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.
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 🙂
$\dagger$: 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 . . .
Best Answer
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.
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.