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A group $G$ is called perfect iff $G’ = G$.

A group $G$ is called complete iff $Z(G) = \{e\}$ and $Aut(G) \cong G$.

Does there exist a non-trivial group $G$, that is both perfect and complete at the same time?

Motivation behind this question:

Both «perfect groups» and «complete groups» are translated to Russian as «совершенные группы» despite those properties being completely different and not implying one another. It would however be interesting to know (despite this interest being purely recreational) whether or not these two properties defined by the same word are completely disjoint, or are there such groups, for which such an ambiguous translation does not cause problems.

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    $\begingroup$ obviously. $G = \{e\}$. $\endgroup$
    – mathworker21
    Jun 16, 2019 at 13:26
  • $\begingroup$ @mathworker21, I meant non-trivial... Thanks for pointing out my typo. $\endgroup$
    – Chain Markov
    Jun 16, 2019 at 13:28

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A way to guarantee the group is perfect with trivial centre is to look at a finite nonabelian simple group. If the group has no outer-automorphisms then we have $\operatorname{Aut}G\cong\operatorname{Inn}G\cong G/Z(G)=G$. We can go through the classification and find such examples, e.g., the exceptional Chevalley group $E_8(p)$ where $p$ is prime, or the Mathieu groups $M_{11},M_{23},M_{24}$.

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  • $\begingroup$ I think you mean $Inn(G) \cong G / Z(G)$ $\endgroup$
    – Sunny
    Jun 16, 2019 at 15:49
  • $\begingroup$ Oops, yes. Thank you. $\endgroup$
    – user10354138
    Jun 16, 2019 at 15:50

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