Originally Posted by

**Swlabr** Let $\displaystyle o(a)=n, <a> \unlhd G$. Then, the question is does $\displaystyle \widehat{<a>} = 1+a+...+a^{n-1} \in Z(RG)$ hold, for $\displaystyle R$ an arbitrary ring?

Clearly, it is enough to show that $\displaystyle \widehat{<a>}g = g\widehat{<a>} \forall g \in G$, that is, $\displaystyle g^{-1}\widehat{<a>}g = \widehat{<a>}$.

So, according to my notes, $\displaystyle g(1+a+...+a^{n-1})g^{-1} = 1+a^g+...+(a^{n-1})^g$, which I understand perfectly well. However, apparently this is just $\displaystyle \widehat{<a>}$ because $\displaystyle <a> \unlhd G$. I don't get why that is true!

My thought is perhaps that the action of $\displaystyle g \in G$ on elements of a (finite) normal subgroup $\displaystyle N$ perhaps induces a unique element of $\displaystyle N$? That is to say, for $\displaystyle \{x_0, x_i,...,x_n\} = N \unlhd G, |N| \leq \infty, g \in G$, then $\displaystyle {x_i}^g \neq {x_j}^g$ for $\displaystyle i \neq j$?

Thanks in advance!