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I just learned about the transfer homomorphism, and I am having trouble internalizing it. I am learning from 'A Course in the Theory of Groups', and I was hoping that perhaps someone had a more intuitive explanation of the transfer to supplement the book's construction.

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2 Answers 2

up vote 9 down vote accepted

The transfer tries to capture conjugacy in a homomorphism. This is a little hard to do so the definition is a little complicated. I give a simpler to verify definition from Kurzweil–Stelmmacher, then the standard product-of-conjugates formula and its specialization to the center and to a Hall subgroup.

Smoother definition

Kurzweil–Stellmacher gives a slightly nicer definition of the transfer: Let $H \leq G$ be a subgroup of finite index in the group $G$. If $R,S$ are transversals of $H$ in $G$, define $$(R|S) = \prod_{(r,s) \in R \times S \atop Hr = Hs} \overline{ rs^{-1} } \in H/[H,H]$$ Then one easily verifies that $$(R|S)^{-1} = (S|R), \quad (R|S)(S|T) = (R|T), \quad (Rg|Sg) =(R|S), \quad \text{and} \quad (Rg|R) =(Sg|S)$$ (At the moment, I couldn't do the last in my head, but $(Rg|R) = (Rg|Sg) (Sg|S) (S|R) = (Sg|S)$ since the outer factors cancel by the previous properties and the fact that $(R|S)$ lives in an abelian group.)

Hence the map $tr:G \to H/[H,H]: g \mapsto (Rg|R)$ is well-defined (by the last property) and a homomorphism since $$(Rgh|R) = (Rgh|Rh)(Rh|R) =(Rg|R)(Rh|R)$$

Some important properties

For g in G, consider the orbits of $\langle g \rangle$ on the cosets $G/H$. Let them have representatives $g_i \in R$ and size $n_i$. Then $$tr(g) = \prod g_i g^{n_i} g_i^{-1}$$ is a product of conjugates of $g$. In particular, if $g \in Z(G)$, then all the conjugates are the same, so $$tr(g) = g^{[G:H]}, \quad \text{if } g \in Z(G)$$ and a similar formula holds if we consider $tr(g)$ mod the subgroup $$H^* = \langle h^{-1} h^g : h,h^g \in H, g \in G \rangle$$ since modulo this subgroup, all $G$-conjugates are the same: $$tr(g) \equiv g^{[G:H]} \mod H^*, \quad \text{if } g \in H$$

In particular, if $G$ does not sufficiently swirl $H$, that is, if $H^* < H$, and if $\gcd([H:H^*],[G:H])=1$ (for instance if $H$ is Hall or Sylow), then the transfer map restricts to an automorphism of $H/H^*$ and in particular $G$ has a proper normal subgroup $K=\ker(tr)$ such that $H \cap K = H^*$. When $H$ is a Sylow $p$-subgroup, this $H^*$ is called the focal subgroup and is equal to $H \cap [G,G]$.

Notice that $H^* \geq [H,H]$ measures how much extra conjugacy (fusion) $G$ causes in $H$. The transfer allows us to pull $H^*$ back into $G$, and so we get a normal subgroup of $G$ that somehow measures fusion in $H$. Again this is most useful and precise when $H$ is a Hall or Sylow subgroup of $G$.

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Other good answers would include the definition from homology of groups, and the definition from character theory. This answer is closest to the material in Robinson, if I recall correctly. –  Jack Schmidt Oct 2 '12 at 21:45

There is a way (in my opinion, quite natural) which leads not only to transfer map, but also to the induced representations.

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Please explain your downvotes. –  Alex W Jun 5 at 16:48

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