Symplectic Forms on Abelian Groups

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If $G$ is finite group such that $G/Z(G)$ is abelian (i.e., $G$ is nilpotent of class at most 2) then the commutator function $$\varphi\colon G/Z(G)\times G/Z(G)\to G^\prime$$ given by $\varphi(g,h)=g^{-1}h^{-1}gh$ satisfies the identities (in additive notation):

  1. $\varphi(x,x)=0$,
  2. $\varphi(x,y)=-\varphi(y,x)$,
  3. $\varphi(x+y,z)=\varphi(x,z)+\varphi(y,z)$,
  4. $\varphi(x,y+z)=\varphi(x,y)+\varphi(x,z)$,
  5. If $\varphi(x,y)=0$ for all $y$ then $x=0$,
  6. If $\varphi(x,y)=0$ for all $x$ then $y=0$.

If $G^\prime$ is cyclic of order $p$ then $G/Z(G)$ is an $\mathbb{F}_p$-vector space and $\varphi$ is a symplectic bilinear form on $G/Z(G)$.

With this in mind, fix an abelian group $A$ and define a symplectic $A$-space to be a finite abelian group $B$ together with a function $\varphi\colon B\times B\to A$ satisfying the above conditions (yes, I realize that conditions 2 and 6 are redundant).

Is there a classification of symplectic $A$-spaces up to isomorphism?


If $A$ is trivial then $B$ is trivial.

If $A$ is cyclic of order $p$ then symplectic $A$-spaces are the same as symplectic $\mathbb{F}_p$-vector spaces. In particular, every symplectic $A$-space has order $p^{2n}$, and there is exactly one symplectic $A$-space of every such order.

In general, if $A$ is cyclic of order $n$ then this answer shows that $B\cong C\times\widehat{C}$ for some finite abelian group $C$ of exponent dividing $n$. This gives a nice classification of symplectic $A$-spaces in the cyclic case.

I'm interested about what happens when $A$ is not assumed to be cyclic.