2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The possibilities for the image of $a/b$ will depend on $\gcd(b,n)$. $\endgroup$ – Gerry Myerson Dec 28 '13 at 14:23
5
$\begingroup$

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 $\Bbb Z/n\Bbb Z$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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