3
$\begingroup$

Recently, I have tried to find out how many group homomorphisms exist from $\Bbb Z$ to $\Bbb Z_n$ .

My argument goes like this: (in the following, $\tau$ and $\phi$ denote the well-known number theoretic functions)

Let, $\psi$ be such a homomorphism. Since $\Bbb Z$ is cyclic, hence $\psi{(\Bbb Z)}$ must be a cyclic subgroup of $\Bbb Z_n$ with the generator $\psi{(1)}$ . Using properties of Cyclic groups, we obtain that $| \psi{(\Bbb Z)}|$ can be any positive integer m such that $ m | n$ , hence there are in total $\tau (n)$ number of possibilities for $| \psi{(\Bbb Z)}|$ and hence for $ o(\psi(1))$.

Let S denote the set of all positive integer divisors of n, i.e. $$ S = \{d \in {\Bbb Z}^+ : d | n \}$$ , then |S| = $\tau(n)$ .

Again, since $\psi{(\Bbb Z)}$ is cyclic it follows that for each $d \in S$, we have $\phi(d)$ number of possible distinct images.

Since for each such image (for each divisor) it defines a homomorphism, the required number of homomorphisms are : $$\sum_{d|n} \phi(d) = n$$ and among these n homomorphisms , there are precisely $\phi(n)$ number of epimorphisms.

Conversely, from $\Bbb Z_n$ to $\Bbb Z$ I get that there exists only one homomorphism namely the 0 homomorphism.

Are my claims and arguments correct? If there is any mistake, please help me by correcting it.

I was also trying to find out how many homomorphisms exist from $\Bbb Z_n$ to $\Bbb Q$ i.e. precisely as group homomorphism and as ring homomorphism. Also, conversely, how many homomorphisms exist from $\Bbb Q$ to $\Bbb Z_n$ again as group homomorphism and ring homomorphism seperately and also how many out of these are epimorphisms (in each case).

Thanks in advance for helping.

$\endgroup$
  • $\begingroup$ It looks fine to me. In particular, after the edit. Have you determined the homomorphisms between $\mathbb{Z}_n$ and $\mathbb{Q}$ in both directions too or is that a part of your question? $\endgroup$ – stressed out Dec 9 '17 at 9:36
2
$\begingroup$

For any group $G$, you have a bijection between $G$ and the set $Hom(\mathbb{Z},G)$ of group homomorphisms $\mathbb{Z}\to G$.

The bijection is as follows: for $x\in G$, set $h_x:\mathbb{Z}\to $G$, \ m\mapsto x^m$.

Then the desired bijection is $G\to Hom(\mathbb{Z},G), \ x\mapsto h_x.$

First of all, you have to check that $h_x$ is indeed a group homomorphism (easy).

For the injectivity part: if $x,y\in G$ are such that $h_x=h_y$, then $h_x(1)=h_y(1)$, that is $x=y$?

For the surjectivity part: the main idea is that any element of $\mathbb{Z}$ is the sum of several copies of $1$ (or $-1$). Thus, in order to define a homomorphism $\varphi: \mathbb{Z}\to G$, it is enough to know $\varphi(1)$.

More precisely, if $\varphi$ is such a morphism, then, for all $m\geq 0$, we have $\varphi(m)=\varphi(1+\cdots+1)=\varphi(1)^m$. If $m<0$, then $\varphi(m)=\varphi(-(-m))=\varphi(-m)^{-1}$ (a morphism repects inverses), so $\varphi(m)=(\varphi(1)^{-m})^{-1}=\varphi(1)^m$.

Finally, $\varphi=h_x$ with $x=\varphi(1)$.

In particular, if $G=\mathbb{Z}_n$, you get $n$ such morphisms.

Concerning your second question, you have only the trivial group/ring morphism. Any element of $\mathbb{Z}_n$ has finite order. But a morphism sends an element of finite order to an element of finite order. Since the only element of $\mathbb{Q}$ of finite order is $0$....

Edit Some answers were given while I was typing. Feel free to delete this post if it seems useless.

$\endgroup$
  • $\begingroup$ Essentially the same argument but in a much more compact form is that $\mathbb Z$ is the free group on one generator, so $\mathbf{Grp}(\mathbb Z, G)=\mathbf{Grp}(F1, G)\cong\mathbf{Set}(1,|G|)\cong |G|$ where $|G|$ is the carrier set of $G$. (This is leverage the characterization of free objects as left adjoints, i.e. $F\dashv |{-}|$.) $\endgroup$ – Derek Elkins Dec 9 '17 at 11:23
  • $\begingroup$ But how to deal with the homomorphisms from $\Bbb Q$ to $\Bbb {Z}_n$ ? $\endgroup$ – reflexive Dec 10 '17 at 2:08
  • $\begingroup$ @ThatIs: For morphisms $\mathbb{Q}\to\mathbb{Z}$, you can notice that any rational $x$ maybe written as $x=ny, y\in\mathbb{Q}$, so you'll get the 0 morphism. $\endgroup$ – GreginGre Dec 10 '17 at 10:10

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.