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There are $n$ homomorphisms from the group $Z/nZ$ to the additive group of rationals $Q$.

how can i find that the above statement is true/false

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Play around with small $n$. Either you'll find a counterexample or you'll (hopefully) see a pattern in which you'll be able to generalize to a proof for all $n$. –  Jason DeVito Nov 24 '12 at 2:46
    
Consider torsion. –  anon Nov 24 '12 at 3:26
    
@rahu I suggest you to change the title by eliminating the first words "There are $n$". –  user26857 Nov 24 '12 at 10:56

2 Answers 2

Hint: Elements of $\mathbb{Q}$ apart from the identity have infinite order, and all elements of $\mathbb{Z}/n\mathbb{Z}$ have finite order. Can a group homomorphism map an element of finite order to an element which does not have finite order?

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Say the image of $1\in Z_n$ is $x\in Q$ then $nx=0$ in the rationals so $x=0$. Thus all homomorphisms are trivial, all elements are sent to $0$.

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