Let $R$ be a ring and $G$ be a group and $RG$ be the group ring. Denote by $[R,R]$, the additive subgroup generated by all lie products $[x,y]=xy-yx , \forall\ x,y\in R$.

Then how is this that $[\sum_{g\in G}\beta(g)g,\sum_{h\in G}\gamma(h)h]=\sum_{g,h}\beta(g)\gamma(h)[g,h]=\sum_{g,h}\beta(g)\gamma(h)(gh-hg)$.

If suppose I take two elements $x=a_1g_1+a_2g_2 , y=b_1g_1+b_2g_2$ in $RG$, then $[x,y]= (a_1g_1+a_2g_2)(b_1g_1+b_2g_2)-(b_1g_1+b_2g_2)(a_1g_1+a_2g_2)=(a_1b_1-b_1a_1)g_1^{2}+(a_2b_2-b_2a_2)g_2^{2}+(a_1b_2-b_1a_2)g_1g_2+(a_2b_1-b_2a_1)g_2g_1$.

But I am unable to put them in above required form. Any help is appreciated.


  • $\begingroup$ The claim is false. If $G$ is the trivial group, then it would boil down to $[R,R]$ being the $R$-linear span of $0$. But this is not the case when $R$ is noncommutative. Maybe whatever source you are using just assumes that $R$ is commutative? $\endgroup$ – darij grinberg Oct 27 '15 at 5:58
  • $\begingroup$ Yes if $R$ is commutative than it is fine but the source is "Units in integral group rings"- Sehgal page $25$ which does not assume $R$ being commutative $\endgroup$ – Bhaskar Vashishth Oct 27 '15 at 6:01
  • $\begingroup$ What do you even mean by the group ring $R[G]$ if $R$ is noncommutative? Are the elements of $R$ and $G$ required to commute? $\endgroup$ – Qiaochu Yuan Oct 27 '15 at 6:31
  • 1
    $\begingroup$ @QiaochuYuan yes elements of R and G commute $\endgroup$ – Bhaskar Vashishth Oct 27 '15 at 8:24

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