Show that ${\rm Var}\, \mathfrak{X}$ is a variety.

Question

This is part of Exercise 2.3.7 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to this MSE search, it is new to MSE.

(Note that "variety" here is in the sense of universal algebra.)

The Details:

Since definitions vary, on page 15, ibid., paraphrased, it states that

A subgroup $N$ of $G$ is normal in $G$ if one of the following equivalent statements is satisfied:

(i) $xN=Nx$ for all $x\in G$.

(ii) $x^{-1}Nx=N$ for all $x\in G$.

(iii) $x^{-1}nx\in N$ for all $x\in G, n\in N$.

On page 56, ibid.,

Let $F$ be a free group on a countably infinite set $\{x_1,x_2,\dots\}$ and let $W$ be a nonempty subset of $F$. If $w=x_{i_1}^{l_1}\dots x_{i_r}^{l_r}\in W$ and $g_1,\dots, g_r$ are elements of a group $G$, we define the value of the word $w$ at $(g_1,\dots,g_r)$ to be $w(g_1,\dots,g_r)=g_1^{l_1}\dots g_{r}^{l_r}$. The subgroup of $G$ generated by all values in $G$ of words in $W$ is called the verbal subgroup of $G$ determined by $W$,

$$W(G)=\langle w(g_1,g_2,\dots) \mid g_i\in G, w\in W\rangle.$$

On page 57, ibid.,

If $W$ is a set of words in $x_1, x_2, \dots$ and $G$ is any group, a normal subgroup $N$ is said to be $W$-marginal in $G$ if

$$w(g_1,\dots, g_{i-1}, g_ia, g_{i+1},\dots, g_r)=w(g_1,\dots, g_{i-1}, g_i, g_{i+1},\dots, g_r)$$

for all $g_i\in G, a\in N$ and all $w(x_1,x_2,\dots,x_r)$ in $W$. This is equivalent to the requirement: $g_i\equiv h_i \mod N, (1\le i\le r)$, always implies that $w(g_1,\dots, g_r)=w(h_1,\dots, h_r)$.

[The] $W$-marginal subgroups of $G$ generate a normal subgroup which is also $W$-marginal. This is called the $W$-marginal of $G$ and is written $$W^*(G).$$

On page 57, ibid.,

A [. . .] class of groups $\mathfrak{X}$ is a class - not a set - whose members are groups and which enjoys the following properties: (i) $\mathfrak{X}$ contains a group of order $1$; and (ii) $G_1\cong G\in\mathfrak{X}$ always implies $G_1\in\mathfrak{X}$.

On page 58, ibid.,

If $W$ is a set of words in $x_1, x_2, \dots $, the class of all groups $G$ such that $W(G)=1$, or equivalently $W^*(G)=G$, is called the variety $\mathfrak{B}(W)$ determined by $W$.

The Question:

If $\mathfrak{X}$ is any class of groups, define ${\rm Var}\,\mathfrak{X}$ to be the intersection of all varieties that contain $\mathfrak{X}$. Prove that ${\rm Var}\,\mathfrak{X}$ is a variety.

Thoughts:

My first impression was that this exercise is relatively easy. The proof ought to run along the same lines as any old proof of an intersection of whatevers being a whatever. Digging a little deeper, though, things are not so simple.

At an extreme case, consider $\mathfrak{X}=\{1\}$, where $1$ is the trivial group up to isomorphism. Since, for any set of words $W$, we have $W(1)=1$, we can deduce that each variety $\mathfrak{B}(W)$ of groups contain $\mathfrak{X}$, which implies ${\rm Var}\, \{1\}$ is simply the variety of all groups.

