "cartesian product", denoted $\mathrm{Cr}_{\lambda\in\Lambda} G_{\lambda}$ to denote what I call the "direct product" of groups. That is, the set of all $\Lambda$-tuples $(g_{\lambda})_{\lambda\in\Lambda}$, with $g_{\lambda}\in G_{\lambda}$, with coordinatewise operations.

He uses "direct product", denoted $\mathrm{Dr}_{\lambda\in\Lambda} G_{\lambda}$ to denote what some people call the "direct sum" of groups, and others (e.g., Hungerford) call the "restricted direct product", which is the subgroup of $\mathrm{Cr}_{\lambda\in\Lambda}$ of elements $(g_{\lambda})_{\lambda\in\Lambda}$ in which $g_{\lambda}=e$ for almost all $\lambda$.

Robinson's definition of the "standard wreath product" is what some people call the "restricted wreath product" (Rotman uses $H\wr_r K$, for example). This is the subgroup of the group I described earlier, $H^{|K|}\rtimes K$, in which we take only the elements $(h_k)\in H^{|K|}$ in which $h_k=e$ for almost all $k\in K$. When $K$ is finite, the two coincide; when $K$ is infinite, the restricted wreath product is a proper subgroup of the unrestricted wreath product.

Now, my argument in the previous problem describes the center of the unrestricted wreath product. A careful look at the argument shows that only elements of the base group can be central (and we used an element of the restricted wreath product to prove this, so that argument still holds in the restricted wreath product), and that the elements in question (constant tuples) are the only ones that centralize $K$. Since the elements of $K$ are also in the restricted wreath product, the only elements of the restricted wreath product that can be central are elements of the "base" that are constant. But the only such element when $K$ is infinite is the trivial element, yielding the result.

Arturo Magidin
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