Let $\mathbb{Q}$ be the group ($\mathbb{Q}$,+) and $\mathbb{Z}$ is a sub-group of $\mathbb{Q}$.
It is quite easy to find all homomorphism from $\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Q}/ \mathbb{Z}$.
However, I couldn't find what would be all homomorphism from $\mathbb{Q}/ \mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$.
Please help me.

  • $\begingroup$ The possibilities for the image of $a/b$ will depend on $\gcd(b,n)$. $\endgroup$ – Gerry Myerson Dec 28 '13 at 14:23

Because there is only the trivial one!

Any morphism $\Bbb Q/\Bbb Z\to\Bbb Z/n\Bbb Z$ is induced by a morphism $\phi:\Bbb Q\to \Bbb Z/n\Bbb Z$ such that $\phi(1)=0$.

Now let $u\in\Bbb Q$ arbitrary, then we must have $$\phi(u)=\phi\left(n\cdot\frac un\right)=n\cdot \phi\left(\frac un\right)=0$$ in $\Bbb Z/n\Bbb Z$.

  • $\begingroup$ I meant $\ker\phi=\Bbb Z$ which is equivalent to saying $\phi(1)=0$. $\endgroup$ – Berci Dec 28 '13 at 23:04
  • $\begingroup$ Ok, that makes sense. $\endgroup$ – Thomas Dec 29 '13 at 0:09

Another approach:

As a homomorphic image of a divisible group, the quotient $\;\Bbb Q/\Bbb Z\;$ is divisible, and thus the image of any homomorphism from this group is also divisible. But the only divisible finite group is the trivial one, and thus the only possible homomorphism $\;\Bbb Q/\Bbb Z\to\Bbb Z/n\Bbb Z\;$ is the trivial one.

  • $\begingroup$ Would it be correct to clam that the only possible homomorphism $\mathbb{Q}/ \mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ is the trivial one since $\mathbb{Q}/ \mathbb{Z}$ is an infinite group while $\mathbb{Z}/n\mathbb{Z}$ is a finite group? $\endgroup$ – Ran Kashtan Dec 28 '13 at 15:35
  • $\begingroup$ No @RanKashtan: each and every finite group is a homomorphic image of, for example, the free group on two generators, which is infinite, of course. $\endgroup$ – DonAntonio Dec 28 '13 at 15:44
  • $\begingroup$ Is there a third way to reach the answer without using "divisible group" (I havn't learn yet about them)? $\endgroup$ – Ran Kashtan Dec 28 '13 at 15:59
  • $\begingroup$ Well, going into cohomogical stuff (you don't really want this), but (1) divisible groups are a pretty simple thing to grasp, and (2) Berci's answer is quite straightforward. $\endgroup$ – DonAntonio Dec 28 '13 at 16:09

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