Lemma 1

Let $H$ be a finite group and $U$ a left $A[H]$-module. Suppose that the $A$-module $U$ has a basis $B$ which is stable under the operations of $H$ in $U$, and put $\Omega = B/H$. For each $w \in \Omega$ let $u_\omega = \sum_{b \in \omega} b$; then

(i) $(u_\omega)_{\omega \in \Omega}$ is a basis of the $A$-module $U^H$.

(ii) For each $\omega \in \Omega$ let $v$, be a point of $\omega$; put $\omega' = w - \{v_\omega\}$ and $B' = \bigcup_{\omega \in \Omega} w'$, then $B'$ is a basis of a supplementary subspace for $U^H$ in $U$.

The union of the set of all $u_\omega$ (for $\omega \in \Omega$) and of $B'$ is a basis of $U$. If $U' = \sum_{\omega \in \Omega} Au$, and $U'' = \sum_{b \in B'} Ab$, we therefore have $U = U' \oplus U''$. On the other hand, we have $u_\omega \in U^H$ for all $w \in \Omega$, hence $U' \subset U^H$. Finally, let $(\alpha_b)_{b \in B}$ be a family of elements of $A$ with finite support and let $x = \sum_{b \in B} \alpha_b b$. If $x \in U^H$, then $\alpha_{hb} = \alpha_b^H$ for all $b \in B$ and all $h \in H$, hence $x \in U'$, and it follows that $U' = u^H$.