Lemma 2

Let G be a group (resp. a topological group), G' a normal subgroup of G, V a finite-dimensional vector space over a commutative field k (resp. over K), $\rho$ a linear representation (resp. a continuous linear representation) of G on V and $\rho' = \rho|_{G'}$. (i) If $\rho$ is semi-simple, $\rho'$ is semi-simple. (ii) If $\rho'$ is semi-simple and every finite-dimensional linear k-representation (resp. continuous linear K-representation) of $G/G'$ (resp. $G/\overline{G}'$) is semi-simple, then $\rho$ is semi-simple. Suppose that $\rho$ is semi-simple; we prove that $\rho'$ is semi-simple. It suffices to consider the case where $\rho$ is simple. Let $V'$ be a minimal non-zero sub-G'-module of V. For all $g \in G$, $\rho(G')\rho(g)V' = \rho(g)\rho(G')V' = \rho(g)V'$, in other words $\rho(g)V'$ is stable under $\rho(G')$; if $V''$ is a sub-$G'$-module of $\rho(g)V'$, then $\rho(g)^{-1}V''$ is a sub-$G'$-module of $V'$ and hence $V''$ is equal to $\{0\}$ or $\rho(g)V'$.

Thus, for all $g \in G$, $\rho(g)V'$ is a simple $G'$-module. But $\sum_{g \in G} \rho(g)V'$ is a non-zero sub-$G$-module of $V$, whence $V = \sum_{g \in G} \rho(g)V'$. Hence $\rho'$ is semi-simple.

Suppose that $\rho'$ is semi-simple. Let $W$ be a non-zero sub-$G$-module of $V$. As $\rho'$ is semi-simple, there exists a projector $f_0$ of $V$ onto $W$ which commutes with $\rho'(G)$. Let $E$ be the set of $f \in \mathscr{L}(V, V)$ which commute with $\rho(G')$, which map $V$ into $W$ and whose restriction to $W$ is a homothety; for $f \in E$, let $\alpha(f)$ denote the ratio of the homothety $f|W$. Then $f_0 \in E$ and $\alpha(f_0) = 1$. Clearly $\alpha$ is a linear form on $E$. Let $F = \mathrm{Ker}\, \alpha$, which is a hyperplane of $E$. For $f \in E$ and $g \in G$, we write $\sigma(g)f = \rho(g) \circ f \circ \rho(g)^{-1}$; then $\sigma(g)f$ maps $V$ into $W$ and its restriction to $W$ is the homothety of ratio $\alpha(f')$; if $g' \in G'$, then

Hence $\sigma(g)f \in E$. Therefore $\sigma$ is a linear $k$-representation (resp. continuous linear $K$-representation) of $G$ on $E$ leaving $F$ stable. Then $\sigma(g) = \mathrm{Id}_E$ for $g \in G'$ and hence for $g \in \overline{G'}$ in the topological case. Suppose that every finite-dimensional $k$-linear representation (resp. continuous $K$-linear representation) of $G/G'$ (resp. $G/\overline{G'}$) is semi-simple. Then there exists in $E$ a supplement of $F$ which is stable under $G$. In other words, there exists $f \in E$ such that $\alpha(f) = 1$, which is invariant under $G$. Then $f$ is a projector of $V$ onto $W$, and, for $g \in G$, $\rho(g) \circ f \circ \rho(g^{-1}) = f$, that is $f$ commutes with $\rho(G)$. Thus $\rho$ is semi-simple.