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In material supplied by my instructor there is a question which asks to pick the incorrect statement among the following:

  1. If $\text{Aut}(G_1)\cong \text{Aut}(G_2)$ and $G_1$ is infinite group then $G_2$ is also infinite

  2. If $\text{Aut}(G_1) \cong \text{Aut}(G_2)$ and $G_1$ is finite group then $G_2$ is also finite

  3. If $G_1$ not isomorphic to $G_2$ then Aut($G_1$) not isomorphic to Aut($G_2$).

$G_1$ and $G_2$ are two groups, Aut(G) is group of their automorphisms and "$\cong$" means isomorphism.


I know all three above are incorrect as Aut($\Bbb Z_3$)=$U_3$ (that is, $\Bbb Z_2$) and I also found this statement on groupprops: "Of the three endomorphisms, two are automorphisms: the identity map and the square map. These form a cyclic group of order two: the square map, applied twice, gives the identity map" and Aut($\Bbb Z$) is isomorphic to $\Bbb Z_2$ so

  1. $G_1=\Bbb Z$ and $G_2=\Bbb Z_3$
  2. $G_2=\Bbb Z$ and $G_1=\Bbb Z_3$
  3. We can easily see $\Bbb Z$ and $\Bbb Z_3$ are not isomorphic but Aut($\Bbb Z$) and Aut($\Bbb Z_3$) are.

But the given answer is that (1.) is only incorrect statement.

Is there any problem in my reasoning?

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    $\begingroup$ Your reasoning is solid. $\endgroup$ – Cheerful Parsnip Dec 22 '18 at 7:26
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You are correct in your logic. It is possible that the instructor made an error and meant that one should pick all statements that are incorrect when writing the problem, which happens to be all of them. Perhaps hastily compiling a solution sheet at some later time, they saw the first was incorrect and seeing that the problem was to pick "the" incorrect one they gave that as the answer without checking the others.

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