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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.

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 $Char (K)=0$.

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 $gHg^{-1}$ 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.

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.

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 $G$.

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 $gHg^{-1}$ 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.

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 $gHg^{-1}$ 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.