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Let $R$ be a ring with identity $1_R\neq 0$, and let $G$ be a finite group with group operation written multiplicatively. Then we can form the group ring $$RG = \{a_1g_1+a_2g_2+\cdots+a_ng_n\ |\ a_i\in R\}.$$ In Dummit & Foote's Abstract Algebra, they state that if $1_G$ is the identity of $G$, then we write $r1_G$ as $r$, and $1_Rg$ as $g$. But then, shouldn't we have: $$1_G = 1_R1_G = 1_R?$$ If that is the case, is there any harm in defining only one identity $1_{RG}$ for $RG$?

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    $\begingroup$ None at all, it's just notational convention. $\endgroup$ – Adam Hughes Aug 5 '14 at 21:26
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There's no harm in this. It's just notation.

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  • $\begingroup$ Great, thanks for the fast reply. One more thing, they say that $G$ "appears" in $RG$ but formally we have: $$G\cong \hat{G}=\{1_Rg\ |\ g\in G\} \leq RG$$ right? $\endgroup$ – chs21259 Aug 5 '14 at 21:27
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    $\begingroup$ Yes, this is the right idea. Be careful of what you mean by $\hat{G} \leq RG$ though! After all, $\hat{G}$ is not a subring! $\endgroup$ – Alex G. Aug 5 '14 at 21:31
  • $\begingroup$ Gotcha, thanks! $\endgroup$ – chs21259 Aug 5 '14 at 21:32

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