Every semisimple Lie algebra of dimension at most 5 is simple.

Question

How does one argue that every semisimple Lie algebra of dimension $\leq 5$ is simple. Since any simple algebra has dimension at least $3$, we have to show that any semisimple algebra of dimension $3,4,5$ is simple.

Answer 1

Let $\mathfrak{g}$ be semisimple, then $\mathfrak{g}=\oplus_{i=1}^n\mathfrak{g}_i,$ where each $\mathfrak{g}_i$ is simple. The dimension of every $\mathfrak{g}_i$ is $\geq3$. If the dimension of $\mathfrak{g}$ is $\leq5$, how many $\mathfrak{g}_i$'s can there be?

Answer 2

In characteristic $p>0$ a semisimple Lie algebra $L$ need not be the direct sum of simple Lie algebras, and there are simple Lie algebras not only of dimension $3$, but also of dimension $5$, i.e., the $p$-dimensional Witt algebra $W(1,1)$ for $p=5$. Probably it is still true that every semisimple Lie algebra of dimension $n\le 5$ is already simple, but I did not find this yet in the literature.

In characteristic zero, every semisimple Lie algebra is the direct sum of simple Lie algebras, which have at least dimension $3$, because every Lie algebra of dimension $n\le 2$ is solvable. Hence there is no semisimple Lie algebra of dimension $4$ and $5$, and it remains to show that every semisimple Lie algebra of dimension $3$ is simple. So let $L$ be semisimple, and $\dim(L)=3$. Let $I$ be a proper ideal of $L$. Then $I$ and $L/I$ are solvable, because both are of dimension at most $2$. Hence $L$ is solvable, a contradiction. Hence $L$ has no proper ideal, and is simple.