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Take the semidirect product of a Lie algebra and a modulewhere $k^2$ is the defining representation of the sl2 Lie algebra. To see directly that this is a Lie algebra, realize $(X,u)$ as the block matrix $\left(\begin{array}{cc}X& u\\ 0& 0\end{array}\right)$. These matrices are closed under the operator commutator, whose Jacobi identity holds in every associative algebra.

Using the standard sl2 triple $e,h,f$, its Lie brackets show $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$: indeed $h=[e,f]$, $e=[h,e]/2$, and $f=-[h,f]/2$. Also $h=\mathrm{diag}(1,-1)$ is invertible on $k^2$. For any $u\in k^2$, put $v=h^{-1}u$; thenIt follows that the derived algebra contains both ${\mathfrak{sl}}_2\oplus 0$ and $0\oplus k^2$, so $[L,L]=L$: this is a perfect Lie algebra.

However, $0\oplus k^2$ is a nonzero ideal of a Lie algebra with zero Lie bracket. It is therefore a solvable Lie algebra and is contained in the solvable radical of $L$. A semisimple Lie algebra has zero solvable radical, soThis illustrates the perfect semidirect product criterion: the Lie brackets internal to the first summand generate it, while its action generates the second summand.

Bilinearity and alternatingness ofare immediate. For three elements, the $\mathfrak{g}$ component of the Jacobi sum vanishes by the Jacobi identity in $\mathfrak{g}$. In the $V$ component, the coefficient of a vector such as $u$ isbecause the action is a Lie algebra representation; the other terms cancel cyclically in the same way. Hence the bracket satisfies Jacobi and defines the semidirect product of a Lie algebra and a module $\mathfrak{g}⋉V$.

For a Lie algebra representation of $\mathfrak{g}$ on $V$, let $\mathfrak{g}V$ denote the linear span of $xv$ for $x\in \mathfrak{g}$ and $v\in V$. The derived algebra of the semidirect product of a Lie algebra and a module is the displayed direct sum of vector spaces. Every Lie bracket has first component in $[\mathfrak{g},\mathfrak{g}]$ and second component in $\mathfrak{g}V$. Conversely, brackets of $(x,0)$ with $(y,0)$ span the former component, and brackets of $(x,0)$ with $(0,v)$ span the latter. Therefore this semidirect product of a Lie algebra and a module is a perfect Lie algebra exactly when $\mathfrak{g}$ is a perfect Lie algebra and $\mathfrak{g}V=V$. When $V\ne 0$, its nonzero abelian Lie algebra ideal prevents it from being a semisimple Lie algebra. The sl2 Lie algebra acting on its two-dimensional defining representation supplies a five-dimensional example.