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Schreier's conjecture states that the outer automorphism group of any finite simple group is solvable. It has been proved as a consequence of the Classification of Finite Simple Groups, by computing all the outer automorphism groups.

Having looked at the list on the atlas (and knowing that sporadics and alternating groups don't really matter here), I am tempted to say that actually every outer automorphism group of any finite simple group is supersolvable (meaning it has a composition series of normal subgroups with cyclic quotients). Is it true? If not, how many counterexamples are there?

the_fox
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FifteenPointOne
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1 Answers1

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No, the outer automorphism groups are not always supersoluble. An example where this fails is $G = O_8^{+}(3)$, where $\operatorname{Out}(G) \cong S_4$. As Derek points out in the comments section, $O_8^{+}(q)$ (or $D_4(q)$) are, in fact, the only exceptions. Indeed, if $q=p^e$ with $p$ odd, then $\operatorname{Out}(O_8^{+}(q)) \cong S_4 \times C_e$ (see Wilson's "The Finite Simple Groups", bottom of p. $75$).

the_fox
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  • In the Chevalley group notation this would be $D_4(3)$, right? I was tempted to say it was the only exception but I was not sure whether the field automorphism $C_f$ or the exceptional $S_3$ acted nontrivially on the diagonal $(C_2 \times C_2)$. – FifteenPointOne Jul 17 '19 at 16:19
  • Yes, that's $D_4(3)$. I am not too sure about the internal structure of $\operatorname{Out}(D_n(q))$, but I think that if $n>4$ is even and $q>2$, then the outer automorphism group is not supersoluble. However, it is most likely true (for all simple groups) that the supersoluble residual of $\operatorname{Out}(G)$ is either trivial or elementary abelian. – the_fox Jul 17 '19 at 16:47
  • It would be interesting to know if $D_n(q)$ is the only exception, theoretically I guess that you could have other groups in which the exceptional permuted $C_d$ and $C_f$ (making it non-supersolvable) – FifteenPointOne Jul 17 '19 at 17:12
  • I think it is the only exception because in all other cases the outer automorphism group is cyclic-by-cyclic-by-cyclic thus supersoluble. Take this with a pinch of salt though. – the_fox Jul 17 '19 at 17:33
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    In fact $D_4(q)$ for odd $q$ are the only exceptions, because only $D_4(q)$ has the graph automorphism of order $3$. – Derek Holt Jul 17 '19 at 17:40
  • So in any other example, such as $D_n(q)$ for $n > 4$ or maybe $G_2(q)$ etc the graph automorphism of order $2$ does not permute the diagonal and the field automorphisms (when they happen to have the same order)? – FifteenPointOne Jul 17 '19 at 17:55
  • Graph and field automorphisms always commute. The non-supersolvability in the case of $D_4(q)$ arises from the action of the graph automorphism of order $3$ on the group of diagonal automorphisms, which is $C_2 \times C_2$ in this case. – Derek Holt Jul 18 '19 at 18:55
  • In particular they are solvable of length two, that is they are metabelian!! – SKH Jan 14 '21 at 05:29