Let $G$ be a (countable) discrete abelian group and denote by $\hat{G}$ its Pontryagin dual, i.e. the compact abelian group of group homomorphisms $\chi:G \longrightarrow \mathbb{T}$. Recall that, for a subgroup $H \subset G$, the annihilator is given by $H^\perp = \{\chi \in \hat{G} \mid \chi(g) = 1~\forall g \in H\}$. 


If $H\_1$ and $H\_2$ are subgroups of $G$, then $H\_1^\perp H\_2^\perp$, the subgroup of $\hat{G}$ generated by $H\_1^\perp$ and $H\_2^\perp$, is contained in $(H\_1 \cap H\_2)^\perp$.


$\textbf{Question 1:}$ Do we actually have $H\_1^\perp H\_2^\perp = (H\_1 \cap H\_2)^\perp$?


More specifially, let $\alpha$ and $\beta$ be commuting, injective group endomorphisms of $G$. These correspond to commuting, surjective group endomorphisms $\hat{\alpha},\hat{\beta}$ of $\hat{G}$, e.g. given by $\chi \mapsto \chi \circ \alpha$. Note that we have $\alpha(G)^\perp = \ker \hat{\alpha}$. $\textbf{Question 1}$ now transforms into:


$\textbf{Question 2:}$ Does $\ker\hat{\alpha}\ker\hat{\beta} = \ker\widehat{\alpha\beta}$ hold?


Comments: We may assume $G$ to be countable or $\alpha(G) \cap \beta(G) = \alpha\beta(G)$ if that helps. Note that a similar statement does hold:
$$\alpha(G)\beta(G) = G \Longleftrightarrow \ker\hat{\alpha} \cap \ker\hat{\beta} = \{1\_\hat{G}\}$$
Here $\alpha(G)\beta(G)$ denotes the subgroup of $G$ generated by $\alpha(G)$ and $\beta(G)$. From this equivalence, one can deduce that the two questions from above have a positive answer in the case where $\alpha$ or $\beta$ has finite cokernel.



Is the annihilator of the intersection of two subgroups of a (countable) discrete abelian group generated by the annihilators of the two subgroups?
The answer to both questions is yes. Indeed, by Pontryagin duality the inclusion $(H\_1\cap H\_2)^\perp \subset H\_1^\perp H\_2^\perp$ you want is equivalent to
$$
(H\_1^\perp H\_2^\perp)^\perp\subset H\_1\cap H\_2
\tag{$\*$}
$$
where for $\Sigma\subset\hat G$ we write $\Sigma^\perp=\{g\in G:\chi(g)=1 \text{ for all }\chi\in\Sigma\}$. Now suppose $a\notin H\_1\cap H\_2$. Without loss of generality, say $a\notin H\_1$. Then there is a character $\eta\in H\_1^\perp$ such that $\eta(a)\neq 1$ (just lift a character of $G/H\_1$ that is nontrivial at $aH\_1$). So $a$ is not in $H\_1^{\perp\perp}$, hence a fortiori not in the smaller set $(H\_1^\perp H\_2^\perp)^\perp$. So we have shown that $a\notin H\_1\cap H\_2 \Rightarrow a\notin (H\_1^\perp H\_2^\perp)^\perp$, whence $(\*)$ by contraposition.