One approach might be to let $G\in{\rm Var}\, \mathfrak{X}$ for some class $\mathfrak{X}$. Then the aim would be to show that, since $G$ is in each variety $\mathfrak{W}$ that contains $\mathfrak{X}$, we have $\mathfrak{W}=\mathfrak{B}(W)$ for some set of words $W$, so that . . . Yeah, this is where I'm stuck. See: I need to construct an appropriate set $W_{\mathfrak{X}}$ of words for which

$${\rm Var}\, \mathfrak{X}=\mathfrak{B}(W_{\mathfrak{X}}).$$

I'm not sure . . .

I have a hunch that something like

$$W_{\mathfrak{X}}=\bigcap_{W: \mathfrak{X}\subseteq \mathfrak{B}(W)}W$$

might work.

Nowhere have I used the definition of a class of groups.

Please help :)

Answer 1

Suppose $\mathbf{V}_1$ is the variety determined by the set of words $W_1$; and $\mathbf{V}_2$ is the variety determined by the set of words $W_2$. So the groups of $\mathbf{V}_1$ are precisely the groups for which $w(G)=\{e\}$ for all $w\in W_1$; and the groups in $\mathbf{V}_2$ are precisely the groups for which $w(G)=\{e\}$ for all $w\in W_2$.

What are the groups in $\mathbf{V}_1\cap\mathbf{V}_2$? Precisely the groups $G$ for which $w(G)=\{e\}$ for all $w\in W_1\cup W_2$. So $\mathbf{V}_1\cap\mathbf{V}_2$ is the variety determined by the set of words $W_1\cup W_2$.

The same is true for an arbitrary intersection: if $\mathbf{V}_i$ is defined by the set of words $W_i$, then $\cap\mathbf{V}_i$ is determined by the set of words $\cup W_i$.

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More generally, what you have here is a Galois connection. Consider the partially ordered set of subsets of $F_{\omega}$, the free group on countably many variables, interpreted as group words. And consider the class (or set in an appropriate universe) of collections of groups.

Given a subset $J$ of $F_{\omega}$, we can consider the collection $J^*$ of all groups $G$ such that $w(G)=\{e\}$ for each $w\in J$; that is, the groups that satisfy every word in $J$. And given a collection of groups $\mathfrak{G}$, we can let $\mathfrak{G}^*$ be the collection of all elements $w$ of $F_{\omega}$ such that $w(G)=\{e\}$ for all $G\in\mathfrak{G}$. Note that for $J\subseteq F_{\omega}$ and $\mathfrak{G}$, we have $$\mathfrak{G}\subseteq J^*\iff J\subseteq\mathfrak{G}^*.$$ Indeed: for every $G\in\mathfrak{G}$ we have $w(G)=\{e\}$ for all $w\in J$, if and only if every $w\in J$ has the property that $w(G)=\{e\}$ for all $G\in\mathfrak{G}$.

Thus we have a Galois connection, which establishes an anti-isomorphism between the “closed” subsets of $F_{\omega}$, and the “closed” collections of groups. The closed collections of groups are the varieties; the closed collections of $F_{\omega}$ are the equational theories. The anti-isomorphism correspondence tells you that the intersection of varieties is the variety corresponding to the union of subsets.

Answer 2

I think you're making this more complicated than it needs to be (per your analysis of $\{1\}$, maybe you're mixing up union and intersection?). Suppose $\{V_i\}_{i\in I}$ is a collection of varieties. Let $E_i$ be the equational theory of $V_i$, so that $V_i$ is exactly the class of models of $E_i$. Then the intersection $V:=\bigcap_{i\in I}V_i$ is exactly the class of models of $\bigcup_{i\in I}E_i$; since the latter is a set of equations, we get that $V$ is a variety.

This indeed uses nothing about groups. Details about groups enter the picture when we look for concrete(ish) descriptions of various varieties of groups, but they play no role in this general structural result.

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The above uses the "model of a set of equations" definition of a variety. If instead we define a variety as a class of structures closed under taking homomorphisms, substructures, and arbitrary products, then the result is a consequence of the more general fact that a collection of classes of structures defined by closure conditions is automatically closed under arbitrary intersections.