I become interested in this problem because $G\cong Aut(G)$ suggests a special symmetry in $G$ (This kind of group describes its own symmetry).

From $G/Z(G)\cong Inn(G)$ we know complete group is the answer for the simplest case, though this class of group itself is quite weird.

What about the case when $Z(G)$ or $Out(G)$ are non-trivial? $G\cong D_8$ is a good example for non-complete group satisfy $G\cong Aut(G)$.

Is there any method to find such non-complete $Aut$-invariant $G$? Maybe we can try apply $Aut$ iteratively to some G and look for fixed points (some interesting results for centerless or non-abelian simple groups can be found at this MO post).

And what other interesting properties does this class of group have?

  • 1
    $\begingroup$ What do you mean by $=$? Unlike $\cong$, usually this is supposed to represent some kind of canonical isomorphism. $\endgroup$ – Alex Degtyarev Mar 16 '14 at 14:48
  • $\begingroup$ What do you mean by $Aut(G)/Out(G)$ ? Out(G) is not a normal subgroup of Aut(G) but a quotient of it. $\endgroup$ – Johannes Hahn Mar 16 '14 at 15:11
  • $\begingroup$ @JohannesHahn Corrected, sorry for that! $\endgroup$ – Yuanzhao Mar 16 '14 at 15:16
  • 1
    $\begingroup$ @Name: that's not true. The dihedral group $D_{12}$ is another example. There are also examples of orders $40$, $48$ and $84$. $\endgroup$ – Derek Holt Mar 16 '14 at 18:08
  • 2
    $\begingroup$ Thank you @DerekHolt for correcting, in fact the dihedral group $D_8$ is the only known $p$-group which is isomorphic to its automorphism group (15.29. UNSOLVED PROBLEMS IN GROUP THEORY THE KOUROVKA NOTEBOOK, arxiv.org/abs/1401.0300) $\endgroup$ – Name Mar 16 '14 at 18:12

Let's call a group $G$ quasicomplete if $Z(G) \ne 1$ and $G \cong {\rm Aut}(G)$.

If $H$ is a complete group with a unique subgroup of index $2$, then $H \times C_2$ is quasicomplete. I have checked that all quasicomplete groups of small order (order less than $768$ so far) apart from $D_8$ have this form. There are lots of $H$ with this property, such as $S_n$ for $n>6$, or ${\rm PGL}(2,p)$ for odd primes $p \ge 5$, or the semidirect product $C_p \rtimes C_{p-1}$ (with faithful action) for odd primes $p$.

Another infinite class of examples is $H \times D_8$, where $H$ is a complete group with no subgroup of index $2$. Again, there are lots of examples of $H$, including some simple groups, such as ${\rm Sp}(2n,2)$ for $n \ge 3$, and there are also complete groups of odd order.

So the question is, are there are any (finite) quasicomplete groups that are not of one of these two types? Of course, this is a more general question than the one about $p$-groups in the Kourovka Notebook.

  • $\begingroup$ should the word pseudocomplete in the second line be quasicomplete? Nice answer by the way. I have absolutely no idea how to answer your question! $\endgroup$ – Nick Gill Mar 18 '14 at 17:29
  • $\begingroup$ Using GAP I managed to compile a complete list of $Aut$-stable groups of order up to 383, a quick check of their center's revealed that all were either centerless or has $Z(G)\simeq\mathbb{Z}_{2}$. Which I've conjectured is a necessary condition for $Aut$-stability. I'm now extremely curious as to why that is happening (or are there counterexamples?). $\endgroup$ – Justin Benfield Mar 7 '16 at 10:01

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.