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This is Exercise 5.2.11 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to this search, it is new to MSE.

This is marked as being referred to later on in the book.

Here are some previous questions of mine about Wreath products:

A question of mine that involves solvable groups is this:

The Details:

(This can be skipped.)

On page 12 of the book,

The exponent of [a] group is [. . .] the least common multiple of all the orders [of its elements].

On page 24 of the book, paraphrased,

An elementary abelian $p$-group is an abelian group of exponent $p$ which is a direct product of cyclic groups of order $p$.

On page 32 and 33 of the book,

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 $X\times Y$ [. . .]

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

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.

For a prime $p$, an infinite $p$-group is an infinite group all of whose elements have orders a power of $p$ (not necessarily the same).

On page 121,

A group $G$ is said to be soluble (or solvable) if it has an abelian series, by which we mean a series $1=G_0\lhd G_1\lhd\dots \lhd G_n=G$ in which each factor $G_{i+1}/G_i$ is abelian.

The Question:

There exist infinite solvable $p$-groups with trivial centre. [Hint: Consider the standard wreath product $\Bbb Z_p\wr E$ where $E$ is an infinite elementary abelian $p$-group.]

I am more concerned with the hint than with providing any (other) example of such a group. Also, $\Bbb Z_p$ is the cyclic group of order $p$.

Thoughts:

Let $G=\Bbb Z_p\wr E$. Then $G$ is nonabelian.

Let $p=2$.

After a quick search online, I found via Wikipedia, that the group $(\mathcal{P}(\Bbb Z),\Delta)$, where $\Delta$ is the symmetric difference $X\Delta Y=(X\setminus Y)\cup(Y\setminus X)$ on arbitrary $X,Y$ in the power set $\mathcal{P}(\Bbb Z)$ of $\Bbb Z$, is an infinite elementary abelian $2$-group, since each nontrivial element has order two.

Therefore, let $E=(\mathcal{P}(\Bbb Z),\Delta)$. Then $G=\Bbb Z_2\wr E$.

Clearly $G$ is infinite.

I need to show $G$ is solvable.

Let $g\in Z(G)$. I need to show $g=e$.

I'm not sure how to handle this case of the hint, let alone the hint in its entirety.

Please help :)

Shaun
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    I suppose that $\mathbb{Z}_p$ means the cyclic group of order $p$? Also, on what sets are these groups supposed to act? Themselves? – Aphelli Dec 23 '21 at 20:53
  • Yes, @Mindlack; it's the cyclic groups of order $p$. Thank you for pointing out the issue of what the groups act on. I have edited the question accordingly. – Shaun Dec 23 '21 at 20:58
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    The lamplighter group $C_2\wr \mathbb Z$ is an example. If $E$ is the direct product of infinitely many copies of $C_2$, then $C_2\wr \mathbb E$ is an example too. – markvs Dec 23 '21 at 21:04
  • Unless I’m mistaken, your $\gamma(y)$ all commute, and they generate a normal subgroup. But the quotient by said subgroup is abelian. No? – Aphelli Dec 23 '21 at 21:09
  • I agree that the $\gamma(y)$ commute, @Mandlack, but I don't see right away how they generate a normal subgroup nor how the quotient is abelian. I'll think about it. Thank you. – Shaun Dec 23 '21 at 21:14
  • @Shaun: the conjugate of $\gamma(y)$ by $\kappa^*$ is of the form $\gamma(y’)$ with $y’=y\kappa^{\pm 1}$, no? As for the quotient (if the first part is correct), it is merely generated by the image of $K$ so it is abelian. – Aphelli Dec 23 '21 at 21:24
  • @Mindlack: I see. Thank you. However, something has escaped my understanding. Are you saying $$1\lhd \langle \gamma(y)\mid y\in K\rangle \lhd H\wr K$$ is an abelian series in general? Because that doesn't seem right. If not, where have you used the properties of $\Bbb Z_p$ and of $E$? – Shaun Dec 23 '21 at 21:36
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    @Shaun: I think that I merely used abelian-ness (so we get your comment’s claim), but it’s very possible that I missed something. I’m not a group theorist by any means so I don’t have any intuition on what wreath products are apart from the formal definition. – Aphelli Dec 23 '21 at 22:44
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    If $G=H\wr K$ (and I seem to remember we ran into an issue of whether Robinson uses the restricted direct product or the cartesian as the base; he uses the restricted product), then by construction the "base group" (the product of copies of $H$ Indexed by $K$) is normal in $G$. If $H$ and $K$ are abelian, then $G$ is solvable, because the base group is abelian and normal, and the quotient is isomorphic to $K$, which is abelian. Then you just need to show that this example has trivial center, which is not hard. – Arturo Magidin Dec 24 '21 at 01:36
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    In fact, you already link to a question showing that $Z(H\wr K)$ is trivial when $H$ is abelian and $K$ is infinite, so aren't you done? – Arturo Magidin Dec 24 '21 at 01:38
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    (If you use the cartesian, then it is no longer true that the center is trivial in this situation, since the subgroup of the base group consisting of constant functions/tuples would be central...) – Arturo Magidin Dec 24 '21 at 01:39
  • Thank you, @ArturoMagidin. Please would you type that up as an answer, so that I may close the question? – Shaun Dec 24 '21 at 12:07

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Recall that Robinson defines the (restricted) wreath product $H\wr K$ to be the split extension of the restricted direct product (I will borrow the notation from Hungerford) ${\prod\limits_{k\in K}}^*H$ of $|K|$ copies of $H$ by $K$. The restricted direct product is the set of (set theoretic) functions $f\colon K\to H$ of finite support, with coordinatewise multiplication. The linked to question contains the explicit description, so I won't repeat it here.

In particular, in the wreath product $H\wr K$, the subgroup ${\prod\limits_{k\in K}}^*H$ is normal, and the quotient of $H\wr K$ by this subgroup is isomorphic to $K$.

In the case at hand, $H$ and $K$ are abelian; thus, $H\wr K$ is metabelian (extension of an abelian group by an abelian group) and thus solvable. If $H\cong \mathbb{Z}_p$ and $K$ is an (infinite) elementary abelian $p$-group, it is not hard to verify that every element has order a power of $p$, so $H\wr K$ is a $p$-group.

When $K$ is infinite and $H$ is abelian, you already proved that $Z(H\wr K)=\{1\}$. Thus, $\mathbb{Z}_p\wr E$ where $E$ is an infinite elementary abelian group is (i) metabelian, hence solvable; (ii) a $p$-group; (iii) with trivial center.

Arturo Magidin
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