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A paper I'm reading claims that the smallest class of monoids which contains $\mathbb{Z}$ and is closed under finite direct product and block product only contains solvable monoids. I think that a proof that solvability of groups is closed under wreath product would be of great help to understand why this is true. Would anyone know where I can find such a proof?

In the "converse direction", I would be really interested in reading a proof of the following fact: Any solvable group is in the variety generated by a wreath product of cyclics (taken from Barrington and Thérien, Non-uniform automata over groups (1987)).

Thank you!

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up vote 6 down vote accepted

A wreath product $G\wr H$ is a semidirect product of $G^H$ and $H$; since a group $G$ with normal subgroup $N$ is solvable if and only if both $N$ and $G/N$ are solvable, this means that $G\wr H$ is solvable if and only if $G^H$ and $H$ are both solvable.

Thus, if $G\wr H$ is solvable, then so are $G^H$ and $H$. The only question then is whether $G$ is solvable if and only if $G^H$ is solvable. This is true.

First, if $G^H$ is solvable, then so is the subgroup corresponding to the $e_H$-coordinate (that is, the subgroup $f\colon H\to G$ such that $f(h)=1$ for all $h\neq e_H$, which is isomorphic to $G$).

Conversely, if $G$ is solvable, then so is any direct power of $G$

This follows because for any direct product of groups, we have $$\left(\prod_{i\in I} K_i\right)^{(n)} \subseteq \prod_{i\in I}K_i^{(n)}$$ where $M^{(n)}$ is the $n$th derived subgroup of $G$. (Simply note that the projection maps $\pi_j\colon \prod_{i\in I} K_i \to K_j$ must map $(\prod K_i)^{(n)}$ into $K_j^{(n)}$, so that gives a map into the product of the $n$th derived subgroups); equality need not hold in general, but the inclusion always does).

Since $G$ is solvable, if and only if there exists $n$ such that $G^{(n)} = \{1\}$; then if $G$ is solvable, we have $$\left(\prod_{i\in I}G\right)^{(n)} \subseteq \prod_{i\in I}G^{(n)} = \prod_{i\in I}\{1\} = \{1\}$$ so $\prod_{i\in I}G$ is solvable.

In particular, taking $I=H$ we get that if $G$ is solvable, then so is $G^H$.

In summary:

  • $G\wr H$ is solvable if and only if $G^H$ and $H$ are both solvable.
  • For any nonempty set $I$, $\prod_{i\in I}G$ is solvable if and only if $G$ is solvable.
  • $G^H \cong \prod_{h\in H}G$, so $G^H$ is solvable if and only if $G$ is solvable.

So: $G\wr H$ is solvable if and only if $G$ and $H$ are both solvable.

Note. It is not true in general that an arbitrary direct product of solvable groups is solvable. What is true is that for fixed $n$, an arbitrary direct product of solvable groups is solvable of length at most $n$ if and only if each direct factor is solvable of length at most $n$.

The converse is a little trickier, but I think the following does it.

First: a group $G$ generates the variety $\mathfrak{V}$ if and only if every group in $\mathfrak{V}$ is a homomorphic image of a subgroup of a direct power of $G$. For example, $\mathbb{Z}$ generates the variety of abelian groups. A set of groups $S$ generates $\mathfrak{V}$ if and only if every group in $\mathfrak{V}$ is a homomorphic image of a subgroup of a direct product of copies of groups in $S$.

Second: a set of groups $S$ discriminates a variety $\mathfrak{V}$ if $S\subset\mathfrak{V}$ and for every finite set $\mathfrak{w}$ of words that are not laws of $\mathfrak{V}$, there is a group $D\in\mathfrak{D}$ in which the equations $w=1$, $w\in\mathfrak{w}$, can be simultaneously falsified.

It is a theorem that a set that discriminates a variety also generates that variety; in fact, a set of groups "discriminates" if and only if it discriminates the variety it generates.

Theorem. (Baumslag, B.H. Neumann, H. Neumann, and P. Neumann). If the group $G$ generates $\mathfrak{V}$, and the set $S$ discriminates $\mathfrak{W}$, then the set $G\wr S = \{ G \wr D\mid D\in S\}$ discriminates $\mathfrak{VW}$, and therefore also generates $\mathfrak{VW}$.

(In fact, one can take the restricted direct product above, rather than the full wreath product).

Now, every group word is equivalent to a group word of the form $x^nc$, where $c$ is a commutator word on and $n\geq 0$. Thus, the only words that are not laws of the variety of abelian groups are words of the form $x^nc$ with $n\gt 1$. All these laws can be simultaneously falsified by $\mathbb{Z}$, by evaluating every letter in the words at $1$;

The argument above was wrong (a bunch of hooey, really). Here's a proper argument: a word $\mathbf{w}$ is not a law of $\mathfrak{A}$ if and only if the exponent sum of at least one variable is not $0$. For abelian groups, the word can be rewritten as a linear equation in the variables with integer coefficients; an evaluation in $\mathbb{Z}$ satisfies the word if and only if it satisfies the corresponding linear equation with integer coefficients.

