Lemma 5

Suppose further that $\sigma$ and $\tau$ are semi-simple and hence $N$ (resp. $P$) is the direct sum of the $N_\delta$ (resp. $P_\delta$). For all $n \in N$ (resp. $p \in P$), let $n^\natural$ (resp. $p^\natural$) be its component in $N_0$ (resp. $P_0$). Let $m_0 \in M_0$. Then for all $n \in N$, $(m_0.n)^\natural = m_0.n^\natural$.

By linearity it suffices to consider the case where $n \in N_\delta$. If $\delta \neq \delta_0$, $n^\natural = 0$ and $m_0.n \in P_\delta$ (Lemma 4), hence $(m_0.n)^\natural = 0 = m_0.n^\natural$. If $\delta = \delta_0$, $n^\natural = n$ and $m_0.n \in P_0$ (Lemma 4), hence $(m_0.n)^\natural = m_0.n = m_0.n^\natural$.