Finite abelian group of type S must be $G=H\times H$ In the following paper by Delaunay http://delaunay.perso.math.cnrs.fr/heuristics_2.pdf, it is claimed that a finite abelian group comes with a unique non-degenerate alternating bilinear pairing
$$\beta: G \times G \rightarrow \Bbb Q/\Bbb Z$$
iff $G=H\times H$. Why?
 A: Suppose that $G$ has such a pairing $\beta$ and write $G = C_{d_1} \oplus \cdots \oplus C_{d_k}$, where each $d_i|d_{i+1}$. We prove by induction on $k$ that $G \cong H \times H$ for some group $H$. The result is true for $k=0$ with $H$ trivial, so assume that $k>0$.
If $d_{k-1} \ne d_k$ then, letting $x$ be the generator of the final direct summand $C_{d_k}$, it is not hard show that $\beta(g,d_{k-1}x) =0$ for all $g \in G$, contradicting $\beta$ being non-degenerate.
Let $x$ and $y$ be generators of the final two direct summands $C_{d_k}$. Non-degeneracy of $\beta$ implies that $\beta(x,y)$ has order $d_k$, so we can assume that $\beta(x,y) = 1/d_k + {\mathbb Z}$. Also, for a generator $z$ of one of the other direct summands of $G$, the orders of $\beta(z,x)$ and $\beta(z,y)$ must divide $d_k$, so they are equal to $a/d_k + {\mathbb Z}$ and $b/d_k + {\mathbb Z}$ for some $a,b \in \{0,1,\ldots,d_k-1\}$. Then, by replacing $z$ by $z -bx + ay$, we can assume that $\beta(z,x)=\beta(z,y)=0$.
Suppose that the restriction of $\beta$ to the subgroup $K := C_{d_1} \oplus \cdots \oplus C_{d_{k-2}}$ is degenerate. Then there exists $g \in K$ with $\beta(g,h) = 0$ for all $h \in K$. Now, since the orders of $\beta(g,x)$ and $\beta(g,y)$ must divide $d_k$, there exist $a,b \in \{0,1,\ldots,d_k-1\}$ with $\beta(g,x) = a/d_k + {\mathbb Z}$ and $\beta(g,y) = b/d_k + {\mathbb Z}$. So, putting, $g' = g - bx + ay$, we have $\beta(g',h)=0$ for all $h \in G$, contradicting non-degeneracy of $\beta$.
So  the restriction of $\beta$ to $K$ is non-degenerate and the result follows by induction applied to $K$.
