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Let $G$ be a finite abelian $p$-group (where $p$ is a prime). Suppose there exists a symmetric bilinear map $\delta\colon G\times G\to \mathbb{Q}/\mathbb{Z}$ such that the induced map $g\to\langle g,\;\rangle$, is an isomorphism from $G$ to $\mathrm{Hom} (G, \mathbb{Q}/\mathbb{Z})$.

Then, is it true that $G$ can be written as $\mathbf{Z}_p^n/\,\mathrm{im} (A)$ for some symmetric matrix (i.e. a linear map) $A\colon\mathbf{Z}_p^n\to\mathbf{Z}_p^n$ for some positive integer $n$? (Here $\mathbf{Z}_p$ denotes the $p$-adic integers)

EDIT2: As user74230 has pointed out, the isomorphism $G\to \mathbf{Z}_p^n/\,\mathrm{im} (A)$ has to respect the evident bilinear forms on both sides. (Otherwise, the answer is trivial and $A$ does not depend on $\delta$ as Amritanshu Prasad has pointed out).

EDIT1: I had this question while reading the following paper: http://arxiv.org/pdf/1402.5129v1. In Theorem 2 of page 4 it says that, given a $p$-group with duality pairing $(G,\delta )$ the probability (w.r.t. Haar measure) that $\mathrm{coker}(A)\cong (G,\delta)$ converges to Cohen-Lenstra type probability measure when $n\to\infty$. The existance of the duality pairing seems relevant for this theorem, but Amritanshu Prasad's comment suggests that $A$ can be chosen independent of the pairing which seems puzzling for me.

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  • $\begingroup$ Is your group commutative? Finite? Finitely generated? $\endgroup$
    – abx
    Commented Jan 26, 2015 at 6:58
  • $\begingroup$ Yes, finite and commutative. Edited, sorry for the confusion. $\endgroup$ Commented Jan 26, 2015 at 7:11
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    $\begingroup$ @abx: I did ask this question in MSE but unfortunately did not get any answer. Could you please elaborate your answer? $\endgroup$ Commented Jan 26, 2015 at 7:41
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    $\begingroup$ @abx: The structure of finite quadratic spaces and its relationship with $p$-adic quadratic spaces is reasonably elementary but not an totally trivial since it does involve input concerning the structure of non-degenerate quadratic spaces over finite fields. (It is not rocket science, but lots of stuff on MO is not rocket scence.) $\endgroup$
    – user74230
    Commented Jan 26, 2015 at 8:02
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    $\begingroup$ @AmritanshuPrasad's comment answers the question as asked, but the OP probably wanted some compatibility between $A$ and the given duality pairing. The OP should spell this out. $\endgroup$ Commented Jan 26, 2015 at 11:22

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For odd $p$ the answer is affirmative. Suppose $p$ is odd, so $\delta$ is valued in $\mathbf{Q}_p/\mathbf{Z}_p$ and we may define the quadratic form $q:G \rightarrow \mathbf{Q}_p/\mathbf{Z}_p$ by $q(g) = (1/2)\delta(g,g)$ so that $(G,q)$ is a "non-degenerate" finite quadratic space with associated symmetric bilinear form $\delta$.

Your question for such $p$ amounts to asking if every non-degenerate finite quadratic space of $p$-power size is the "discriminant form" of a quadratic lattice over $\mathbf{Z}_p$. This is part of the assertion of Theorem 1.9.1 of Nikulin's paper "Integral symmetric bilinear forms and some of their applications" in Math. USSR Izv. Vol. 14 No. 103 (1980), which also gives uniqueness aspects (under a suitable minimality requirement) and also provides analogues when $p=2$. The result there for $p=2$ probably also solves your question for $p=2$ but it would require some care to unravel the passage between bilinear forms and quadratic forms in the 2-adic setting.

Nikulin's proof amounts to the systematic study of primary parts under orthogonality and inductive knowledge built from the classification of non-degenerate quadratic spaces over finite fields.

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    $\begingroup$ The question asks for a presentation of $G$ as a group (which is trivial), not as a quadratic space. This is why I downvoted. You are certainly right in interpreting the question as you do, but this is not what is written. $\endgroup$
    – abx
    Commented Jan 26, 2015 at 8:35
  • $\begingroup$ @abx: Ah, OK. The OP has identified a gap in the proof in the cited reference now given, but in the authors' defense they seem to have had in mind applications only to cases where their finite quadratic space arises in the given form, so for their purposes the gap in the proof of their result as written has no effect. $\endgroup$
    – user74230
    Commented Jan 26, 2015 at 15:12

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