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Suppose that $(G,+)$ is a locally compact abelian group.

Let $G_n:=\{g\in G : ng=0_G\}$ and let $\chi:G\rightarrow S^1$ be a continuous character.

Assuming that $\chi(G_n) = 1_{S^1}$. Is it necessarily true that there exists a character $\tau:G\rightarrow S^1$ such that $\chi = \tau^n$?

Some progress: If $G=(\mathbb{R}/\mathbb{Z},+)$ the answer is yes. Every character of $\mathbb{R}/\mathbb{Z}$ is given by $x\mapsto mx$ for some $m\in\mathbb{N}$. The assumption implies that $n$ divides $m$ which means that there's an $n$'th root ($x\mapsto \frac{m}{n}x$).

If $G$ is a finite group, then the answer is still yes. Here's a proof when $n$ is a prime:

Write $G=\mathbb{Z}/{n^{k_1}}\mathbb{Z}\oplus...\oplus \mathbb{Z}/{n^{k_l}}\mathbb{Z} \oplus T$ for some torsion group $T$ whose order is co-prime to $n$. Hence $G_n= \mathbb{Z}/n\mathbb{Z}^l \leq \mathbb{Z}/{n^{k_1}}\mathbb{Z}\oplus...\oplus \mathbb{Z}/{n^{k_l}}$.

Any character on $T$ has an $n$'th root. (because if $(n,m)$ are co-prime, any generator of $\mathbb{Z}/m\mathbb{Z}$ has an $n$'th fraction). So let's assume that $T$ is trivial. Also we can deal with each $\mathbb{Z}/{n^{k_i}}\mathbb{Z}$ separately so let's assume there's only one.

Now we're done because any character $\chi:\mathbb{Z}/{n^k}\mathbb{Z}\rightarrow S^1$ that is trivial on $\mathbb{Z}/n\mathbb{Z}$ can be extended to a character on $\mathbb{R}/\mathbb{Z}$ which is trivial on $C_n$.

The theorem holds for some non-compact groups as well, such as $\mathbb{Z}$ or $\mathbb{R}$ in which cases we have $\hat {\mathbb{Z}} = S^1$,$\hat{\mathbb{R}}=\mathbb{R}$ which are divisible. In these cases $G_n$ is trivial (for any $n$) and we see that we always have $n$'th roots. More generally if $G$ is torsion free then $\hat G$ is connected and so divisible.

tomasz
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Yanko
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    $\chi(G_n)=0_G$ doesn't make sense, you mean $=1_{S^1}$ or $=0_{S^1}$, according to whether you write $S^1$ additively or multiplicatively (probably multiplicatively since you write $\tau^n$) – YCor Mar 31 '19 at 16:26
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    @YCor: If $\lvert G\rvert=2$, then $G_2=G$. $G_n$ is not the $n$-th powers, it is the $n$-torsion. – tomasz Apr 02 '19 at 11:31
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    Thanks indeed I made a confusion. I posted a hopefully more useful answer. – YCor Apr 02 '19 at 13:15

1 Answers1

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The answer is yes in a quite large generality ((a) below, including discrete groups) and false in general, even among $\sigma$-compact locally compact abelian groups.

a) It's true if $G\stackrel{n}\to G$ is a strict homomorphism, in the sense that its image $G^n$ is closed and $G\stackrel{n}\to G^n$ is open (if $G^n$ is closed the latter assumption is automatic if $G$ is $\sigma$-compact).

We have the exact sequence $0\to G_n\to G\stackrel{n}\to G^n\to 0$. Pontryagin duality says that the induced exact sequence $0\to\mathrm{Hom}(G^n,S^1)\stackrel{n}\to\mathrm{Hom}(G,S^1)\to\mathrm{Hom}(G_n,S^1)\to 0$ is exact too. Exactness at the middle precisely means the desired statement.

b) So one should find an example for which it's not strict. For instance, choose $n=p$ prime, and consider the semirestricted power of $\mathbf{Z}_p$ over $p\mathbf{Z}_p$, defined below.

Here, if a $G$ is a group and $H$ a subgroup and $I$ a set, the semirestricted power $G^{[I,H)}$ is the subgroup of $G^I$ consisting of those $(g_i)$ such that $g_i\in H$ for all but finitely many $I$. It contains the subgroup $H^I$, as well as $G^{(I)}=G^{[I,\{1\})}$. Here we take $G=\mathbf{Z}_p$, $H=p\mathbf{Z}_p$, $I$ infinite, and endow the resulting semirestricted power $G$ with the group topology for which the compact group $(p\mathbf{Z}_p)^I$ is a compact open subgroup (the discrete quotient being $\mathbf{Z}/p\mathbf{Z})^{(I)}$).

So $G$ is torsion-free and we have an exact sequence $$0 \to(\mathbf{Z}/p\mathbf{Z})^{I}\to \mathrm{Hom}(G,S^1)\to C_{p^\infty}^{(I)}\to 0.$$ In this dual, clearly multiplication by $p$ has a countable image, so is not surjective.

YCor
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