Weyl's theorem states
Let $\mathfrak{g}$ be a finite dimensional semisimple Lie-algebra over an algebraically closed field with characteristic 0. Then all finite dimensional representations are semisimple.
One can also look at the other direction
Let $\mathfrak{g}$ be a finite dimensional Lie-algebra over an algebraically closed field with characteristic 0. If all finite dimensional representations are semisimple then $\mathfrak{g}$ is semisimple.
I want to prove that direction. For that I considered the adjoint representation giving us a direct sum of irreducible subrepresentations. These correspond to the ideals of $\mathfrak{g}$. Now, I would like to show that these ideals are simple (or semisimple). But now, I am stuck showing this. I have tried to look at abelian ideals of these ideals and tried to show that they are 0, but it has not worked. Any hints would be appreciated.
Best Answer
Update: I have expanded this answer to try and provide more details in response to the comment below.
Let $\mathfrak g$ be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero, and let $\mathrm{Rep}_{fd}(\mathfrak g)$ denote the category of finite-dimensional representations of $\mathfrak g$.
Claim: If $\mathrm{Rep}_{fd}(\mathfrak g)$ is semisimple (that is every object of $\mathrm{Rep}_{fd}(\mathfrak g)$ is semisimple) then $\mathfrak g$ is a semisimple Lie algebra.
Definition: Let $\mathfrak g$ be a Lie algebra over a field of characteristic zero. We say that $\mathfrak g$ is reductive if its radical (i.e. its largest solvable ideal) is equal to its centre. If the radical $\mathrm{rad}(\mathfrak g)$ is equal to the centre $\mathfrak z(\mathfrak g)$, then the Levi decomposition of $\mathfrak g$ will split as a direct sum of a semisimple Lie algebra and the centre of $\mathfrak g$.
Claim 1: If the adjoint representation $(\mathfrak g, \mathrm{ad})$ is completely reducible, then $\mathfrak g$ is reductive.
If $(\mathfrak g, \mathrm{ad})$ is completely reducible, we may write $\mathfrak g$ as a direct sum of irreducible subrepresentations,. Suppose this decomposition takes the form $$ \mathfrak g = \bigoplus_{i=1}^k \mathfrak p_i \oplus \mathfrak g^{\mathfrak g}, $$ where for a $\mathfrak g$-representation $(V,\rho)$, we write $V^{\mathfrak g} = \{v \in V: \rho(g)(v)=0, \forall g\in \mathfrak g\}$, hence when $V$ is semisimple, $V^{\mathfrak g}$ is the isotypical summand of $V$ corresponding to the trivial representation. Now each $\mathfrak p_i$ is an irreducible $\mathfrak g$-representation, and hence a minimal ideal in $\mathfrak g$, thus each $\mathfrak p_i$ is a Lie algebra which has no proper ideals.
Now because the ideals $\mathfrak p_i$ form a direct sum, $[\mathfrak p_i,\mathfrak p_j]\subseteq \mathfrak p_i \cap \mathfrak p_j = \{0\}$. Thus as the $\mathfrak p_i$ are, by definition, not contained in $\mathfrak g^{\mathfrak g}$ we must have $[\mathfrak p_i,\mathfrak p_i]\neq 0$ and hence, by minimality of $\mathfrak p_i$, it follows that $[\mathfrak p_i,\mathfrak p_i]= \mathfrak p_i$.
Thus we see that $\mathfrak g = \mathfrak s \oplus \mathfrak g^{\mathfrak g}$, where $\mathfrak s$ is a direct sum of non-abelian simple Lie algebras (and hence is semisimple) and $\mathfrak g^{\mathfrak g}$ is an abelian ideal on which $\mathfrak g$ acts by zero, i.e. it is the centre of $\mathfrak g$, and the claim is established.
Claim 2: If $\mathfrak g$ is any Lie algebra which has $\mathfrak{gl}_1$ as a quotient, then $\mathrm{Rep}_{fd}(\mathfrak g)$ is not semisimple.
Proof: The point here is that the quotient map $q\colon \mathfrak g \to \mathfrak{gl}_1$ can be used to pull back any representation of $\mathfrak{gl}_1$ to become a representation of $\mathfrak{g}$. In particular, if $\mathfrak{gl}_1$ has any representations which are not semisimple then so will $\mathfrak g$. But the map $t\mapsto t\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)$ gives a representation of $\mathfrak{gl}_1$ on $\mathsf k^2$ which has a unique proper submodule (the span of $(1,0)^\intercal$) and hence cannot be semisimple.
Finally, to establish the original Claim, notice that if $\mathrm{Rep}(\mathfrak g)$ is semisimple then certainly $(\mathfrak g, \mathrm{ad})$ is a semisimple representation and so by Claim 1, $\mathfrak g$ is reductive. But then if $\mathfrak z(\mathfrak g)\neq 0$, then since $\mathfrak g = \mathfrak s \oplus \mathfrak z(\mathfrak g)$ where $\mathfrak s$ is a semisimple ideal, it follows that $\mathfrak g/\mathfrak s \cong \mathfrak z(\mathfrak g)$, and as $\mathfrak z(\mathfrak g)$ is abelian, if $\lambda \in \mathfrak z(\mathfrak g)^*$ is a nonzero linear functional on $\mathfrak z(\mathfrak g)$, then $\lambda \colon \mathfrak z(\mathfrak g) \to \mathfrak{gl}_1$ is a surjective homomorphism of Lie algebras. Since $q\colon \mathfrak g \to \mathfrak g/\mathfrak s \cong \mathfrak z(\mathfrak g)$ is also surjective, it follows that $\lambda \circ q$ is a surjective homomorphism from $\mathfrak g$ to $\mathfrak{gl}_1$.
Therefore, it follows from Claim $2$ that if $\mathrm{Rep}_{fd}(\mathfrak g)$ is semisimple we must have $\mathfrak z(\mathfrak g)=\{0\}$ and $\mathfrak g$ is semisimple as required.
[Note that an abelian Lie algebra $\mathfrak a$ is just a $\mathsf k$-vector space equipped with the zero Lie bracket. Thus $\mathfrak a$ can be written as a direct sum of one-dimensional subspaces which are also subalgebras. The only one-dimensional Lie algebra is the trivial one because the alternating property of the bracket forces it to vanish, and this Lie algebra is usually denoted $\mathfrak{gl}_n$.]