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.