How many homomorphisms are there from $\mathbb Z_n$ to $\mathbb Q$ ?

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    $\begingroup$ Is there an element $g$ of $\mathbb{Q}$ such that the following holds? $$\underbrace{g+\cdots+g}_{n}=0$$ $\endgroup$ – angryavian Aug 18 '14 at 15:17
  • $\begingroup$ @angryavian: yes !!! , there is no such non-zero $g$ in $Q$ , thanks $\endgroup$ – Souvik Dey Aug 18 '14 at 15:22

Hint If $\eta:G\to H$ is a morphism of groups, and if $x\in G$ has finite order say $|x|=k$, then $(\eta x)^k=\eta(x^k)=\eta 1=1$ so $\eta x$ has finite order, and $|x| \mid |\eta x|$. Now look at the elements of finite order of $\Bbb Q$ (is this additive or multiplicative here?)

  • $\begingroup$ (I'm not sure if the OP is thinking $\Bbb Q$ as an additive of multiplicative group. In the former case the only torsion elt is $0$, in the latter, there are two torsion elts.) $\endgroup$ – Pedro Tamaroff Aug 18 '14 at 15:19

Assuming you are taking the operation to be addition in $\mathbb{Z}_n$ and $\mathbb{Q}$, then $\phi(0) = 0$, as is true for all homomorphisms. Then:

  • $\phi(1) = \text{something}$

  • $\phi(1+1) = \phi(1) + \phi(1) = \text{something + something}$

  • $\phi(1+1+1) = \cdots$

What happens when this process is iterated $n$ times?

  • $\begingroup$ Iterating $n$ times will results in $\phi(0)=$something, which is non-zero. Contradiction. So there is no homomorphism. Right? $\endgroup$ – Kushal Bhuyan Apr 18 '16 at 5:39
  • $\begingroup$ @KushalBhuyan;No there is the zero homomorphism i.e. $\phi(x)=0\forall x$ and this is the only one $\endgroup$ – Learnmore Dec 8 '17 at 6:29
  • $\begingroup$ "No nontrivial homomorphism" as they say $\endgroup$ – Kaj Hansen Dec 8 '17 at 12:35

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