Independence of some events with respect to Haar measure

Question

In order to fulfill details in a proof, I came across with the following fact which, whenever true, allows me to conclude the line of reasoning.

Here is the stage: I have an abelian group $G=\Bbb{Q}^n/\Bbb{Z}^n$ (here $n$ is a positive integer) and it is equipped with the discrete $\sigma$-algebra. Furthermore, let us denote by $\Gamma_n$ the dual of $G$, i.e. the group of all group homomorphisms form $G$ to the complex unit circle $T=\{z\in\Bbb{C}\colon \lvert z\rvert=1\}$ and denote by $\mu$ the probability Haar measure on $\Gamma_n$.

What I want to prove is the following: the events $A=\{\psi\in\Gamma_n\mid \psi(a+\Bbb{Z}^n)=0\}$ and $B=\{\psi\in\Gamma_n\mid \psi(b+\Bbb{Z}^n)=0\}$ are $\mu$-independent whenever $a,b\in\Bbb{Q}^n\setminus \Bbb{Z}^n$ are linearly independent over $\Bbb{Z}$, i.e. $$\mu(A)\cdot \mu(B)=\mu(A\cap B).$$

Although I spent on this much time, I got the least in the sense that I'm stuck. In some sense, that I'm not able to make more precise, I expect all the sets $A$, $B$ and $A\cap B$ have $0$ measure wrt $\mu$.

Any suggestion or contribution is appreciated.

Edit As @AlexRavsky remarked below, the correct (and more general) notion of independence to be used here is the following: $a,b$ are lineraly independent iff $ma+kb=0$ implies $m=k=0$, not the one I used in my original question.

Answer 1

Since the group $G=\Bbb Q^n/\Bbb Z^n$ is torsion, it has no pairs $a,b$ of independent elements. Even for any element $a\in G$ there exists $k$ such that $ka=0$. So we formulate the following condition on linear independence of elements $a$ and $b$ instead: for all integers $m,k$ such that $ma=kb$ we have $ma=kb=0$.

Let $a,b$ be any elements of $G$, $\langle a\rangle$, $\langle b\rangle$, and $\langle a,b\rangle$ be subgroups of $G$ generated by $a$, $b$, and $\{a,b\}$ respectively. Then sets $A$, $B$, and $A\cap B$ be the annihilators of groups $\langle a\rangle$, $\langle b\rangle$, and $\langle a,b\rangle$, respectively.

Let $H$ be a finite subgroup of $G$. Its annihilator $H^\perp$ is a closed subgroup of $\Gamma_n$ and so $\mu(H^\perp)^{-1}$ equals to the index $|\Gamma_n: H^\perp|$. According to the duality theory (see, for instance Theorem 37 in [Pon]), $H$ is the group of characters of the group $\Gamma_n/H^\perp$. So $\Gamma_n/H^\perp$ is isomorphic to the group of characters of the group $H$ (see, for instance, Theorem 37 in [Pon]). But, since $H$ is a finite group, the group of its characters is isomorphic to $H$, see [Pon, §36.G]. That is, $|\Gamma_n: H^\perp|=|\Gamma_n/H^\perp|=|H|$.

Thus $\mu(A)=|\langle a\rangle|^{-1}$, $\mu(B)=| \langle a,b\rangle |^{-1}$, $\mu(A\cap B)=| \langle a,b\rangle |^{-1}$. Thus $\mu(A)\mu(B)= \mu(A\cap B)$ iff $|\langle a\rangle|\cdot |\langle a\rangle|=|\langle a,b\rangle|$. A homomorphism $h:\langle a\rangle\times \langle b\rangle\to \langle a,b\rangle$, $(x,y)\mapsto xy$ for any $x\in \langle a\rangle$ and $y\in \langle b\rangle$, is surjective. So $|\langle a\rangle|\cdot |\langle b\rangle|=|\langle a,b\rangle|$ iff $h$ is injective, which holds iff the kernel of $h$ is zero, that is iff for all integers $m,k$ such that $ma=kb$ we have $ma=kb=0$.

References

[Pon] Lev Pontrjagin, Continuous groups, 2nd ed., M., (1954) (in Russian).

Answer 2

Here we answer a question by mathworker21

what do elements of $\Gamma_n$ look like?

It is well-known that the group $\Bbb Q/\Bbb Z$ is a direct sum of Prüfer groups $\Bbb Z(p^\infty)$, see, for instance, this thread. Thus $G$ is $n$-th power of this sum. By [DPS, Proposition 3.1.2], its group of characters $\Gamma_n$ is a Cartesian product of the group of characters of Prüfer groups. By Example 61, [Pon], the group of characters of the group $\Bbb Z(p^\infty)$ is the group $\Bbb Z_p$ of $p$-adic integers, so $\Gamma_n$ is topologically isomorphic to $\prod_{{\Large p}\mbox{ is prime} } \Bbb Z_p^n$.

References

[DPS] Dikran N. Dikranjan, Ivan R. Prodanov, Luchezar N. Stoyanov. Topological Groups, Marcel Dekker, New-York, 1990.

[Pon] Lev Pontrjagin, Continuous groups, 2nd ed., M., (1954) (in Russian).