A finite set of words that are not laws of $\mathfrak{A}$ correspond to a finite system of linear equations with integer coefficients; an element of $\mathbb{Z}^n$ ($n$ the number of variables) satisfies at least one of the equations if it lies in the union of the solution sets of the individual equations. Since the equations are nontrivial, the solution sets are proper submodules of $\mathbb{Z}^n$; since $\mathbb{Z}^n$ is not a union of finite many proper submodules (same argument as for vector spaces over infinite fields), it follows that given a finite collection of words that are not laws of $\mathfrak{A}$ there is always an evaluation in $\mathbb{Z}$ that falsifies all of them simultaneously.

Therefore, $\mathbb{Z}$ discriminates $\mathfrak{A}$, the variety of all abelian groups. Also, $\mathbb{Z}$ generates the variety of all abelian groups.

Therefore, applying the B+3N theorem, we get that $\mathbb{Z}\wr\mathbb{Z}$ generates $\mathfrak{A}^2$, the variety of solvable groups of length at most $2$; then it follows that $(\mathbb{Z}\wr\mathbb{Z})\wr\mathbb{Z}$ discriminates, and hence generates, $\mathfrak{A}^3$. Continuing this way, the iterated wreath product of $\mathbb{Z}$ discriminates, and hence generates, the variety of solvable groups of length at most $n$. Thus, every solvable group lies in a variety generated by an iterated (restricted) wreath product of cyclic groups.

Presumably, if you know a bit more about $G$ some of the copies of $\mathbb{Z}$ could even be replaced by suitable finite cyclic groups.

Note. It is false in general that if $S$ generates $\mathfrak{V}$ and $T$ generates $\mathfrak{W}$, then $S\wr T = \{G\wr H\mid G\in S, H\in T\}$ generates $\mathfrak{VW}$. But if $S$ generates $\mathfrak{V}$, then $S\wr F_{\infty}(\mathfrak{W}) = \{ G\wr F_{\infty}(\mathfrak{W})\mid G\in S\}$ generates $\mathfrak{VW}$, where $F_{\infty}(\mathfrak{W})$ is the countably generated relatively free group in $\mathfrak{W}$, and the wreath products are restricted wreath products. This is a consequence of the B+3N theorem above, but it was actually proven earlier by Baumslag.

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If you don't answer the converse, I'll be forced to ask it as a separate question. :P –  Jack Schmidt Mar 23 '11 at 22:03
    
@Jack: It was a bit trickier than I thought, and I had to rely on Baumslag and the Neumanns, but I think this does it. –  Arturo Magidin Mar 23 '11 at 22:30
    
Thanks! I like your very specific generators for each A^n variety. I'll see if I can prove Z wr Z generates the metabelian groups without using the full B+3N. –  Jack Schmidt Mar 23 '11 at 22:59
    
@Jack: My argument for why $\mathbb{Z}$ discriminates is wrong; I'm fixing it. –  Arturo Magidin Mar 24 '11 at 0:10

Solvable groups of derived length at most n form a variety. The direct product of copies of a single solvable group is therefore solvable. Any extension of a solvable group by a solvable group is solvable, so wreath products of solvable groups are solvable (see Rotman's Introduction to the Theory of Groups, Theorem 5.17, page 103ish). In fact H wr K has derived length at most the sum of the derived lengths of H and K.

In the converse direction, any extension of H by K is contained as a subgroup of H wr K (see Rotman's Introduction to the Theory of Groups, theorem 7.37, page 187ish). I'm not sure about the infinite solvable groups, but for finite solvable, they are iterated extensions of cyclic groups, so contained as subgroups of iterated wreath products of cyclic groups.

Since every abelian group is a quotient of a direct product of cyclics, it seems believable this works in general for solvable groups.

I'm pretty sure the variety generated by wreath products H wr K of cyclic groups H and K does not contain all solvable groups. In particular, H wr K has derived length 2 and so is contained in the variety of all metabelian groups. In other words, they must mean iterated wreath products, H wr K wr L, etc.

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In Scott's Group Theory, "wreath of solv is solv" is 9.2.5, page 215. –  Jack Schmidt Mar 23 '11 at 21:50
    
I don't quite see the converse for infinite solvable groups. If X ≤ G and Y ≤ H, then X wr Y ≤ G wr H, but if X is a section of G and Y is a section of H, is X wr Y a section G wr H? I don't understand how to take quotients willy-nilly in the middle. –  Jack Schmidt Mar 23 '11 at 22:03

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