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For the nonzero rational number multiplicative group $Q^\times$, how to calculate the automorphisms group $Aut(Q^\times)$ ?

  1. First,suppose $\phi :Q^\times \to Q^\times$ is an automorphism,it must send 1 to 1, and -1 to -1, I think the question is to determine the primes to be sent what?
  2. But I have trouble in determining this thing. I guess this group is $ Z_2\oplus \oplus_{p \ primes} Z $. Any help will be greatly appreciated, thanks!
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Hint: $Q^\times$ is isomorphic to $\Bbb Z_2+\sum_{n\in\Bbb N}\Bbb Z$.

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  • $\begingroup$ @MiloBrandt Right, thanks. $\endgroup$ – David C. Ullrich Mar 11 '18 at 15:26
  • $\begingroup$ Could you explain how did you get this result? I don’t know how to construct such automorphisms in a systematic way. $\endgroup$ – Jiabin Du Mar 11 '18 at 16:05
  • $\begingroup$ Well I haven't found the auutomorphisms. The reason $Q^\times$ is isomorphic to that sum is that if $r$ is a non-zero rational then $r$ has a unique prime factorization $r=\pm p_1^{n_1}p_2^{n_2}\dots p_N^{n_N}$. $\endgroup$ – David C. Ullrich Mar 11 '18 at 16:09
  • $\begingroup$ I know this fact, so the question is how to send primes. It not necessarily send primes to integral numbers, moreover, it’s not reasonable to arrange primes at random. Say, send 2 to $1/2$) ,3 to $1/3$ and 5 to $1/6$ , but here troubles come, since this map must sent 6 to $1/6$. So I can’t see a reasonable way. $\endgroup$ – Jiabin Du Mar 11 '18 at 16:41
  • $\begingroup$ Any element of order two must map to an element of order two, so you can forget about the $\Bbb Z_2$. For a warmup, try finding the automorphisms of $\Bbb Z^2$... $\endgroup$ – David C. Ullrich Mar 11 '18 at 17:00

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