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If $\operatorname{Aut} (G)$ is isomorphic to $\operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic to $H$?

My answer is no. $\operatorname{Aut} (\mathbb{Z)}$ is isomorphic to $Z_2$ and $\operatorname{Aut} (Z_3)$ is also isomorphic to $U(3)$, which is isomorphic to $Z_2$. But $\mathbb Z$ is not isomorphic to $Z_3.$ Correct? Thanks

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  • $\begingroup$ Sorry I meant$Z_3$ $\endgroup$
    – ramanujan
    Dec 17, 2018 at 11:22
  • $\begingroup$ @Derek Holt I edited it to$Z_3$. Am I correct now? $\endgroup$
    – ramanujan
    Dec 17, 2018 at 11:23
  • $\begingroup$ Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(\mathbb{Z}_3)$ has exactly two elements, thus it has to be $\mathbb{Z}_2$. $\endgroup$
    – freakish
    Dec 17, 2018 at 11:24
  • $\begingroup$ @freakish thanks $\endgroup$
    – ramanujan
    Dec 17, 2018 at 11:25

3 Answers 3

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Besides your example, there is even an example with finite groups, as $$ {\rm Aut}(S_3)\cong {\rm Aut}(C_2\times C_2)\cong S_3, $$ but $S_3$ is of course not isomorphic to $C_2\times C_2$.

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As another example, both the trivial group $Id$ and the cyclic group of order two $C_2$ have trivial automorphism group: $$\operatorname{Aut}(Id)\cong Id\cong\operatorname{Aut(C_2)}$$

This is the smallest possible example...

(These are the only two groups $G$ with $\operatorname{Aut}(G)\cong Id$. See here for a proof.)

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You can even use non-isomorphic finite groups of the same order. The smallest example is $Aut(C_4\times C_2)\cong Aut(D_8)\cong D_8$.

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