Let $G$ be a group and $I_G$ be the augmentation ideal of the group ring $\mathbb{Z}G$, i.e. $I_G$ consists of formal linear combinations $\sum n_i g_i$ ($n_i\in\mathbb{Z}$, $g_i\in G$) such that $\sum n_i=0$. Now if $N$ is a normal subgroup of $G$, and $\mathcal{N} $ is augmentation ideal of $\mathbb{Z}N$ then why $\mathbb{Z}G\otimes_{\mathbb{Z}N} \mathcal{N}$= $\mathcal{N}^G$ is a non trivial idempotent ideal in $\mathbb{Z}G$.
What does $(\mathbb{Z}G\otimes_{\mathbb{Z}N} \mathcal{N})\cdot(\mathbb{Z}G\otimes_{\mathbb{Z}N} \mathcal{N})$ look like? Is it $(\mathbb{Z}G)^2 \otimes (\mathcal{N})^2?$ Which would mean $(\mathbb{Z}G)^2 \otimes (\mathcal{N})^2= \mathbb{Z}G\otimes_{\mathbb{Z}N} \mathcal{N}$. Is it?