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Is it true that if $G$ is centerless, so is $\operatorname{Aut}(G)$?

A centerless $G$ can be embedded into $\operatorname{Aut}(G)$ via $G/Z(G)≅\operatorname{Inn⁡}(G)⊲\operatorname{Aut}(G)$. It will be sufficient to show that $\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn⁡}(G)$ is centerless (according to this question). Any clues for continuation?

Roy Sht
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NOTE: The below does not provide a counterexample, it only shows that the method of proof in showing that it suffices to show that the outer automorphism group is centerless will not work (as pointed out in the comments), but the question has been solved in a previous post that has now been linked.

Consider $S_6$, we know that $S_6$ is centerless, and there is an outer automorphism of $S_6$. It is a fact that the outer automorphism group of $S_6$ is cyclic of order $2$ which is not centerless, so the claim that if $G$ is centerless then $\text{Out}(G)$ is centerless is a false claim.

Steven Creech
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  • So what is the meaning of this article's first lines? – Roy Sht Jan 01 '22 at 23:18
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    @Steven Creech I think this is valid proof – Roy Sht Jan 01 '22 at 23:46
  • See here, StevenCreech. According to the paper cited, there are $720$ outer automorphisms of $S_6$. – Shaun Jan 01 '22 at 23:56
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    @Shaun That is true, but the outer automorphism group is not defined to be the set of all outer automorphisms, it is defined to be all automorphisms modulo the group of inner automorphisms. As there are $720$ inner automorphisms (one for each element of $S_6$), and the size of the outer automorphism group is $2$, this would imply that there are $720$ outer automorphisms, so everything seems to check out. – Steven Creech Jan 02 '22 at 00:04
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    Ah, I see. Sorry, and thank you. – Shaun Jan 02 '22 at 00:08
  • I'm not sure what to believe. Both @RoySht and Steven make sense. (Granted, I'm somewhat sleep deprived.) – Shaun Jan 02 '22 at 01:27
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    @Shaun I think I figured it out, and I edited my answer. I did not provide a counterexample to the claim as $\text{Aut}(S_6)$ will be centerless, I believe I just showed that the method that the original poster asked for the proof cannot be done that is you cannot prove this fact by showing that the outer automorphism group is centerless. – Steven Creech Jan 02 '22 at 01:38