7
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Sorry I meant$Z_3$ $\endgroup$ – ramanujan Dec 17 '18 at 11:22
  • $\begingroup$ @Derek Holt I edited it to$Z_3$. Am I correct now? $\endgroup$ – ramanujan Dec 17 '18 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 '18 at 11:24
  • $\begingroup$ @freakish thanks $\endgroup$ – ramanujan Dec 17 '18 at 11:25
4
$\begingroup$

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 $C_2\times C_2$.

$\endgroup$
2
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.