How do I translate functions on the Prufer 2-group between functions on the $2^n$ roots of unity and the dyadic fractions modulo 1?

Question

How do I translate functions on the Prufer 2-group between functions on the $2^n$ roots of unity and the dyadic fractions modulo 1?

Let the Prufer 2-group be denoted $\mathbb{Z}(2^{\infty})_2$ and think of it as the dyadic fractions modulo $1$.

And let the Prufer 2-group denoted $\mathbb{Z}(2^{\infty})$ be defined as the $2^n$ roots of unity.

What is the morphism $$\mathbb{Z}(2^{\infty})_2\to\mathbb{Z}(2^{\infty})$$

Additionally, on the dyadic fractions i have the function $\lvert\cdot\rvert_2$ which I need to translate into $\mathbb{Z}(2^{\infty})$

It looks like it's going to be something like $\lvert\cdot\rvert_2\to e^{2i\theta}$ or possibly even $\lvert\cdot\rvert_2\to e^{\sqrt{i\theta}}$

This function has an interesting property on this group; which is something like "Measured by this metric, the group contains its own reflection." Is there some known, precise, statement about this function in relation this group that stands out as interesting? I am imagining this statement may be related to the concept of the Pontyragin dual but that may be way off.

A slightly descriptive answer is likely to be more useful to me than one only in algebra.

Answer 1

I assume you are asking for an isomorphism

$$f: (\Bbb{Z}[\frac{1}{2}]/\Bbb{Z}, \; +) \simeq (\mu_{2^\infty}(\Bbb{C}), \; \cdot)$$

(left hand side: dyadic fractions modulo 1, right hand side: all $2^n$-th roots of unity in $\Bbb{C}$)

No big deal. Try the most obvious one defined by $$f:\displaystyle \frac{1}{2^n} +\Bbb{Z} \mapsto exp\left(\frac{2\pi i}{2^n}\right).$$

(There are others: $$f_k = \displaystyle \frac{1}{2^n} +\Bbb{Z} \mapsto exp\left(k\frac{2\pi i}{2^n}\right)$$ works for every odd $k$ ($f$ is just $f_1$), actually it works for every $k \in \Bbb{Z}_2^*$. Try $k=3$ to see what happens. This is the actual place where the Pontryagin dual makes its entrance.) _{See "Added" below for more detail.}

As for "translating", I assume you want to pull the $2$-adic absolute value on the LHS over to the RHS. Formally, you define $|\cdot|'_2$ on $\mu_{2^\infty}(\Bbb{C})$ via $|\xi|'_2 := |f^{-1}(\xi)|_2$. Again, nothing spectacular happens: $$|\xi|'_2 = 2^n \Leftrightarrow \xi \text{ is a }primitive \:2^n\text{-th root of unity}.$$

Added: To say a bit more about the "$k$" above. For every $r\in \Bbb{Z}$, one has an endomorphism of the group to itself $x \mapsto rx$ (on the LHS) resp. $x\mapsto x^r$ (on the RHS). So far that is true for any abelian group. Now it turns out that in this case, one can even take $r$ from the much bigger set of $2$-adic integers $\Bbb{Z}_2 \supsetneq \Bbb{Z}$. Namely, this still makes sense because for each $x$ in the LHS (resp. RHS) there is some $n$ such that $2^nx = 0$ (resp. $x^{2^n} =1$), so to multiply (resp. exponentiate) that $x$ with $r\in \Bbb{Z}_2$ is just defined to be the same as $r'x$ (resp. $x^{r'}$) for an $r' \in \Bbb{Z}$ with $r' \equiv r$ mod $2^n$. One finds that all endomorphisms of the group are of that kind,

$$End(\Bbb{Z}[\frac{1}{2}]/\Bbb{Z}) = Hom(\Bbb{Z}[\frac{1}{2}]/\Bbb{Z}, \Bbb{Z}[\frac{1}{2}]/\Bbb{Z}) \simeq \Bbb{Z}_2 \qquad (*)$$

This is (a version of) Pontryagin duality.

Now, not all those endomorphisms are bijective (=automorphisms). In fact, exactly the ones with $r\in \Bbb{Z}_2^*$ are. (In fact, the identification (*) is a ring isomorphism, where composition in $End(\Bbb{Z}[\frac{1}{2}]/\Bbb{Z})$ corresponds to multiplication in $\Bbb{Z}_2$.) Whereas for $r \notin \Bbb{Z}_2^*$, the map $x \mapsto rx$ (resp. $x\mapsto x^r$) is not injective. Fun exercise: Find the kernel for $r=2^n$. Look at the example $r=2$. Second fun exercise: On the RHS, the endomorphism corresponding to $r=-1$ does the same as complex conjugation.

That all being said, we now have that for each $k\in \Bbb{Z}_2^*$, the map $\alpha_k: x \mapsto kx$ (resp. $\beta_k: x\mapsto x^k$) is a group automorphism -- i.e. an isomorphism of the group to itself -- on the LHS (resp. RHS). Now of course composing an isomorphism with an automorphism of the domain or codomain gives an isomorphism again. Now just notice that the above $f_k$ is nothing else than

$$ f_k = f\circ \alpha_k = \beta_k \circ f.$$

Final note: All this works, mutatis mutandis, if one replaces $2$ by any prime $p$. Also compare https://math.stackexchange.com/a/178771/96384 for the $p$-free version.