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

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 form: $$\omega(x \oplus \chi, \, y \oplus \eta) = \chi(y) - \eta(x).$$ In the case of finite abelian groups the "dual role" is played by the group of characters, as usual. [1]: http://arxiv.org/abs/0708.2758

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 form: $$\psi((x, \, \chi), \, (y, \, \eta))=\chi(y)\eta(x)^{-1}.$$ 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 form: $$\omega(x \oplus \chi, \, y \oplus \eta) = \chi(y) - \eta(x).$$ In the case of finite abelian groups the "dual role" is played by the group of characters, as usual.

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 ][1]

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. [1]: http://arxiv.org/abs/0708.2758

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

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 form: $$\omega(x \oplus \chi, \, y \oplus \eta) = \chi(y) - \eta(x).$$ In the case of finite abelian groups the "dual role" is played by the group of characters, as usual. [1]: http://arxiv.org/abs/0708.2758

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  

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 ][1]

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. [1]: http://arxiv.org/abs/0708.2758