1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $s=\frac{ord(H)}n$ $$ \epsilon(\alpha)=\sum_{h\in H}a_h =\sum_{i=1}^s\sum_{k=1}^na_{h_ih_0^k} =\sum_{i=1}^sna_{h_i} = n\sum_{i=1}^s a_{h_i} $$

$\endgroup$

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.