Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
$Hom(G,\mathbb{Z})$ is torsion free. –  user641 Apr 7 '12 at 10:53

1 Answer 1

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 $\tau=0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.