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Let $G$ be a finitely generated abelian group (written additively), let $\langle \cdot,\cdot\rangle\colon G\times G\to \mathbb Z$ be a bilinear form and let $\sigma\colon G\rightarrow G$ be an automorphism satisfying $\langle g,h\rangle=-\langle h,\sigma(g)\rangle$ for all $g, h \in G$. I unfortunately was unable to prove the following:

If $\varphi\colon G\rightarrow \operatorname{Hom}_{\mathbb{Z}}(G,\mathbb{Z})\colon g \mapsto \langle g,\cdot\rangle$ is an isomorphism, then $G$ has no torsion.

I would be thankful for any ideas.

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$Hom(G,\mathbb{Z})$ is torsion free. – user641 Apr 7 '12 at 10:53
up vote 2 down vote accepted

If $\tau$ is a torsion element of order $n$, then $n\varphi(\tau)=\varphi(n\tau)=0$ implies $\varphi(\tau)$ is the zero map. But by hypothesis $\varphi$ is an isomorphism, and therefore should have trivial kernel, ergo $\tau=0$.

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