Where did we used the condition $\langle , \rangle: A \times A \rightarrow \mathbb{Q}/\mathbb{Z},$ is alternative to prove $A$ has square cardinatiry?

Question

Lemma 4. Let $A$ be a finite abelian group. If there is a non-degenerate alternating bilinear pairing $$\langle , \rangle: A \times A \rightarrow \mathbb{Q}/\mathbb{Z},$$ then, $A \cong S \times \hat{S}$ for some subgroup $S$ of $A$, where $\hat{S}$ is the character group of $S$.

Proof. Let $S$ be the subgroup of $A$ such that $\langle s, s' \rangle=0$ for all $s, s' \in S$ and $S$ is maximal with respect to this property. Since $\langle 1,1 \rangle = 0$, there is at least one subgroup with the property. Consider the character $\chi_a(s) = \langle a, s \rangle$ of $A$, for each $a \in A$. They are all distinct, since the pairing $\langle , \rangle$ is non-degenerate. Consider $A/S$. For each $a \in A/S$, we have a character of $S$ defined by $\chi_a(s) = \langle a, s \rangle$. By the definition of $S$, they are distinct. Therefore, $\hat{S} \cong A/S$. Therefore, $A \cong S \times \hat{S}$. (cf. lemma 4 in https://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0998-10.pdf )

My question: This uses the condition that $\psi$ as an nondegenerate, but it appears that the condition of $\psi$ being alternative is not used. Where in this lemma is the condition that $\psi$ is alternative actually used?

Answer 1

The proof in question is very poorly written and has a lot of misstatements and gaps. Here is how I would rewrite it.

Let $S$ be a maximal subgroup of $A$ with the property that $\langle s,s'\rangle=0$ for all $s,s'\in S$ (at least one such subgroup exists, since the trivial subgroup has this property). Note that then $\langle,\rangle:A/S\times S\to\mathbb{Q}/\mathbb{Z}$ is a well-defined pairing, since if $s\in S$ then $\langle s',s\rangle=0$ for all $s'\in S$. I claim this pairing is nondegenerate, so that $A/S\cong \hat{S}$. It is nondegenerate in the second coordinate since the original pairing on $A$ was nondegenerate. To see that it is nondegenerate on the first coordinate, suppose $a\in A$ is such that $\langle a,s\rangle=0$ for all $s\in S$. Then since $\langle a,a\rangle=0$ as well (here we use the fact that the original pairing is alternating), $\langle a,x\rangle=0$ for any $x$ in the subgroup $S'$ of $A$ generated by $S$ and $a$. It follows that $S'$ satisfies $\langle s,s'\rangle=0$ for all $s,s'\in S'$, so by maximality of $S$, $S'=S$. Thus $a\in S$ and its image in $A/S$ is $0$.

(The proof then concludes that $A\cong S\times \hat{S}$, but this conclusion is actually incorrect: the short exact sequence $0\to S\to A\to A/S\to 0$ may not split. For instance, consider $A=(\mathbb{Z}/4\mathbb{Z})^2$, with the pairing $A\times A\to\mathbb{Z}/4\mathbb{Z}$ given by the matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$. Then one choice of maximal subgroup $S$ as in the proof would be the subgroup generated by $(2,0)$ and $(0,2)$, and for this $S$ it is not true that $A\cong S\times\hat{S}$. There is another choice of $S$ for which this would be true (for instance, the subgroup generated by $(1,0)$), so at least the statement of Lemma 4 still is true in this example. A correct proof of that full statement would evidently need to choose $S$ more carefully, though.)