# order of $h_0$ divides augmentation of $\alpha\in \Bbb{Z}H$

Let $H$ be an abelian group and $\Bbb{Z}H$ be its integral group ring. Now let $\alpha=\sum_{h\in H}a_h.h\in \Bbb{Z}H$ and $\alpha(1-h_0)=0$ for some $h_0\in H$.

Why does this imply that order of $h_0$ divides $\epsilon(\alpha)$ where $\epsilon(\alpha)=\sum_ha_h$.

After trying out some basic stuff, I could not see any connection between order of element of a group and augmentation of an element of group ring. Any help is appreciated. Thanks!

## 1 Answer

from $\alpha(1-h_0)=0$ $$\sum_{h\in H}a_h\cdot h = \sum_{h\in H} a_h\cdot hh_0$$ also by relabelling $$\sum_{h\in H}a_h\cdot h = \sum_{hh_0 \in H} a_{hh_0}\cdot hh_0$$ by linear independence $\forall h\in H$ $$a_h = a_{hh_0}$$ this implies $\forall k$ $$a_{hh_0^k}=a_h$$ let $n$ be the order of $h_0$. we now have $$\sum_{j=1}^n a_{hh_0^k} = na_h$$ hence, setting $s=\frac{ord(H)}n$ $$\epsilon(\alpha)=\sum_{h\in H}a_h =\sum_{i=1}^s\sum_{k=1}^na_{h_ih_0^k} =\sum_{i=1}^sna_{h_i} = n\sum_{i=1}^s a_{h_i}$$