Let $G \subset SL(V)$ be a connected algebraic group, acting irreducible on $V$, where $V$ is a complex vectorspace of dimension $n$
I want to show that the Lie algebra $\mathfrak{g}$ of $G$ is semisimple and acts irreducible on $V$.
Below is the full answer to this problem. Thanks goes to Callum and Bart Michels for their hints, that greatly helped to getting this solution.
Irreducibility:
We have $\mathfrak{g} \subset \mathfrak{sl}(V)$ a Lie subalgebra. Now note, that by assumption $G$ acts irreducible on $V$ via an representation say $\Phi$. Let $\phi$ be the associated representation for $\mathfrak{g}$. To show: $\Phi$ irreducible $\Leftrightarrow$ $\phi$ irreducible:
So if we have a subspace of $V$, invariant
under the action of every $X \in \mathfrak{g}$, it is invariant under $\phi(X)^n$ for all $n$, since $V$ is finite-dimensional. Hence it must be invariant under $exp(\phi(X))$. Using that, if $G$ is a connected Lie group, it is generated by exponentials of elements of $\mathfrak{g}$, we get one implication. Conversely, if a subspace is invariant under $G$, it must be invariant under each $\Phi(exp(hX))$ and hence also under the limit
$$lim_{h \to 0}\left(
\frac{\Phi(exp (hX)) − \Phi(I)}{h}\right)
= \phi(X).$$
Semi-simplicity:
To show, that $\mathfrak{g}$ is semisimple, we need to prove that the center of $\mathfrak{g}$ is trivial. That is $$\mathfrak{z}(\mathfrak{g})=\{x \in \mathfrak{g}\mid[x,s]=0 ,\forall s \in \mathfrak{g}\}.$$ Since we consider a Lie-subalgebra of $\mathfrak{sl}(V)$, we have $[x,s]=xs-sx$. Now, by the irreducibility of the representation together with Schur´s lemma, we have that all elements in the center are multiples of the identity matrix. But since $\mathfrak{g} \subset \mathfrak{sl}(V)$, and hence the traces need to be zero, we conclude that the center is trivial.
Edit:
Let $\mathfrak{h}$ be an abelian ideal and $V = \oplus_{\lambda \in\Lambda} V_{\lambda}$ the decomposition of $V$ into joint eigenspaces
$$V_{\lambda} = \{v \in V : \forall X \in \mathfrak{h}, (X – \lambda(X))v=0\}.$$
First we show, that forall $\lambda$ we have that $V_{\lambda}$ is $\mathfrak{g}$-invariant. Let $v \in V_{\lambda}$, $X \in \mathfrak{h}$ then for any $Y \in \mathfrak{g}$ we note by the comment below
$$(X – \lambda(X))Yv = [X,Y]v -\lambda(X)Yv + YXv = [X,Y]v$$
as $v \in V_{\lambda}$. Observe that $[X,Y] \in \mathfrak{h}$ and so $[X,Y]v \in V_{\lambda}$, so we have
$$(X – \lambda(X))^2Yv = (X – \lambda(X))[X,Y]v=0.$$
Recall that $X \in End(V)$ is said to be diagonalizable, if $V$ has a basis of eigenvectors for $X$. But this is equivalent to our decomposition $V = \oplus_{\lambda \in\Lambda} V_{\lambda}$.
So $X$ is diagonalizeable. Then, since the generalized eigenspaces coincide with the normal eigenspaces, this imples $Yv \in V_{\lambda}$.
Since by the first part (hopefully), $\mathfrak{g}$ acts irreducible on $V$, there can only be trivial $\mathfrak{g}$-invariant subspaces. So with the decomposition above we conclude $V = V_{\lambda}$ for some $\lambda \in \Lambda$. Hence $\forall v \in V, \forall X \in \mathfrak{h}: (X – \lambda(X))v=0$, i.e. $\forall v \in V, h \in \mathfrak{h}: Xv=\lambda(X) v$. So $\mathfrak{h}$ acts by scalars on $V$. But then again, the elements of $\mathfrak{h}$ being scalar multiples of the identity map, contradicts $\mathfrak{h} \subset \mathfrak{g} \subset \mathfrak{sl}(V)$ as the traces need to be $0$. So $\mathfrak{h}=0$.
Best Answer
You have only shown that the center is trivial. That is not sufficient to conclude that $\mathfrak g$ is semisimple.
What is sufficient is that $\mathfrak g$ has no abelian ideal. So suppose $\mathfrak h$ is an abelian ideal.
Hint 1:
Hint 2:
Conclude that $\mathfrak h$ acts by scalars and therefore $\mathfrak h = 0$.