Let $G$ is an arbitrary group and and $U=U_1(\mathbb{Z}G)$ is the set of normalized units of $ZG$ i.e. $U_1(\mathbb{Z}G)$ here means units of $\mathbb{Z}G$ with augmentation $1$, i.e. set of all elements $u=\sum_{g\in G} a(g)g \in U(\mathbb{Z}G)$ such that $\sum_{g \in G} a(g)=1$.
Now if $U$ contains $G$ and [$U:G$] is finite, then why does it imply that $G$ contains a subgroup $H$ which is normal and of finite index in $U$.
If I take $H=G$, then index point is satisfied but does $U$ normalize $G$? I do not think so, but if it does how to prove it?