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I'm asking this question to review the second answer, the one with 7 votes, posted in response to this question. Since the thread is very old, I'm not sure if I'll get a response to the comment I posted.

Isn't the proof given in the answer incorrect? The last part of the proof assumes that $\mathbb{1}_\mathbb{Z} = 1$. Isn't this wrong since in the additive group, $(\mathbb{Z}, +)$, isn't the identity element, $\mathbb{1}_\mathbb{Z}$, equal to $0$?

Junaid Aftab
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  • Maybe you should link to the answer itself rather than the question, or just give the author's name. – Jair Taylor May 30 '17 at 21:25
  • The notation can be rather deceptive. Of all 1s mentioned in the proof, none is assumed to be the group identity. – Ivan Neretin May 30 '17 at 21:25
  • I think here $1_\mathbb{Z}$ is not referring to the additive identity for $\mathbb{Z}$. It looks like it's just $1$, but considered as an element of $\mathbb{Z}$ rather than as an element of $\mathbb{Q}$, for clarity. – Jair Taylor May 30 '17 at 21:26
  • @JairTaylor Yes, I thought so, too. See the comment I posted below. I guess I'll remove this question as it's wasn't very constructive either way. Also, I wasn't sure how to link the specific question within the post. – Junaid Aftab May 30 '17 at 21:29

1 Answers1

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He doesn't use the fact that $1_{\Bbb Z}$ is the identity element in $\Bbb Z$ because is clearly false. He uses the fact that an isomorphism is a bijection, then if there exists an isomorphism $\phi$ between $\Bbb Q$ and $\Bbb Z$, then there necessarily exists a rational $r\in\Bbb Q$ such that $\phi(r)=1_{\Bbb Z}$. Finally he uses the homomorphism property to get an absurdness.

InsideOut
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