Let $G$ be the multiplicative group of all complex $2^n$th roots of unity, $n=0, 1, 2,\dots$. Prove that [. . .] every subgroup is marginal.

Question

This is part of Exercise 2.3.3 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE. However, the first part of the exercise is discussed here.

The Details:

On page 57 of Robinson's book:

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 Question:

Let $G$ be the multiplicative group of all complex $2^n$th roots of unity, $n=0, 1, 2,\dots$. Prove that [. . .] every subgroup is marginal.

Thoughts:

With respect to what collection of words $W$ is each subgroup of $G$? Would it be (some union of) $$W_m=\{x_1^{2^m}\}$$ for some $m\in\Bbb N\cup\{0\}$? I don't yet know where to go from here.

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The exercise is at the end of the chapter on free groups and presentations, and, since I am studying for a postgraduate research degree in combinatorial group theory, it concerns me that I can’t do it yet.

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A related concept is that of a verbal subgroup; indeed, the first part of the exercise is about them. I have encountered these before in, say, Magnus et al.'s, "Combinatorial Group Theory: [. . .]".

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The type of answer I am looking for is a detailed hint or a full solution, with an eye to understanding marginal subgroups in general.

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Please help :)

Answer 1

If $G$ is an abelian group, then its verbal subgroups are of the form $W(G)$ where $W=\{x^m\}$ for some nonnegative integer $m$. For example, this follows from B.H. Neumann’s old result that every group word is equivalent to a commutator word and a power word. You can see a sketch of that argument here. Now, because the Prüfer groups are divisible, that means that they are verbally simple: the only verbal subgroups are the whole group and the trivial group.

What is the marginal subgroup associated to the word $w(x)=x^m$? It consists precisely of the elements $z\in G$ such that $w(x)=w(zx)=w(xz)$. That means, since $G$ is abelian, that we must have $$x^m = z^mx^m=x^mz^m$$ which means that the marginal subgroup of $G$ associated to $w(x)=x^m$ consists precisely of $\{z\in G\mid z^m=e\}$.

Here, every proper subgroup of $G$ is finite and cyclic of order a power of $2$. Thus, the subgroup of order $2^k$ will be precisely the marginal subgroup associated to $x^{2^k}$. The whole group corresponds to the marginal subgroup associated to $x^0$.

When $m$ is relatively prime to $2$, you get the elements that satisfy $x^m=e$, which is just the trivial group.

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Marginal subgroups in general are a bit tricky. The $i$th marginal group associated to the word $w(x_1,\ldots,x_n)$, $w_i^*(G)$ consists of the $z\in G$ such that $$\begin{align*} w(x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n) &= w(x_1,\ldots,x_{i-1},zx_i,x_{i+1},\ldots,x_n)\\ &= w(x_1,\ldots,x_{-1},x_iz,x_{i+1},\ldots,x_n). \end{align*}$$ The marginal subgroup associated to $w$ is then $\cap_{i=1}^n w_i^*(G)$. Even Philip Hall messed it up at one point: his initial paper introducing marginal subgroups says that the marginal subgroup associated to $w(x,y)=[x,y]$ is the center (true), and that the marginal subgroup associated to $w(x)=x^n$ consists of the central elements of exponent $n$; but this is not necessarily true, since if $G$ is of exponent $n$, then the marginal subgroup is everything (since the verbal subgroup is trivial).

But for abelian groups, since every word amounts to a power word $x^k$, it will just be the elements of exponent $k$. So the marginal subgroups of an abelian group $G$ are precisely the subgroups $G_{(k)} = \{g\in G\mid g^k=e\}$.