Transfer homomorphism

Remarks

The product is taken in the abelian group $H/H'$, so its order does not matter. One proves that $V$ is a group homomorphism and that it does not depend on the choice of $t_1, \dots, t_n$.

Example: if $G$ is abelian, then $H' = \set{e}$ and $h_i(g) = t_{\sigma_g(i)}^{-1} g t_i$. Since $\sigma_g$ permutes $\set{1, \dots, n}$, the factors $t_i$ cancel, and $V(g) = g^n$. For $G = \Z$ and $H = n\Z$, the transfer is $V(k) = nk$.

Non-example: take the product $h_1(g) \cdots h_n(g)$ in $H$ itself, without passing to $H/H'$. In general it depends on the order of the factors and on the choice of the $t_i$, so it does not give a well-defined homomorphism to $H$.

A main application is Burnside's normal $p$-complement theorem. Let $P$ be a Sylow $p$-subgroup of a finite group $G$ that lies in the centre of its normaliser. Then $G$ has a normal subgroup $K$ with $G = KP$ and $K \cap P = \set{e}$.