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Let $G$ be a group .

Does a subring of the integral group ring $\mathbb{Z}[G]$ has the form $\mathbb{Z}[H]$ for a subgroup $H$ of $G$?

Thanks in advance.

M.Ramana
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1 Answers1

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Not necessarily.

Let $G=\{e,g\}$ have two elements. Consider, say $$R=\{ae+bg:a,b\in\Bbb Z:a-b\in2\Bbb Z\}.$$

Angina Seng
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  • @ Lord Shark the Unknown Thank you for interesting answer. How about retracts of $\mathbb{Z}[G]$? i.e. if $S$ is a retract of $\mathbb{Z}[G]$, then is there a retract $H$ of $G$ so that $S=\mathbb{Z}[H]$? – M.Ramana Oct 05 '17 at 05:19
  • @ Common If $G$ is a group and $R$ is any ring with an identity element, the group ring $RG$ is defined to be the set of all formal sums $\sum_{x\in G} r_x x$ where $r_x \in R$ and $r_x = 0$ with finitely many exceptions. – M.Ramana Oct 05 '17 at 05:24
  • @ Common $RG$ is a ring together with the rules of addition and multiplication $(\sum_{x}r_x x)+(\sum_{x}r'{x}x)=\sum{x}(r_x +r'x )x$ and $(\sum_x r_x x)(\sum_x r'_x x)=\sum_x (\sum{yz=x}r_y r'_z )x$, respectively. – M.Ramana Oct 05 '17 at 05:28