0

Every infinite cyclic group is isomorphic to the additive group $\Bbb{Z}$ and every finite cyclic group of order $m$ is isomorphic to the additive group $\Bbb{Z}_m$.

Proof: If $G=\langle a\rangle$ is a cyclic group then the map $\alpha:\Bbb{Z} \to G$ given by $k\mapsto a^k$ is an epimorphism by Theorems 1.9 and 2.8. If $\text{Ker}\ \alpha = 0$, then $\Bbb{Z}\cong G$ by Theorem 2.3 (i). Otherwise $\text{Ker}\ \alpha$ is a nontrivial subgroup of $\Bbb{Z}$ (Exercise 2.9) and hence $\text{Ker}\ \alpha =\langle m\rangle$, where $m$ is the least positive integer such that $a^m=e$ (Theorem 3.1). For all $r,s\in \Bbb{Z}$, $$a^r=a^s\Leftrightarrow a^{r-s}=e \Leftrightarrow r-s\in \text{Ker}\ \alpha =\langle m\rangle \Leftrightarrow m| r-s \Leftrightarrow \bar{r}=\bar{s}\text{ in }\Bbb{Z}_m,$$ (where $\bar{k}$ is the congruence class of $k\in \Bbb{Z}$). Therefore the map $\beta : \Bbb{Z}_m\to G$ given by $\bar{k}\mapsto a^k$ is a well-defined epimorphism. Since $$\beta (\bar{k})=e \Leftrightarrow a^k=e=a^0 \Leftrightarrow \bar{k}=\bar{0}\text{ in }\Bbb{Z}_m,$$ $\beta$ is a monomorphism (Theorem 2.3(i)), and hence an isomorphism $\Bbb{Z}_m\cong G$.

Question: How does “$\text{Ker}\ \alpha =\langle m\rangle$ where $m$ is smallest natural number such that $a^m=e$“ lead to a straight forward contradiction? My approach to prove injectivity: Suppose $\alpha (k)=\alpha (l)$, for some $k,l\in \Bbb{Z}$. Then $a^k=a^l$. Assume towards contradiction $k\neq l$. WLOG $k\gt l$. Then $a^{k-l}=e$. So $|a|=n\leq k-l$. By division algorithm, it is easy to see $\langle a\rangle =\{e,a,…,a^n\}$. So $G=\langle a\rangle$ is finite. Thus we reach contradiction. Hence $k=l$. My proof uses the notion of order of $a$. Author defined order of $a$ after theorem 2 section 1.3 and used this theorem to immediately conclude lots of properties of $|a|$ which is essentially what we are trying to prove in injectivity.

Why we need to show $\beta$ is well defined map? In the process we used $\text{Ker}\ \alpha$. Which is technically $\{0\}$, not $\langle m\rangle$.

user264745
  • 4,143
  • 1
    $\mathrm{Ker}(\alpha) = \langle m \rangle$ doesn't lead to a contradiction? The argument is just saying that either the kernel is trivial or it is not. If it is trivial, we have the infinite case of the theorem, and if the kernel is not trivial we get the other. – Jakob Streipel Mar 27 '23 at 17:20
  • @prets We have to show ker $\alpha$ can’t be non trivial. – user264745 Mar 27 '23 at 17:22
  • 1
    Why do you say that? In the quoted proof, $\alpha$ is not the bijection between the groups in the second case, $\beta$ is. – Jakob Streipel Mar 27 '23 at 17:23
  • @prets if it’s non trivial, then $\alpha$ is not injective. Therefore $\Bbb{Z}$ is not isomorphic to $G=\langle a\rangle$. – user264745 Mar 27 '23 at 17:25
  • 1
    Absolutely, but you're not trying to prove that $\mathbb Z$ is isomorphic to $G$, you're trying to show that $\mathbb Z$ or $\mathbb Z_m$ is isomorphic to $G$ (for some $m$). – Jakob Streipel Mar 27 '23 at 17:26
  • 1
    @prets ohhh….. I thought we’re proving theorem in two cases: (1) $G$ is infinite, and (2) $G$ is finite. But author killed two birds with one stone. – user264745 Mar 27 '23 at 17:28

0 Answers0