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Can two groups of automorphisms over non-isomorphic groups be isomorphic? If $G$ and $G'$ are non-isomorphic groups and $\text{Aut}(G)$ and $\text{Aut}(G')$ be their group of automorphisms, then can they be isomorphic to each other or not?

ke shradha
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Aut$(\Bbb Z_3)$ and Aut$(\Bbb Z_4)$ are both isomorphic to $\Bbb Z_2$.

Angina Seng
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A non-abelian example, $Aut(Q) \cong S_4 \cong Aut(S_4)$. Here $Q$ is the quaternion group of $8$ elements.

Nicky Hekster
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  • Are they always same...can u give an example where they are not same but isomorphic – ke shradha Jul 30 '18 at 07:32
  • What do you exactly mean? Two isomorphic groups with non-isomorphic automorphism groups? Those do not exist! – Nicky Hekster Jul 30 '18 at 08:59
  • I mean two non-isomorphic groups whose automorphism groups are only isomorphic they should not be same...like ur example both are $S_4$ – ke shradha Jul 30 '18 at 10:17
  • OK (also have a look at https://math.stackexchange.com/questions/586471/show-that-if-gcdg-h-1-then-textautg-times-h-cong-textau), let $p \geq 5$ be prime. Then $Aut(Q \times C_p) \cong S_4 \times C_{p-1} \cong Aut(S_4 \times C_p)$. Now you have three non-isomorphic non-abelian groups. – Nicky Hekster Jul 30 '18 at 20:54
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As a more complicated example, we have $\operatorname{Aut}(C_2\times C_8)\cong C_2\times D_8$ and $$\operatorname{Aut}(C_2\times C_{12})\cong\operatorname{Aut}(C_3)\times\operatorname{Aut}(C_2\times C_4)=C_2\times D_8.$$

Jianing Song
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