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I want to know when an abelian group of even order is admissible (or has a complete map)? And when a nonabelian group of even order is admissible (or has a complete map)?

Thanks.

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To clarify: a complete map $\phi$ is a permutation of $G$ such that the map $g \mapsto g\phi(g)$ is also a permutation. –  Colin Reid Sep 19 '10 at 8:22
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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!) –  Mariano Suárez-Alvarez Oct 14 '13 at 3:29
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2 Answers

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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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. –  Victor Protsak Sep 19 '10 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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