Let $\cal{U}$ be the unit group of group ring $\Bbb{Z}G$ then the Normalizer Problem (NP) states that $N_{\cal{U}}(G)=G\frak{z}$ where $\frak{z}=\cal{Z(U)}$.

Now why (NP) is equivalent to saying that $Aut_Z(G)=Inn(G)$ where $Aut_Z(G)$ denotes the automorphisms of $G$ induced by conjugation with units of $\Bbb{Z}G$.

If $u\in N_{\cal{U}}(G)$ then it certainly induces an automorphism of $G$ and if (NP) holds then conjugation by $u$ is certainly inner too. But I do not see how to deal with converse?



Take an element $u \in N_{\mathcal{U}}(G)$, then we need to prove that $ u \in G \mathcal{Z}(\mathcal{U})$ (the other inequality is of course Always true). Now due to the definition of the normalizer we know that conjugation by $u$ on $G$ is an automorphism of $G$. Hence we find a $h \in G$ such that for any $g \in G$ we have: $$ u^{-1} g u = h^{-1}gh $$ We can rephrase this to find for any $g \in G$: $$ hu^{-1} g uh^{-1} = g $$ Thus we know that $uh^{-1} \in \mathcal{U}$ centralizes $G$, but then it also centralizes $\mathbb{Z}G$ (since $\mathbb{Z}$ is commutative). Hence we have that $uh{-1} \in \mathcal{Z}(\mathcal{U})$. Thus we find that $$ u = uh^{-1} h \in \mathcal{Z}(\mathcal{U}) G $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.