If $G$ is abelian and $H \leq G$ of index $n$ , then show that transfer map is just $g \to g^n$.
If i follow the definition from issac, transfer map will be same as pretranfer map as G is abelian i.e. $v : G \to H$ where $g$ goes to $\Pi_{t \in T} (tg)(t.g)^{-1}$, so if i open it up, there will be $g^n$, and once all the elements of transversal $T$ will have a product, say $t_1t_2...t_n$ , but what about $\Pi_{t \in T} (t.g)^{-1}$ any why $(t_1t_2...t_n)(\Pi_{t \in T} (t.g)^{-1})$ will vanish to become $e$. These $(t.g)$ will also be elements $t_1,t_2,.....t_n$ of $T$ in some order, but what can we say about there inverses?
Thanks!
If I understood correctly what you meant, we have the following: in the product
$$\prod_{t\in T}(tg)(t\cdot g)^{-1}$$
the order is unimportant as everything's abelian here. Remember that $\;(g\cdot t)\in T\;$ by definition, so when we re-arrange the factors in the above product, each element of $\;T\;$ will reduce to the unit when multiplied by its inverse, and in the end you will remain with only
$$\prod_{t\in T}g=g^{|T|}=g^n$$
and everything's fine.
You may want to see it this way: if $\;T:=\{t_1,...,t_n\}\;$, then if $\;g\in G\;$, the product $\;g\cdot t_1\;$ , for example, is defined to be the unique element $\;t_{i_1}\in T\;$ s.t. $\;t_{i_1}\in H(t_1g)\;$ ,and re-arranging:
$$\prod_{i=1}^m(t_ig)(t_i\cdot g)^{-1}=\prod_{i=1}(t_ig)t_{k_i}^{-1}=\prod_{i=1}^ng=g^n$$
Everything's explained pretty nicely in Isaacs' book, but you may want to read it also in Robinson's, Rotman's or any other good book in group theory.