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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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