Let $S$ be the subgroup of $A$ such that $\psi(s,s')=0$ for all $s,s'\in S$ and $S$ is maximal with respect to this property.

Question

Let $A$ be a finite abelian group, and let

$$ \psi : A \times A \to \mathbb{Q}/\mathbb{Z} $$

be an alternating, non-degenerate bilinear form on $A$.

Let $S$ be the subgroup of $A$ such that $\psi(s,s')=0$ for all $s,s'\in S$ and $S$ is maximal with respect to this property.

My question is, why can we take such $S$?

I think it is difficult to prove the existence of $S$ without constructing it concretely.

cf. This definition appears in lemma 4 of this.

https://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0998-10.pdf

Answer 1

The word "the" here is incorrect--there is not necessarily a unique such subgroup $S$. However, there does always exist some such subgroup. Since $A$ is finite, you can just take a subgroup on which $\psi$ is zero of maximal cardinality. (At least one such subgroup exists, namely the trivial subgroup.)