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Does every finite non-trivial complete group have even order?

I checked three well known classes of complete groups, and this statement is true for them all:

1) Symmetric groups: All symmetric groups have even order (a well known fact)

2) Automorphism groups of non-abelian simple groups: All non-abelian simple groups are of even order by Feit-Thompson theorem. Thus they have elements of order 2. And, as all non-abelian simple groups are centreless, this element is not in its centre. Thus, the conjugation by it is an automorphism of order 2. That means, that all automorphism groups of non-abelian simple groups have even order.

3) Holomorphs of cyclic groups of odd order (here is the proof, why they are complete: Is the statement that $ \operatorname{Aut}( \operatorname{Hol}(Z_n)) \cong \operatorname{Hol}(Z_n)$ true for every odd $n$?): All cyclic groups are abelian and thus all cyclic groups of even order have automorphism of order 2, that maps all their elements to their inverse. Thus both their automorphism group and their holomorph are of even order.

However, I do not know, how to prove this statement in general.

Any help will be appreciated.

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    $\begingroup$ No. Try searching for "odd order complete group". $\endgroup$ – Derek Holt Sep 5 '18 at 17:57
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No. There was at one time a conjecture that this is true, but an example of a complete group of order $3\cdot 19\cdot 7^{12}$ was produced by R.S. Dark in “A complete group of odd order”.

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