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.

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".

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