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Let $A$ be a finite, Abelian, additive group. Let $A^{*} = Hom(A, \mathbb{Q}/\mathbb{Z})$ denote the group of homomorphisms $f$ from $A$ to $\mathbb{Q}/\mathbb{Z}$. Take for granted that $A^{*}$ is an Abelian group (I have already proved this). Prove that $A$ is isomorphic to $A^{*}$ if $A$ is cyclic.

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What's the group operation on $A^*$? $(f \cdot g)(a) = f(a) \cdot g(a)$? –  Clive Newstead Oct 30 '12 at 18:30
That is indeed the case! –  user44069 Oct 30 '12 at 18:36

1 Answer 1

up vote 2 down vote accepted

Hint: If $A$ is cyclic, and $a$ generates it, then every homomorphism $f:A\to G$ is completely determined by $f(a)$.

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