Show that the group of all complex $2^n$th roots of unity, $n=0,1,2,...$, has a fully-invariant subgroup which is not verbal

Question

This is the last part of Exercise 2.3.3 of Robinson's "A Course in the Theory of Groups (Second Edition)". The following are the other parts.

1. Verbal subgroups of the group of all $2^n$th roots of unity.

2. 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.

The Details:

On page 28 of Robinson's book, slightly edited, is:

A subgroup $H$ of a group $G$ is said to be fully-invariant if $\alpha (H)\le H$ for all $\alpha \in{\rm End}(G)$.

Here ${\rm End}(G)$ is the set of all endomorphisms of $G$; that is, all homomorphisms from $G$ to $G$.

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 $G$. 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 $W$,

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

The Question:

Let $G$ be the multiplicative group of all complex $2^n$th roots of unity, $n=0,1,2,\dots$. [. . .] Show [. . .] that $G$ has a fully-invariant subgroup which is not verbal.

Thoughts:

By part (1) of the exercise, it would be sufficient to find a nontrivial, proper subgroup of $G$ that is fully-invariant.

The subgroup cannot be cyclic${}^\dagger$ but, since $G$ is abelian, it must be abelian as well.

Beyond these simple observations, I'm stuck.

Further Context:

As I said in my question on part (2): the exercise is at the end of the chapter of Robinson's book about presentations and free groups; so, basically, it's about combinatorial group theory; and my postgraduate studies are in & around that discipline.

The type of answer I'm looking for is not just a naming of the subgroup. I want to understand as much as possible. I would like to know what thought processes went into finding the subgroup.

Please help :)

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$\dagger$: From the comments . . . My reasoning is that the order of a cyclic subgroup of $G$ would be a power of two, say $2^m$, so that $W=\{ x^{2^m}\}$ would make it verbal . . .

Answer 1

As we've discussed before, the verbal subgroups correspond to words $x^m$, and since this group is divisible, the only verbal subgroups are the trivial (when $m=0$), and the whole group (when $m\neq 0$).

So this question is really just asking you to find a proper nontrivial subgroup that is fully invariant.

The proper subgroups are precisely the subgroups of the form $$H_n = \{x\in G \mid x^{2^n}=e\}$$ for $n=0,1,\ldots$, with the proper nontrivial ones being those with $n\neq 0$.

Note that if $x\in H_n$ and $\varphi\colon G\to G$, then $e = \varphi(x^{2^n}) = (\varphi(x))^{2^n}$, hence $\varphi(x)\in H_n$.

So all of the subgroups are fully invariant.

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You got turned around with the definition of verbal subgroup. The verbal subgroup associated to $w(x)=x^k$ is the subgroup generated by all values of the word. that is, it would be $G^{k} = \{g^k\mid g\in G\}$, rather than the set $\{x\in G\mid w(x)=1\}$.

(A subgroup for which $w(x)=1$ for all $x$ would of course lie in the variety determined by $w$, in this case, "groups of exponent $k$")

Note that in any group $G$, for any $k\gt 0$, the subgroup generated by all elements of exponent $k$ is fully invariant, by the argument above.