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5 added 534 characters in body; added 6 characters in body

Actually, one can show the following stronger result:

Proposition

Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $A$ has a lagrangian decomposition, i.e. there exists a subgroup $G$, isotropic for $\psi$, such that

$A \cong G \times \widehat{G}$,

where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.

Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following shape:

$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.

An easy proof, by induction on the order of the group, can be found in Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758

Remark. It is interesting to notice the analogy with symplectic vector spaces. In fact, any symplectic vector space $(V, \omega)$ can be written as $V = W \oplus W^{*}$, where $W$ is a lagrangian (=isotropic of maximal dimension) subspace for $\omega$. In particular, $\dim V = 2 \dim W$. Moreover, with respect to this decomposition, $\omega$ has the following shape:

$\omega(x \oplus \chi, y \oplus \eta) = \chi(y) - \eta(x)$.

In the case of finite abelian groups the "dual role" is played by the group of characters, as usual.

4 added 54 characters in body

Actually, one can show the following stronger result:

Proposition

Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $A$ has a lagrangian decomposition, i.e. there exists a subgroup $G$, isotropic for $\psi$, such that

$A \cong G \times \widehat{G}$,

where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.

Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following shape:

$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.

For a

An easy proof, see by induction on the order of the group, can be found in Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758

3 added 110 characters in body; deleted 1 characters in body

Actually, one can show the following stronger result:

Proposition

Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $G$ A$has a lagrangian decomposition, i.e. there exists a subgroup$G$, isotropic for$\psi\$, such that

$A \cong G \times \widehat{G}$,

where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.

Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following shape:

$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.

For a proof, see Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758

2 added 160 characters in body; deleted 12 characters in body

Actually, one can show the following stronger result:

Proposition

Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form . $\psi$. Then $G$ has a lagrangian decomposition, i.e. there exists a finite abelian group subgroup $G$ G$, isotropic for$\psi\$, such that

$A \cong G \times \widehat{G}$,

where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$. Moreover, in such a presentation the form $\psi$ take the following shape:

$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.

For a proof, see Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758

1

Actually, one can show the following stronger result:

Proposition

Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form. Then $G$ has a lagrangian decomposition, i.e. there exists a finite abelian group $G$ such that

$A \cong G \times \widehat{G}$,

where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.

For a proof, see Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758
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