A reductive lie algebra $L$ satisfies that $[L,L]$ is semisimple 
Let $L$ be a reductive Lie algebra (i.e., $\mathrm{Z}(L) = \operatorname{Rad}(L)$). Then $\mathrm{ad} \colon L \to \mathfrak{gl}(V)$ is completely reducible, $L = [L, L] \oplus \operatorname{Z}(L)$ and $[L, L]$ is semisimple.

I was able to prove that $\mathrm{ad}$ is a completely reducible representation, and if I assume the second part then $[L, L] \cong L/\mathrm{Z}(L)$ which is semisimple.
My attempt of the second part:
We have that since $L/\mathrm{Z}(L)$ is semisimple, so
$$
  [L, L]/\mathrm{Z}(L)
  \cong
  [L/\mathrm{Z}(L), L/\mathrm{Z}(L)]
  =
  L/\mathrm{Z}(L) \,.
$$
And therefore for every $z \in L$ there exist $y \in [L, L]$ and $c \in \mathrm{Z}(L)$ s.t. $z = y + c$. What I’m stuck with is proving that $[L, L] \cap \mathrm{Z}(L) = \{ 0 \}$.
Any help would be appreciated.
 A: By complete reducibility, the subrepresentation $Z(L)$ has a complement, say
$$L \simeq A \oplus Z(L)$$
(as $L$-modules). In particular, $A$ is a subalgebra. $A \simeq L/Z(L)$ is semisimple by the assumption $Z(L)=rad(L)$, hence it is perfect (i.e. $[A,A]=A$).
But then $[L,L]=[A+Z(L), A+Z(L)]=[A,A]=A$, i.e. $A$ was the derived subalgebra all along.
A: Let $L$ be a finite-dimensional Lie algebra whose adjoint representation is completely reducible.
This means that we have a decomposition
$$
 L = L_1 ⊕ \dotsb ⊕ L_n
 \tag{$\ast$}
$$
into irreducible subrepresentations $L_j$.
Each $L_j$ is thus an ideal of $L$, and $(\ast)$ is a decomposition of $L$ into ideals.
This implies that the Lie bracket of $L$ is calculated summand-wise with respect to the decomposition $(\ast)$;
i.e., we have
$$
 [x_1 + \dotsb + x_n,\; y_1 + \dotsb + y_n]
 =
 [x_1, y_1] + \dotsb + [x_n, y_n]
$$
for all $x_j, y_j ∈ L_j$.
This has the following consequences.

*

*We have $[L, L] = [L_1, L_1] ⊕ \dotsb ⊕ [L_n, L_n]$.


*We have $\mathrm{Z}(L) = \mathrm{Z}(L_1) ⊕ \dotsb ⊕ \mathrm{Z}(L_n)$.


*The $L$-subrepresentations of $L_j$ are precisely the $L_j$-subrepresentations of $L_j$.
But $L_j$ is irreducible as an $L$-representations, so it is irreducible as an $L_j$-representation.
This means that $L_j$ is either simple as a Lie algebra, or one-dimensional and abelian.
We may rearrange the summands $L_1, \dotsc, L_n$ in such a way that $L_1, \dotsc, L_m$ are simple and $L_{m + 1}, \dotsc, L_n$ are one-dimensional and abelian.
We find that
\begin{align*}
 [L, L]
 &=
 [L_1, L_1] ⊕ \dotsb ⊕ [L_m, L_m] ⊕ [L_{m+1}, L_{m+1}] ⊕ \dotsb ⊕ [L_n, L_n]
 \\
 &=
 L_1 ⊕ \dotsb ⊕ L_m ⊕ 0 ⊕ \dotsb ⊕ 0
 \\
 &=
 L_1 ⊕ \dotsb ⊕ L_m
\end{align*}
is a sum of simple Lie algebras, and therefore semi-simple.
We also find that
\begin{align*}
 \mathrm{Z}(L)
 &=
 \mathrm{Z}(L_1) ⊕ \dotsb ⊕ \mathrm{Z}(L_m) ⊕ \mathrm{Z}(L_{m+1}) ⊕ \dotsb ⊕ \mathrm{Z}(L_n)
 \\
 &=
 0 ⊕ \dotsb ⊕ 0 ⊕ L_{m+1} ⊕ \dotsb ⊕ L_n
 \\
 &=
 L_{m+1} ⊕ \dotsb ⊕ L_n \,,
\end{align*}
and therefore
$$
 L
 =
 L_1 ⊕ \dotsb ⊕ L_m ⊕ L_{m+1} ⊕ \dotsb ⊕ L_n
 =
 [L, L] ⊕ \mathrm{Z}(L) \,.
$$
