Cup products and the transfer map

Question

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 $\tau(x_1 \cup \cdots \cup x_k) \neq 0$?

Answer 1

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.

Answer 2

Since $i$ is a map of rings, if $\tau$ were a map of rings, then $\tau\circ i$ would also be a map of rings. But the latter maps $1\in H^0(G_2)$ to $[G_2:G_1]$ which, in general, is not $1$.