Equivalent characterizations of reductive Lie algebras

Assume condition 1 and write $\mathfrak{z}=Z(\mathfrak{g})$. By [L1], $\mathfrak{q}=\mathfrak{g}\mathrm{/}\mathfrak{z}$ is semisimple. It acts on $\mathfrak{g}$ through adjoints, and $\mathfrak{z}$ is a trivial submodule. By [L2] there is a $\mathfrak{q}$-submodule $\mathfrak{s}$ with $\mathfrak{g}=\mathfrak{z}\oplus \mathfrak{s}$. Invariance says $[\mathfrak{g},\mathfrak{s}]\subseteq \mathfrak{s}$, so $\mathfrak{s}$ is an ideal, and projection identifies it with $\mathfrak{q}$. Hence it is semisimple by [L3]. Since $\mathfrak{z}$ is central and $\mathfrak{s}$ is perfect by [L4], $[\mathfrak{g},\mathfrak{g}]=\mathfrak{s}$. This is condition 2.

Assume condition 2. The adjoint module is the direct sum of the trivial module $Z(\mathfrak{g})$ and the adjoint module of the semisimple ideal $[\mathfrak{g},\mathfrak{g}]$. The latter is completely reducible by [L2], so the whole adjoint module is completely reducible. This proves condition 3.

Assume condition 3. By [L5], the adjoint module is a finite direct sum of irreducible submodules. Every submodule $W$ therefore has an invariant complement: starting with $C=0$, if $W\oplus C\mathrm{\ne }\mathfrak{g}$, some irreducible summand in the displayed finite decomposition is not contained in $W+C$; its intersection with $W+C$ is then zero by irreducibility, so adjoining it strictly enlarges $C$ while preserving $W\cap C=0$. Finite dimensionality makes this process terminate with $\mathfrak{g}=W\oplus C$. Write $\mathfrak{r}=\mathrm{rad}(\mathfrak{g})$ and use this observation to choose an invariant complement $\mathfrak{s}$. Both are ideals, so $[\mathfrak{r},\mathfrak{s}]\subseteq \mathfrak{r}\cap \mathfrak{s}=0$. Apply the observation again to the characteristic ideal ${\mathfrak{r}}^{\mathrm{\prime }}=[\mathfrak{r},\mathfrak{r}]$, choosing an invariant complement $\mathfrak{c}$ in $\mathfrak{g}$. If $\mathfrak{u}=\mathfrak{r}\cap \mathfrak{c}$, then the decomposition $\mathfrak{g}={\mathfrak{r}}^{\mathrm{\prime }}\oplus \mathfrak{c}$ restricts to $\mathfrak{r}={\mathfrak{r}}^{\mathrm{\prime }}\oplus \mathfrak{u}$. The summands ${\mathfrak{r}}^{\mathrm{\prime }}$ and $\mathfrak{u}$ are ideals and commute.

From step 1.3, $\mathfrak{r}={\mathfrak{r}}^{\mathrm{\prime }}\oplus \mathfrak{u}$ and $[{\mathfrak{r}}^{\mathrm{\prime }},\mathfrak{u}]=0$. Hence ${\mathfrak{r}}^{\mathrm{\prime }}=[\mathfrak{r},\mathfrak{r}]=[{\mathfrak{r}}^{\mathrm{\prime }},{\mathfrak{r}}^{\mathrm{\prime }}]+[\mathfrak{u},\mathfrak{u}]$. The second summand lies in both $\mathfrak{u}$ and ${\mathfrak{r}}^{\mathrm{\prime }}$, so it is zero; therefore ${\mathfrak{r}}^{\mathrm{\prime }}=[{\mathfrak{r}}^{\mathrm{\prime }},{\mathfrak{r}}^{\mathrm{\prime }}]$. But ${\mathfrak{r}}^{\mathrm{\prime }}$ is solvable, and a nonzero perfect algebra cannot be solvable, so ${\mathfrak{r}}^{\mathrm{\prime }}=0$. Thus $\mathfrak{r}$ is abelian and, because it also commutes with $\mathfrak{s}$, is central in $\mathfrak{g}$. Conversely the center is an abelian ideal and lies in the radical. This proves condition 1 and closes the equivalence. For $\mathfrak{g}=0$, all three conditions hold.