0
$\begingroup$

Here is my idea to prove this: I've already known that every subgroup of a cycle group is also cycle. Let $H=[a] $ be a subgroup of the infinite group $G=[b]$. For any $x\in H$, I can find an integer $p$ such that $x=a^p$. Let's suppose that $H$ is finite with $o(x)=n$. Then we get $x^n=e\Rightarrow (a^p)^n=e\Rightarrow a^{np}=e$, which means that $o(a) =np$, but since $a\in G$ this implies that $o(a)$ is finite in $G$, a contradiction. Therefore, $H$ is infinite.

$\endgroup$
4
  • $\begingroup$ You might want to say that the order of $a$ divides $np.$ The gist of your argument is that if an element of the subgroup $H$ has finite order in $H$ then it must have finite order in $G.$ Unless I'm missing something silly, your argument seems to work just fine. $\endgroup$ Feb 2 at 21:53
  • $\begingroup$ Looks good to me. $\endgroup$
    – aschepler
    Feb 2 at 21:54
  • $\begingroup$ The other answer from Matthew is correct, so this is just serving as extra intuition. An infinite cyclic group is isomorphic to the integers $\mathbb{Z}$. Play around with the integers and you will quickly see it cant have a finite non-trivial subgroup. $\endgroup$
    – JimmyJummy
    Feb 2 at 21:55
  • $\begingroup$ Yes, considering the isomorphism definition, it is easy, but I forgot to mention the book which I take this question from only teaches isomorphism in a later section. That's why I didn't think this way $\endgroup$ Feb 3 at 12:14

3 Answers 3

3
$\begingroup$

Your proof is correct. I'd give essentially the same proof, although it unwraps easier (at least in my head) if I write every element as a power of the generator:

Let $C$ be an infinite cyclic group, $c$ its generator. Let $H$ be a non-trivial subgroup. Then $H$ contains $c^n$ for some $n \neq 0$. Since $H$ is a subgroup, it contains $\langle c^n \rangle$ (the subgroup generated by $c^n$). And this subgroup of $H$ is infinite, or else $c^{kn} = e$, but that would imply the generator for $C$ had finite order.

$\endgroup$
0
1
$\begingroup$

The proof provided by yourself and rewritten in multiplicative notation by Matthew Niemiro works excellently. I will however elaborate a little bit one the idea first mentioned by bagggggggs. As he points out, any infinite cyclic group is isomorphic to the additive group of integers (and any finite cyclic group is isomorphic to $\mathbb{Z}/n \mathbb{Z}$). So one way to prove this is to show that every non-trivial additive subgroup of the integers is infinite. However, one can actually do better than this. Using for example the well-ordering axiom of the integers, one can show that the non-trivial subgroups of the integers are exactly the groups $n \mathbb{Z}$ for $n \geq 1$.

$\endgroup$
0
$\begingroup$

Your proof is fine.

Here's an alternative:

The group of integers under addition is free. Any nontrivial subgroup of a free group is also free. Nontrivial free groups are infinite.

$\endgroup$
1
  • $\begingroup$ (Admittedly, my proof like nuking a fly.) $\endgroup$
    – Shaun
    Feb 2 at 22:29

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.