Let $G\_1$ be a finite-index subgroup of $G\_2$. Let $i : H^{\ast}(G\_2) \rightarrow H^{\ast}(G\_1)$ be the induced map of rings. There is then a transfer homomorphism $\tau : H^{\ast}(G\_1) \rightarrow H^{\ast}(G\_2)$ whose key property is that $\tau(i(x)) = [G\_2:G\_1] \cdot x$ for all $x \in H^{\ast}(G\_2)$. I have two questions.


1. If $\tau$ a map of rings? In other words, if $x,y \in H^{\ast}(G\_1)$, then must we have $\tau(x \cup y) = \tau(x) \cup \tau(y)$? My guess is that the answer is "no".
2. Assuming that the answer to the first question is "no", does there exist explicit examples of groups $G\_1$ and $G\_2$ as above and elements $x\_1,\ldots,x\_k \in H^1(G\_1)$ such that $\tau(x\_i)=0$ for all $i$ but $\tau(x\_1 \cup \cdots \cup x\_k) \neq 0$?



Cup products and the transfer map
What is true is that $\tau$ is a map of *modules*; that is,
$$\tau(i^\*(x)\cup y) = x\cup \tau(y)$$
for $x\in H^\*(G\_2)$ and $y\in H^\*(G\_1)$.


In particular, the kernel of $\tau$ is a sub-$i^\*(H^\*(G\_2))$-module of $H^\*(G\_1)$.


For an example, consider $G\_1=C\_p$ (cyclic group) and $G\_2=\Sigma\_p$ (symmetric group), where $p$ is an odd prime. The generator $x\in H^2(C\_p)$ satisfies $\tau(x)=0$ (since $H^2(\Sigma\_p)=0$), but $\tau(x^{p-1})\neq 0$. 


*Added.* As Neil points out, I'm using cohomology with mod $p$ coefficients here. 


