Proposition 6

*Let M be a semi-simple g-module. Then the submodule $M_0$ of invariant elements of M admits one and only one supplement stable under the $x_M$, namely the submodule $M_1$ generated by the $x_M.m$ ($x \in g, m \in M$).

Let $M'$ be a submodule of M which is stable under the $x_M$ and a supplement of $M_0$ in M. For all $m \in M, m = m_0 + m'$ with $m_0 \in M_0, m' \in M'$, and hence $x_M m = x_M m' \in M'$. Hence $M_1 \subset M'$. Let $M_2$ be a submodule of $M'$ stable under the $x_M$ and supplementary to $M_1$ in $M'$. For all $m \in M_2$,

for all $x \in g$, hence $m \in M_0$ and hence $m = 0$. Hence $M_2 = \{0\}$, which proves that $M_1 = M'$.