There exist infinite solvable $p$-groups with trivial centre. (Use a hint.)

group-theoryinfinite-groupsp-groupssolvable-groupswreath-product

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 🙂

Best Answer

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.

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