This is Exercise 2.3.4 of Robinson's "A Course in the Theory of Groups (Second Edition)". My universal algebra is a little rusty, so this question is not what I'm looking for; besides, I ought to be able to use tools given in Robinson's book.
The Details:
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 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:
Prove that every variety is closed with respect to forming subgroups, images, and subcartesian products.
Thoughts:
This must have made sense to me at least two times (although it took some searching to confirm it):
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When I read Chapter 11 of "A Course in Universal Algebra" by Burris and Sankappanavar several years ago; this is tackled, as I said, here.
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When I read Chapter 12 of Roman's "Fundamentals of Group Theory: An Advanced Approach" a few months ago. This is how I found out it's part of Birkhoff's Theorem. In the proof there, however, is the ever-infuriating "It is clear"; also: images are not covered.${}^\dagger$
I don't know why I've been stuck on it the last few days, but I have, so here I am.
My rough understanding is that the processes of taking of subgroups, images, and subcartesian products mean they each inherit the words in $W$, but this "understanding" is nothing more, I fear, than a restatement of the question.
The chapter in Robinson's book is on free groups and presentations, so I've looked in many of my combinatorial group theory books, like Magnus et al., but the theorem is nowhere obvious.
(I hope this is enough context.)
I'm looking for something more than "it is clear" and less than a deep dive into universal-algebra.
Please help 🙂
$\dagger$ But quotients are, which, I suppose, is equivalent . . . I don't know.
Best Answer
Let $W$ be a set of words, let $\mathfrak{W}$ be the corresponding variety.
The key observation is that for any word $w(x_1,\ldots,x_n)$, any groups $G$ and $K$, and any morphism $f\colon G\to K$, $$f(w(g_1,\ldots,g_n))=w(f(g_1),\ldots,f(g_n)).$$ This follows because $w(x_1,\ldots,x_n)$ is just an element of the free group $F_n$, and the value of $w$ is the image of that element under the unique homomorphism $F_n\to G$ induced by the assignment $x_i\mapsto g_i$; while $f(w(g_1,\ldots,g_n))$ is therefore the value of $w$ under the unique morphism $F_n\to K$ induced by the composite assignment $x_i\mapsto g_i\mapsto f(g_i)$.
In particular, for any groups $G$ and $K$ and any morphism $f\colon G\to K$, we have $f(W(G))\subseteq W(K)$, since the image of the generators of $W(G)$ lie in $W(K)$.
Note that any class of groups that is closed under subgroups will be closed under subcartesian products if and only if it is closed under cartesian (arbitrary unrestricted) products. Indeed, if $\{G_{\lambda}\}_{\lambda\in\Lambda}$ is a family of groups in the class, the cartesian product is a special case of the subcartesian product, and the subcartesian product is a subgroup of the cartesian.
So it suffices to show that: (i) If $\{G_{\lambda}\}_{\lambda\in\Lambda}$ is a family of groups in $\mathfrak{W}$, then $\mathop{\mathrm{Cr}}\limits_{\lambda\in\Lambda} G_{\lambda}$ lies in $\mathfrak{W}$; (ii) that if $G\in\mathfrak{W}$ and $H\leq G$, then $H\in\mathfrak{W}$; and (iii) that if $G\in\mathfrak{W}$ and $\varphi\colon G\to K$ is a surjective homomorphism of groups, then $K\in\mathfrak{W}$.
Assume that $\{G_{\lambda}\}_{\lambda\in\Lambda}$ is a family of groups with $G_{\lambda}\in\mathfrak{W}$ for every $\lambda$; that means that $W(G_{\lambda})=\{e_{\lambda}\}$ for every $\lambda$. Let $w(x_1,\ldots,x_n)\in W$, and let $\mathbf{y}_1,\ldots,\mathbf{y}_n\in \mathop{\mathrm{Cr}}\limits_{\lambda\in\lambda}G_{\lambda}$. We want to show that $w(\mathbf{y}_1,\ldots,\mathbf{y}_n)=e$. Indeed, recall that the operations are componentwise; thus, if $\pi_{\lambda}$ is the projection on the $\lambda$ coordinate, which means that $\pi_{\lambda}(\mathbf{y}_i) = \mathbf{y}_i(\lambda)\in G_{\lambda}$ (since the elements of the cartesian product are functions from $\Lambda$ to $\cup G_i$ with the value at $\lambda$ in $G_{\lambda}$). Therefore, for each $\lambda$, $$\begin{align*} \pi_{\lambda}(w(\mathbf{y}_1,\ldots,\mathbf{y}_n)) &= w(\pi_{\lambda}(\mathbf{y}_1),\ldots,\pi_{\lambda}(\mathbf{y}_n))\\ &= w(\mathbf{y}_1(\lambda),\ldots,\mathbf{y}_n(\lambda))\\ &\in W(G_{\lambda}). \end{align*}$$ But $W(G_{\lambda})=\{e_{\lambda}\}$ because $G_{\lambda}\in\mathfrak{W}$, so $\pi_{\lambda}(w(\mathbf{y}_1,\ldots,\mathbf{y}_n) = e_{\lambda}$. As this occurs for every $\lambda$, we conclude that $w(\mathbf{y}_1,\ldots,\mathbf{y}_n)=e$. Thus, $W(\mathop{\mathrm{Cr}} G_{\lambda}) \in \mathfrak{W}$, as desired.
Let $G\in\mathfrak{W}$, $H\leq G$. If $w(x_1,\ldots,x_n)\in W$, and $h_1,\ldots,h_n\in H$, then $h_i\in G$, so $w(h_1,\ldots,h_n)\in W(G)=\{e\}$; hence $W(H)=\{e\}$, and thus $H\in\mathfrak{W}$. Alternatively, the inclusion map $\iota\colon H\to G$ has $\iota(W(H))\subseteq W(G)=\{e\}$, hence $W(H)=\{e\}$.
If $G\in\mathfrak{W}$ and $f\colon G\to K$ is surjective, I claim that $f(W(G))=W(K)$. Indeed, let $w(x_1,\ldots,x_n)\in W$, and let $k_1,\ldots,k_n\in K$. Then there exist $g_i\in G$ such that $f(g_i)=k_1$, and hence $$\begin{align} w(k_1,\ldots,k_n) &= w(f(g_1),\ldots,f(g_n))\\ & = f(w(g_1),\ldots,w(g_n))\\ &\in f(W(G)). \end{align}$$ Since the words $w(k_1,\ldots,k_n)$ generete $W(K)$, and $f(W(G))$ is a subgroup, it follows that $W(K)\subseteq f(W(G))$. But as we observed above, $f(W(G))\subseteq W(K)$ , giving equality. Thus, $W(K) = f(W(G))=f(\{e\}) = \{e\}$, so $W(K)=\{e\}$ and hence $K\in\mathfrak{W}$, as desired.