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I want to prove this lemma

Let $G$ be a group, $H$ an invariant subgroup, $k$ a field; consider the canonical map $\varphi \colon k[G] \to k[G/H]$; then the kernel of $\varphi$ is the ideal generated by the elements $(g − g')$, where $gH = g'H$.

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On one hand for $gH = g'H$ we have $g-g' \in \ker \let\phi\varphi\phi$ as $$ \phi(g-g') = gH -g'H = 0 $$ On the other hand, pick a set $A \subseteq G$ of representatives for the left cosets in $G/H$, suppose $\phi\bigl(\sum a_g g\bigr) = 0$, that is $$ 0 = \phi\left(\sum a_g g\right) = \sum a_g gH = \sum_{a \in A} \sum_{g \in aH} a_ggH = \sum_a \left(\sum_{g \in aH} a_g\right)aH $$ As the $aH$ are linearly independent over $k$, we have $\sum_{g \in aH} a_g = 0$ for each $a\in A$, that is $$ \sum a_g g = \sum_{a\in A} \sum_{g\in aH} a_g g = \sum_{a \in A} \sum_{g\in aH} a_g(g-a) \in \left< g-g' \mid gH =g'H \right>$$

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  • $\begingroup$ Are you assuming that $H$ has finite index in $G$? $\endgroup$ – Tobias Kildetoft Jun 25 '14 at 11:10
  • $\begingroup$ I do not think so. All but finitely many $a_g$ are equal to $0$, and hence for all but finitely many $a$ we have $\sum_{g \in aH} a_g = 0$, so all sums above are finite sums. Did I overlook something (besides the dirty notation)? $\endgroup$ – martini Jun 25 '14 at 11:13
  • $\begingroup$ Ahh, right. There does seem to be something missing in the second displayed equation (it has no mention of $\varphi$ it seems). $\endgroup$ – Tobias Kildetoft Jun 25 '14 at 11:14
  • $\begingroup$ You're right, will add the $\phi$, thanks. $\endgroup$ – martini Jun 25 '14 at 11:15

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