I know how to prove that for $\mathbb{Z}/n$ where $n$ is odd, but what should we do in case of $\mathbb{Z}$?
Asked
Active
Viewed 143 times
0
-
The proof is actually very similar to the case of $\mathbb{Z}/n$, how do you prove that fact? – Captain Lama Dec 03 '19 at 01:07
-
6https://math.stackexchange.com/a/256173/337971 – bsbb4 Dec 03 '19 at 01:07
-
@CaptainLama The same way it's done here https://math.stackexchange.com/a/2555701/731194 – die Schizophren Dec 03 '19 at 01:12
-
1Well then it's the same thing, $G$ has an automophism of order 2, so its automorphism group is not $\mathbb{Z}$. – Captain Lama Dec 03 '19 at 01:19
-
@dieSchizophren the proof in the link you give is incomplete (there is a lot of work wrapped up in the comment below the post!). The arguments that imply that abelian groups have automorphisms of order two the Axiom of Choice. Without Choice, it is consistent that there exists a group $G$ with infinite cyclic automorphism group. – user1729 Dec 03 '19 at 07:46
-
@CaptainLama As far as I see, this proof is using Lagrange theorem, but if group is infinite we can't use it, right? – die Schizophren Dec 07 '19 at 13:04
-
The point is that there is an automorphism of order $2$. If you want to conclude from that that the automorphism group cannot have odd order, you use Lagrange theorem. If you want to conclude that it cannot be $\mathbb{Z}$, it's even easier: elements of $\mathbb{Z}$ have infinite order. – Captain Lama Dec 07 '19 at 15:59