Let $G$ be a finite abelian group. A *quadratic form* on $G$ is a map $q: G \to \mathbb{C}^\*$ such that $q(g) = q(g^{-1})$ and the symmetric function $b(g,h):= \frac{q(gh)}{q(g)q(h)}$ is a bicharacter, i.e. $b(g\_1g\_2, h) = b(g\_1, h)b(g\_2, h)$ for all $g, g\_1, g\_2, h \in G.$


The quadratic form $q$ is called *non-degenerate* if the corresponding bicharacter $b$ is non-degenerate.


**Question**: Is there a non-degenerate quadratic form on every finite abelian group?


Motivation: it is used to make pointed braided/modular tensor categories, see Chapter 8 of [this book](http://www-math.mit.edu/%7Eetingof/egnobookfinal.pdf) (in particular Sections 8.4, 8.13 and 8.14).



Is there a non-degenerate quadratic form on every finite abelian group?
Thanks to the Fundamental Theorem of Abelian Groups, let
$$G:=\prod\_{k=1}^{n}\{z:z^{m\_k}=1\,,z\in\mathbb{S}\}\,,$$
and let $\chi(m)=2$ if $m$ is odd and $\chi(m)=1$ if $m$ is even. Then define
$$q\colon G\to\mathbb{C}^\*\,,~\,~\,~\,(e^{2\pi i\frac{a\_k}{m\_k}})\_k\mapsto \exp(\pi i\sum\_k \chi(m\_k)\frac{a\_k^2}{m\_k})\,.$$
We observe that $q$ is a quadratic form because
$$e^{\chi(m)\pi i\frac{(a+m)^2}{m}}=e^{\chi(m)\pi i\frac{a^2}{m}}e^{2\pi i(a\chi(m)+\frac{m\chi(m)}{2})}= e^{\chi(m)\pi i\frac{a^2}{m}}\,.$$
The associated bi-character then becomes
$$B((e^{2\pi i\frac{a\_k}{m\_k}})\_k,(e^{2\pi i\frac{b\_k}{m\_k}})\_k)= \exp(2\pi i\sum\_k \chi(m\_k)\frac{a\_kb\_k}{m\_k})\,.$$
Since the degeneracy of the bi-character $B$ is equivalent to $\sum\_k\chi(m\_k)\frac{a\_kb\_k}{m\_k}\in\mathbb{Z}$ say for all admissible $(b\_k)\_k$, this forces $a\_k\equiv 0\mod m\_k$; thus $B$ is non-degenerate!


