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 2010 at 8:22

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 2010 at 9:35