Let $G \subset GL(V)$ be a subgroup for a complex vectorspace $V$. Let $H=\overline{G}$ be its Zariski-closure. Then I want to check if the following statement (1) implies statement (2):
- $G$ acts irreducibly on $V$.
- $H$ acts irreducibly on $V$.
My first idea was to show this by contradiction, i.e. assuming that $H$ is not irreducible. But I don't know, if my idea is OK. So anyways, I would be very thankful if someone could look at this proof for some feedback, or give me a hint/help on this question.
$(1) \Rightarrow (2):$
Assume $H$ does not act irreducible on $V$, hence $H$ is reducible. Then there exists a proper, non-trivial $H$-invariant subspace $W \subset V$, i.e. we have $hW \subset W$ for all $h \in H$. But now $H=\overline{G}$ is the Zariski-closure of $G$. So in particular we also have $\forall g \in G: gW \subset W$. Hence $W$ is also a non-trivial $G$-invariant subspace, contradicting the irreducibility of $G$.