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(added) Definition. A complete map of a group is a permutation $\phi$ such that $g\mapsto g\phi(g)$ is also a permutation. A group is admissible if it admits a complete map.

I want to know when an abelian group of even order is admissible? And when is a nonabelian group of even order admissible?

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    $\begingroup$ To clarify: a complete map $\phi$ is a permutation of $G$ such that the map $g \mapsto g\phi(g)$ is also a permutation. $\endgroup$
    – Colin Reid
    Commented Sep 19, 2010 at 8:22
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    $\begingroup$ It would not hurt to explain what admissible means in this context. (And someone should enforce a ban on using words like admissible, normal and regular for —say— some 50 years to name anything!) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 14, 2013 at 3:29
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    $\begingroup$ @MarianoSuárez-Álvarez although I agree, the terminology can be found in this 1950 paper by L.J. Paige (more than 50 years before the question was asked!) $\endgroup$
    – YCor
    Commented Oct 24, 2022 at 18:02
  • $\begingroup$ @YCor, while I am usually annoyed by requests for context, without context it is simply impossible to know what admissible means. I bet you a beer that someone else gave a different and completely unrelarted defintion of what an admissible group is in some other paper at least a couple of decades ago :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 24, 2022 at 21:53
  • $\begingroup$ @MarianoSuárez-Álvarez, call a paper admissible if it was written more than 50 years after the last usage of the term "admissible", and so can use that term again …. $\endgroup$
    – LSpice
    Commented Oct 25, 2022 at 4:08

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There is a conjecture of Hall and Paige on this subject, in the paper

Hall, Marshall; Paige, L. J. Complete mappings of finite groups, Pacific J. Math. 5 (1955), 541–549

Let $G$ be a finite group of even order. Then $G$ admits a complete map if and only if its $2$-Sylow subgroups are non-cyclic.

By the looks of it, the conjecture hasn't quite been resolved yet, but a lot of progress has been made. See, for instance:

Stewart Wilcox, Reduction of the Hall–Paige conjecture to sporadic simple groups, J.Algebra, 321:5, 1407–1428

This paper and references ought to give a good idea of what is known.

Hall and Paige proved their conjecture in the soluble case, so the answer to your question in the abelian case is that a finite abelian group admits a complete map exactly if its $2$-Sylow subgroup is either trivial or non-cyclic.

(Aside: the identity is a complete map for any finite group of odd order, hence the focus on groups of even order in the question.)

(Aside 2: this answer brought to you by the power of Google; I don't actually know much about the subject myself.)

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    $\begingroup$ In the abelian case, it's possible to give a direct proof. An abelian group $G$ decomposes into a direct product of an abelian 2-group $A$ and an abelian group $B$ of odd order. A necessary condition for the existence of a complete map is that the sum of all elements of $G$ is 0, which implies that $A$ is not a cyclic group. Conversely, if $A$ is a non-cyclic abelian 2-group and $\varphi:A\to A$ has the property that both $\varphi$ and $a\to\varphi(a)+a$ are bijections then $(a,b)\to (\varphi(a),b)$ is a complete map of $G.$ The map $\varphi$ can be chosen to be a suitable matrix polynomial. $\endgroup$
    – Victor Protsak
    Commented Sep 19, 2010 at 9:35
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The only even groups that are admissable, are the finite fields of the form $GF(2^n)$, that is the fields of characteristic 2. The Hall-Paige conjecture has actually been laid to rest (proven in 2009), although I'm not sure if the final result has actually been published yet.

Complete mappings are closely related to orthomorphisms, so if you're looking through the literature, be sure to search for both of these. I define an orthomorphism $\theta$, as a permutation of a group $G$, such that $\theta(x) - x$ is also a permutation. A complete mapping is a permutation such that $\theta(x) + x$ is also a permutation, so you can see why the existence of an orthomorphism implies the existence of a complete mapping, and vice versa. A linear orthomorphism is of the form $\theta(x) = ax$, for some $a \in G$, $a \neq 0,1$.

For a complete set of (n-3) orthogonal orthomorphisms over any field $GF(n)$, you can always achieve this with linear orthomorphisms, since there are always $n - 3$ choices for $a$, and each of these orthomorphisms are orthogonal to each other.

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  • $\begingroup$ Can you post any reference for the even groups case? I've been trying to find without success so far. $\endgroup$
    – Sudarsan
    Commented Jul 20, 2014 at 9:19
  • $\begingroup$ A. B. Evans, Orthomorphism graphs of groups, Lect. Notes in Math. 1535, Springer, 1992. This contains a very detailed summary of all things Orthomorphisms/Complete Mappings. There's plenty of constructions over finite fields in the literature, and these work in even fields. Note: The only even groups that admit Orthomorphsims/Complete Mappings are fields. $\endgroup$
    – Dave Fear
    Commented Jul 27, 2015 at 1:27

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