Let $G$ be a finite group. I want to prove that if $u\in \mathbb{Z}G$ be a torsion unit (i.e. for some $n$, $u^n=1$) of integral group ring $\mathbb{Z}G$ such that $u$ normalizes $G$, then $u$ is a trivial unit i.e. $u=\pm g$ for some $g\in G$.
In simple language , I need to prove that for a finite group $G$, all torsion units in $N_U(G)$ are trivial (i.e. of form $\pm g$) where $U$ is unit group of $\mathbb{Z}G$
It looked easy, but after several attempts and using known results on units in Integral group rings I was getting nowhere.
Any help is appreciated! Thanks
Let $u$ be a unit of finite order in $N_U(G)$ and $^*$ be the standard involution on $\mathbb{Z}G$.
From An Introduction to Group Rings by Milies and Sehgal we need
Now, by hypothesis for each $r \in G$ there exists an $s \in G$ such that $u=rus$. Then $uu^*=(rus)(rus)^*=ruu^*r^{-1} \quad \forall r \in G$, i.e. $uu^*$ is central, which implies $ uu^* uu^*=uu u^*u^*$ which in turn implies $uu^*=u^*u$. But now $u$ satisfies the conditions of the corollary, thus $u=\pm g$ for some $g \in G$.