Let $i : H \to G$ be a subgroup of finite index. The transfer map is a special homomorphism $V(i) : G^\mathrm{ab} \to H^\mathrm{ab}$. The usual ad hoc definition uses a set of representatives of $H$ in $G$ and then you have to check that it is independent from this choice and that it is a homomorphism at all. I think this definition is not enlightening at all (although it is, of course, useful for explicit calculations). A better one uses group homology. Namely, for a $G$-module $A$ there is a natural transformation $A_G \to \mathrm{res}^{G}_{H} A_H$, $[a] \mapsto \sum_{Hg \in H/G} [ga]$, which extends to a natural transformation $H_\*(G;A) \to H_\*(H;\mathrm{res}^{G}_{H} A)$ (usually called corestriction or transfer). Now evaluate at $A = \mathbb{Z}$ and $* = 1$ to get $G^\mathrm{ab} \to H^\mathrm{ab}$. One can then calculate this map using the explicit isomorphisms and homotopy equivalences involved; but now you know by the general theory that it is a well-defined homomorphism.

It also follows directly that the transfer is actually a functor $V : \mathrm{Grp}_{mf} \to \mathrm{Ab}^{\mathrm{op}}$ with object function $G \mapsto G^{\mathrm{ab}}$, where $\mathrm{Grp}_{mf}$ is the category whose objects are groups and whose morphisms are monomorphisms of finite index.

I would like to know if there is an even more "abstract" definition. To be more precise: Is there a categorical characterization of the functor $V$ which only uses the adjunction $\mathrm{Grp} {\longleftarrow \atop \longrightarrow} \mathrm{Ab}$?

Edit: There are many interesting answers so far which give, in fact, very "enlightening" definitions of the transfer. But I would also like to know if there is a pure categorical one, such as the one given by Ralph.

Edit: A very interesting note by Daniel Ferrand is A note on transfer. There a more general statement is proven (even in a topos setting): Let $G$ act freely on a set $X$ such that $X/G$ is finite with at least two elements. Then there is an *isomorphism* of abelian groups $(\mathrm{Ver},\mathrm{sgn}) : {\mathrm{Aut}_{G}(X)}^{\mathrm{ab}} \cong G^{\mathrm{ab}} \times \mathbb{Z}/2$. It is natural with respect to $G$-isomorphisms. Here again I would like to ask if it is possible to characterize this isomorphism by its properties (instead of writing it down via choices, whose independence has to be shown afterwards).

Proposition 7.1. in this paper includes the interpretation via determinants mentioned by Geoff in his answer, actually something more general: For w.l.o.g. abelian $G$ there is a commutative diagram

$\begin{matrix} {\mathrm{Aut}_{G}(X)}^{\mathrm{ab}} & \cong & \mathrm{Aut}_{\mathbb{Z}G}{\mathbb{Z}X}^{\mathrm{ab}} \\\\ \downarrow & & \downarrow \\\\ G \times \mathbb{Z}/2 & \rightarrow & (\mathbb{Z} G)^{x} \end{matrix} $

Thus we may think of transfer and signature as the embedding the standard units into the group ring.

De gustibus non est disputandum, I suppose. $\endgroup$4more comments