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

Question

Hi: Let $G$ be the multiplicative group of all complex $2^n$th roots of unity, $n=0,1,2,...$ Prove that $1$ and $G$ are the only verbal subgroups of $G$.

Let $F$ be a free group on a countably finite set $\{x_1,x_2,...\}$. I must prove that for any nonempty subset $W$ of $F$, either $W(G)=G$ or $W(G)=1$. Let's try with $W=\{x_1^p\}$, $p$ a prime, $p\neq 2$. I denote the roots of unity by there argument alone. $W(G)=\langle g^p:g\in G\rangle=\{g^p:g\in G\}$ because $G$ is abelian. I'll show that $z=2\pi/2^n\in W(G)$ by proving there is a $p$th root of unity $z_1$ of $z$ that belongs to $G$ and so $z=z_1^p$ will belong to $W(G)$.

As $(2^n,p)=1$ there exist $l,k$ such that $pl-2^nk=1$, $(1+2^nk)/p=l$ and $z_1:=z^{1/p}=2\pi/(2^np)+2k\pi/p=2\pi l/2^n\in G$.

Although I proved it I cannot understand why a $p$th root of an element of $G$ belongs to $G$. It does not matter. What I am unable to prove is the general case, i.e. $W$ any subset of $F$. Any suggestion?

Answer 1

According to groupprops, the verbal subgroups of an abelian group $G$ are the subgroups $G_n=\{g^n\mid g\in G\}$, where $n\in\mathbb{N}$.

You can easily see that for the group $G=\{\exp(a/2^b)\mid a,b\in\mathbb{N}\}$ and for any nonzero $n$, the map $$G\to G, ~g\mapsto g^n$$ is onto, so that $G_n=G$. Indeed write $n=2^km$ with odd $m$: exponentiation by $2^k$ is onto since $\exp(a/2^b)=\exp(a/2^{b+k})^{2^k}$, and exponentiation by $m$ is onto (bijective, actually): indeed, multiplication by $m$ is an automorphism of $\mathbb{Z}/2^b\mathbb{Z}$ which is isomorphic to the group of $2^b$-th roots of unity. So if $mc=a\mod 2^b$, then $$ \exp(c/2^{b+k})^{n}= \exp(c/2^{b})^{m}= \exp(a/2^b). $$

Answer 2

Careful: it is not true that a $p$-th root of an element of $G$ is necessarily in $G$; what is true is that every element of $G$ has a $p$-th root in $G$. That’s a different statement.

Indeed, note that if you take $z$ to be a primitive sixth root of unity, then its cube is a primitive square root of unity. Thus, $z^3\in G$. That is, $z$ is a cubic root of an element of $G$. However, $z$ is not in $G$, because it is not a $2^n$th root of unity for any $n$. So it is false that a $p$th root of an element of $G$ is necessarily in $G$.

However, if $g\in G$, then $g$ has order $2^n$ for some $n$. Since $\gcd(p,2^n)=1$, then we can write $1=2^nr + ps$ for some integers $r$ and $s$. Then $$g = g^1 = g^{2^nr+ps} = (g^{2^n})^rg^{ps} = (g^s)^p.$$ Thus, $g^s$ is a $p$th root of $g$, and it clearly lies in $G$ because $g$ does.

So every element of $G$ has a $p$th root in $G$; which is not the same as saying that every $p$th root of an element of $G$ is also in $G$. That’s not true.

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Every word is equivalent to a power word and a commutator word; that is, the verbal subgroup determined by a single word $w$ is equivalent to the verbal subgroup determined by two words, $w_1,w_2$, where $w_1(x)=x^n$ is a power word, and $w_2$ is an element of the commutator subgroup of $F_{\infty}$.

But commutator words evaluate to the trivial element in any abelian group. And the verbal subgroup determined by a collection of power words $\{x^{a_1},\ldots,x^{a_k},\ldots\}$ is the verbal subgroup determined by the word $x^m$ with $m$ the gcd of $a_1,a_2,\ldots$. Thus, the only verbal subgroups of an abelian group $G$ are the power subgroups, $G^n = \{g^n\mid g\in G\}$.

Now, I interpret your group as being the group of all complex numbers $z$ for which $z^{2^k}=1$ for some $k$; because if you fix $n$, the corresponding group is simply cyclic of order $2^n$, and then $G^2$ is a proper subgroup of $G$ that is both nontrivial and verbal, if $n\gt 1$.

That means that your group is isomorphic to the Prufer $2$-group, $C_{2^{\infty}}$. It is well known that the subgroups of this group are either finite or the whole group. Since the kernel of the map $g\mapsto g^k$ is finite (it consists of the $k$th roots of unity among the elements of $G$), the image must be infinite, and hence must be the whole group. Thus, every nontrivial power verbal subgroup of $G$ gives the whole group. The verbal subgroup determined by $[x,y]$ gives the identity.

If you are not familiar enough with the Prufer $2$-group to use the argument above, it is enough to show that for any $g\in G$ and any $n$, there is an element $x$ in $G$ such that $x^n=g$.

If $\gcd(n,2)=1$, then this follows as above: take $|g|=2^m$, write $1=2^mr + sn$, and then note that $$g = g^{2^mr+sn} = (g^{2^m})^rg^{sn} = (g^s)^n,$$ so $g$ is an $n$th power.

Otherwise, $n=2^rk$ with $\gcd(2,k)=1$. We already know there is an element $x\in G$ such that $x^k=g$, so it suffices to find a $2^r$th root of $x$. That is, we just need to show that you can always find a square root of an arbitrary element of $G$ in $G$ (by simply iterating).

So let $x\in G$. We need to find an element $y$ of $G$ such that $y^2=x$. But here it does work: if $y$ is either complex square root of $x$, then $y^2=x$, and if $|x|=2^n$ then $y^{2^{n+1}}=(y^2)^{2^n} = x^{2^n} = 1$. So $y$ is a $2^{n+1}$th root of unity, and so it lies in $G$.

Thus, $G^k=G$ for all $k\neq 0$, which is what you want to prove.