If $G$ is compact, $H \leq G$ open, $V$ an irreducible $H$-rep, is $\text{Ind}_H^G$ semisimple? Let $G$ be a compact group, $H$ a normal open subgroup, and $K$ a $p$-adic field (so that not all $G$-reps with coefficients in $K$ are semisimple).  Let $V$ be a finite-dimensional topological $K$-vector space with continuous action of $H$, such that $V$ is irreducible as an $H$-representation.  Is the induced representation $\text{Ind}_H^G(V)$ semisimple?
In this specific context, I'm thinking of $G$ as the Galois group of a field, if that's helpful.
Thanks in advance!
 A: The answer is yes. I think it should be an exercise in  any book on representation theory. Since $H$ has finite index in $G$, and $V$ is finite dimensional, so is the representation $W=Ind _H^G (V)$ induced to $G$. In characteristic zero, this is equivalent to proving that the Zariski closure of $G$ in $GL(W)$ is reductive.
The connected component ${\mathcal G}^0$ of the Zariski closure ${\mathcal G}$ of $G$ in $GL(W)$ is unchanged if we replace $G$ by any finite index subgroup $K$. We take $K$ to be the intersection $\cap gHg^{-1}$ as $g$ varies over $G$; this is a finite intersection since $G/H$ is finite. 
Restricted to $K$, the representation is semi-simple since $W$ restricted to $K$ is the span of the restriction to $K$ of the semi-simple $gV$ as $g$ varies over $G/H$.  Therefore, the unipotent radical of ${\mathcal G}^0$ acts trivially on $gV$ for every $g$ and hence on $W$. Thus, ${\mathcal G}^0$ has no unipotent radical. 
[Edit] I should have said that I am assuming $Char (K)=0$.  
