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I want to show that if for some $n \in \mathbb{Z}$, the map $f_n\colon (\mathbb{Z},+) \rightarrow (\mathbb{Z},+), x \mapsto nx$ is a group isomorphism, then $n \in \{-1,1\}$, without using anything besides the group structure. So especially I don't want to use that there are the rationals or that $\mathbb{Z} $ is an ordered set ( this would make the proof of course very trivial).

I.e. it boils down to show that if $nx=1$, then $n \in \{-1,1\}$ without using rationals or the fact that the integers are ordered.

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  • $\begingroup$ So what is your question? $\endgroup$ – Derek Holt Nov 15 '14 at 22:11
  • $\begingroup$ how can I show this without referring to anything but the group structure of $\mathbb{Z}$. I mean it is trivial to solve this over $\mathbb{Q}$ if you know that there must be something that gets mapped to $1$ or if you know that $\mathbb{Z}$ is an ordered set(by monotony) but if I don't use any of these two things, then I think it is not that easy $\endgroup$ – user66906 Nov 15 '14 at 22:13
  • $\begingroup$ @MikeMiller well, that $-1$ and $1$ do it, is pretty clear, the question is rather: How do I exclude the other ones? $\endgroup$ – user66906 Nov 15 '14 at 22:16
  • $\begingroup$ Ah! I misread your question. Apologies. $\endgroup$ – user98602 Nov 15 '14 at 22:16
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    $\begingroup$ So are you allowed to use the fact that there is no integer $x$ such that $nx = 1$ if $n\ne \pm 1$? If not, I'm not sure how you're going to make any progress. $\endgroup$ – rogerl Nov 15 '14 at 22:22
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Well, there are three (relatively simple) things that you must show:

  • If $n\in\{-1,1\},$ then for any $j,k\in\Bbb Z,$ we have $f_n(j+k)=f_n(j)+f_n(k).$ (That is, $f_n$ is a homomorphism.)
  • If $n\in\{-1,1\},$ and if $j,k\in\Bbb Z$ such that $f_n(j)=f_n(k),$ then $j=k.$ (That is, $f_n$ is injective.)
  • If $n\in\{-1,1\},$ and if $k\in\Bbb Z,$ then there is some $j\in\Bbb Z$ such that $f_n(j)=k.$ (That is, $f_n$ is surjective.)

It turns out that $f_n$ is a homomorphism regardless of our choice of $n\in\Bbb Z,$ and is injective for all $n\in\Bbb Z$ except for $n=0.$ As you've determined, however, $f_n$ is surjective if and only if $n\in\{-1,1\}.$

To show this, you should use the fact that $(\Bbb Z,+)$ is a cyclic group, generated by $1$ or $-1,$ but nothing else. A homomorphic image of a cyclic group will again be cyclic, generated by the element to which its generator is mapped--in this case, $f_n(1).$ In particular, the image of $f_n$ will be a cyclic subgroup of $(\Bbb Z,+).$ In order to be all of $\Bbb Z,$ however, $f_n(1)$ must be one of the two generators of $(\Bbb Z,+).$

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  • $\begingroup$ no, I want to show that -1 and 1 are the only ones that do it, not that -1,1 do it, this is clear to me. Of course the problem is, that for $n \neq -1,1$ this map is not surjective, but I don't see how to show this without using the fact that the set is ordered or by using the rationals. $\endgroup$ – user66906 Nov 15 '14 at 22:18
  • $\begingroup$ I, too, misread your question. My apologies! I have updated my answer. $\endgroup$ – Cameron Buie Nov 15 '14 at 22:21
  • $\begingroup$ How can you show, for example, that $f_2$ is not injective? (Not just with the group structure, but by any means.) $\endgroup$ – Cameron Buie Nov 15 '14 at 22:22
  • $\begingroup$ sorry, I meant of course not surjective. Well, I would need to show that nothing maps to 1. $\endgroup$ – user66906 Nov 15 '14 at 22:24
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    $\begingroup$ Excellent! Let me know if you run into any more trouble with the proof, or even if you'd just like me to double-check it for you. $\endgroup$ – Cameron Buie Nov 15 '14 at 22:41
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Hint: Consider the composition of $f_n$ with $g_m\colon {\bf Z}\to {\bf Z}/m{\bf Z}$ for various natural $m\geq 2$. Conclude that $f_n$ is not onto for $n\notin \{-1,1\}$.

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