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This is the first part of Exercise 1.6.15 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this Google search, it is new to MSE.

The Details:

Robinson calls the restricted wreath product $H\wr K$ the "standard wreath product". There is cause for confusion, as discussed in this answer to a previous question of mine.

For clarity's sake, then, I'll share Robinson's definition here . . .

On page 32 and 33, ibid.,

Let $H$ and $K$ be permutation groups on sets $X$ and $Y$ respectively. [. . .]

If $\gamma\in H$, $y\in Y$, and $\kappa\in K$, define

$$\gamma(y):\begin{cases} (x,y) \mapsto (x\gamma, y) & \\ (x,y_1) \mapsto (x,y_1) & y_1\neq y,\end{cases}$$ and

$$\kappa^\ast: (x,y)\mapsto (x,y\kappa).$$

[. . .] The functions $\gamma\mapsto \gamma(y)$, with $y$ fixed, and $\kappa\mapsto \kappa^\ast$ are monomorphisms from $H$ and $K$ to ${\rm Sym}(X\times Y)$: let their images be $H(y)$ and $K^\ast$, respectively. Then the wreath product of $H$ and $K$ is the permutation group on $X\times Y$ [. . .]

$$H\wr K=\langle H(y), K^\ast\mid y\in Y\rangle.$$

[. . .]

[T]he base group $B$ of the wreath product [is]

$$B=\underset{y\in Y}{{\rm Dr}}\, H(y).$$

Here $\underset{y\in Y}{{\rm Dr}}\, H(y)$ is defined as in this question.

Standard Wreath Products:

On page 41 of the book,

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

The Question:

Prove that the [standard/restricted] wreath product $\Bbb Z\wr \Bbb Z$ is finitely generated.

Thoughts:

We have two copies of $\Bbb Z$ given by, say, the two presentations

$$\langle a\mid\rangle \quad\text{and}\quad\langle b\mid\rangle.$$

I am led to believe here that $\Bbb Z\wr \Bbb Z$ is not finitely presented, so I'm not sure what to do with the presentations of $\Bbb Z$.

It appears that obtaining a presentation for a wreath product is, in general, somewhat of an active field of research; indeed, see again this Google search: there are papers as recent as 2018.

The reason I mentioned the two presentations is that perhaps the generators in question relate to $a$ and $b$ somehow.

My Background:

I'm used to finitely presented groups as part of my postgraduate research degree, which I am still undergoing. I don't have much experience, though, with nonfinitely presented groups.

A while ago, I asked this question:

Describing the Wreath product categorically.

It concerns wreath products for semigroups.

The type of answer I would prefer is a conceptual explanation of why $\Bbb Z\wr\Bbb Z$ is finitely generated, not necessarily a proof of the fact, because I think I ought to be able to prove it myself.

Besides, I want to move on to Robinson's chapter on free groups and presentations.

Please help :)

Shaun
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    https://berstein2015.wordpress.com/2015/06/15/wreath-products-and-finiteness/ – Eric Towers Apr 01 '21 at 15:15
  • That's an excellent find, @EricTowers. Thank you! – Shaun Apr 01 '21 at 15:17
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    I’m not sure (I don’t know that much about group theory), but don’t you have (when $\mathbb{Z}$ acts on itself by translation) $n(m)=m^n(0)(m^)^{-1}$, $n(0)=1(0)^n$, and $m^=(1^)^m$, so that $1^*$ and $1(0)$ generate all your generators and hence the wreath product? – Aphelli Apr 01 '21 at 15:20
  • Maybe. What do you mean by $x^*$, @Mindlack? – Shaun Apr 01 '21 at 15:27
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    The notation from Robinson’s restricted wreath product, I think. – Aphelli Apr 01 '21 at 15:37
  • Oh, of course! Thank you, @Mindlack. – Shaun Apr 01 '21 at 15:38
  • Why not upgrade your comment to an answer, @Mindlack? – Shaun Apr 01 '21 at 15:39
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    It is true more generally that, if $G$ and $H$ are finitely generated groups, then so is the restricted wreath product $G \wr H$. – Derek Holt Apr 01 '21 at 15:42
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    Intuitively, you only need to get enough elements in the $e_K$-indexed copy of $H^{|K|}$ to generate $H$; and then enough elements of $K$ to generate all of $K$, since then you can “shift” the copy of $H$ around to any position using just the elements that generate $K$. So you can certainly generate $H\wr_rK$ with at most $m+n$ elements, where $m$ is the number of elements you need to generate $H$, and $n$ is the number of elements needed to generate $K$. – Arturo Magidin Apr 03 '21 at 23:38

1 Answers1

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I’m turning my comment into an answer (with more generality), as Shaun suggested.

Assume we’re in the setting of a standard wreath product, with groups $H$ and $K$ acting by right translation on themselves. With the Robinson notation, it’s not hard to see that for any $h \in H,y \in K$, we have $h(y)=y^*h(e)(y^{-1})^*$.

We can also see that $y \in K \longmapsto (y^{-1})^*$, and $h \in K \longmapsto (h^{-1})(e)$ are group homomorphisms. Thus the standard wreath product of $H$ and $K$ is generated by the $k^*$ and the $h(e)^*$, where $k$ runs through a system of generators for $K$ and $h$ runs through a set of generators for $H$.

In particular, if $H$ and $K$ are finitely generated, the standard wreath product is finitely generated.

Aphelli
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