# Is a reductive subalgebra of a semisimplie Lie algebra semisimple?

I have the following doubt. Say that $\mathfrak g$ is a semisimple Lie algebra, $\mathfrak k$ a reductive subalgebra, and suppose further that any Cartan subalgebra in $\mathfrak k$ is a Cartan subalgebra in $\mathfrak g$. The reasoning that leads me to say that $\frak k$ is semisimple as well is the following: since $\frak k$ is reductive, we can write it as a direct sum \begin{equation} \frak k = \frak s + \frak l \end{equation} with $\frak s$ Abelian and $\frak l$ semisimple. As far as I understand, $[\frak s,\frak l]=0$.

Now, $\frak s$ can be made to sit in some Cartan subalgebra of $\frak k$, right? And I would say that we can express $\frak l$ via a system of subroots of the root system of $\frak g$ - the bottom line is that I don't see how $[\frak s,\frak l]=0$ could hold. But also, this reasoning is far from crystal clear.

So I'd be happy to receive feedback, counterexamples, references.

• Well, an obvious counter example is taking the Cartan subalgebra itself. This is clearly reductive (being abelian) and its Cartan is a Cartan of the original algebra. – Tobias Kildetoft Jun 20 '16 at 19:03
• @TobiasKildetoft Thanks. Gosh, should have seen that – nelv Jun 20 '16 at 20:18