2 added 737 characters in body

Edit: As Martin remarked, there is a gap in the proof below. It can be closed by replacing axiom 1 by 1'. However, this isn't very satisfying, as it lowers the categorial flauvor of the characterization. Perhaps one should further investigate, if axiom 1 couldn't be used anyway.

$\hspace{5pt}$1'. If $G=\langle H,x \rangle, n=(G:H)$ and $h \in \cap_{i=0}^{n-1}x^iHx^{-i}$, then $t^G_H(h[G,G])$ is represented $\hspace{10pt}$ $\hspace{12pt}$ by $(hx)^nx^{-n}$.

Futhermore axiom 3 should be

$\hspace{5pt}$3. If $f: G \to G'$ is a homomorphism, $H' \le G, H = f^{-1}(H')$ and $(G:H) = (G':H'),$ $\hspace{5pt}$ $\hspace{12pt}$then the diagram commutes.

As far as I can see, the answers above are all concerned with an explicit construction of the transfer. Here I will go the other direction and characterize the transfer by its properties. Let $V$ denote the usual transfer.

Suppose for each pair $H \le G$ with $(G:H) < \infty$ there is a homomorphism $t^G_H: G_{ab} \to H_{ab}$ satisfying the subsequent properties. Then $t^G_H = V^G_H$.

1. The composition $G_{ab}\hspace{1pt} \xrightarrow{ t } \hspace{1pt} H_{ab} \hspace{1pt} \xrightarrow{\bar{i}} \hspace{1pt} G_{ab}$ is multiplication by $(G:H)$.

2. If $H \le K \le G$ then $t^K_H \circ t^G_K = t^G_H$

3. If $(G:H) = (G':H')$ and $f: G \to G'$ is a homomorphism with $f(H) \le H'$ then the following diagram commutes: $$\begin{array}{ccc} G_{ab} & \xrightarrow{\bar{f}} & G_{ab}' \newline t \downarrow & & \downarrow t' \newline H_{ab} & \xrightarrow[\bar{f}]{} & H_{ab}' \end{array}$$

Proof: a) It's well-known that $V$ satisfies $1.-3.$.

b) By 1., $t^G_H$ and $V^G_H$ agree on $\bar{x}$ for $x \in H$.

c) Suppose $G = \langle H, x \rangle$ and $(G:H) = n$. Let $f: \mathbb{Z} \to G, 1 \mapsto x$. By 1. we have $t: \mathbb{Z} \to n\mathbb{Z}, 1 \to n$. Now 3. implies $t^G_H(\bar{x}) = \bar{x}^n = V^G_H(\bar{x})$. In particular $t^G_G = id|G_{ab} = V^G_G$.

d) We show by induction on $n=(G:H)$ that $t^G_H = V^G_H$ for all $H \le G$. The case $n=1$ was shown in c). Suppose $n>1$ and $t^G_H = V^G_H$ holds for all $H \le G$ with $(G:H) < n$. Let $x \in G$. If $G = \langle H, x \rangle$ then $t^G_H(\bar{x}) = V^G_H(\bar{x})$ by c). So assume $K := \langle H, x \rangle$ is a proper subgroup of $G$. Because of b) we may assume $x \notin H$. Thus $(G:K),(K:H) < n$ and we conclude from 2. and the induction hypothesis and a) that $t^G_H = V^G_H$. q.e.d.

1

As far as I can see, the answers above are all concerned with an explicit construction of the transfer. Here I will go the other direction and characterize the transfer by its properties. Let $V$ denote the usual transfer.

Suppose for each pair $H \le G$ with $(G:H) < \infty$ there is a homomorphism $t^G_H: G_{ab} \to H_{ab}$ satisfying the subsequent properties. Then $t^G_H = V^G_H$.

1. The composition $G_{ab}\hspace{1pt} \xrightarrow{ t } \hspace{1pt} H_{ab} \hspace{1pt} \xrightarrow{\bar{i}} \hspace{1pt} G_{ab}$ is multiplication by $(G:H)$.

2. If $H \le K \le G$ then $t^K_H \circ t^G_K = t^G_H$

3. If $(G:H) = (G':H')$ and $f: G \to G'$ is a homomorphism with $f(H) \le H'$ then the following diagram commutes: $$\begin{array}{ccc} G_{ab} & \xrightarrow{\bar{f}} & G_{ab}' \newline t \downarrow & & \downarrow t' \newline H_{ab} & \xrightarrow[\bar{f}]{} & H_{ab}' \end{array}$$

Proof: a) It's well-known that $V$ satisfies $1.-3.$.

b) By 1., $t^G_H$ and $V^G_H$ agree on $\bar{x}$ for $x \in H$.

c) Suppose $G = \langle H, x \rangle$ and $(G:H) = n$. Let $f: \mathbb{Z} \to G, 1 \mapsto x$. By 1. we have $t: \mathbb{Z} \to n\mathbb{Z}, 1 \to n$. Now 3. implies $t^G_H(\bar{x}) = \bar{x}^n = V^G_H(\bar{x})$. In particular $t^G_G = id|G_{ab} = V^G_G$.

d) We show by induction on $n=(G:H)$ that $t^G_H = V^G_H$ for all $H \le G$. The case $n=1$ was shown in c). Suppose $n>1$ and $t^G_H = V^G_H$ holds for all $H \le G$ with $(G:H) < n$. Let $x \in G$. If $G = \langle H, x \rangle$ then $t^G_H(\bar{x}) = V^G_H(\bar{x})$ by c). So assume $K := \langle H, x \rangle$ is a proper subgroup of $G$. Because of b) we may assume $x \notin H$. Thus $(G:K),(K:H) < n$ and we conclude from 2. and the induction hypothesis and a) that $t^G_H = V^G_H$. q.e.d.