group algebra and kernel

Question

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 $gH = g'H$.

Answer 1

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 $$ \sum a_g g = \sum_{a\in A} \sum_{g\in aH} a_g g = \sum_{a \in A} \sum_{g\in aH} a_g(g-a) \in \left< g-g' \mid gH =g'H \right>$$