Derived algebra of a lie algebra contained in an ideal

Question

Let $\mathfrak{g}$ be a Lie algebra over $\mathbb{R}$ or $\mathbb{C}$. Assume $\mathfrak{i}$ is an ideal with $\mathfrak{g/i}$ abelian. Then the derived algebra $[\mathfrak{g},\mathfrak{g}]\subseteq \mathfrak{i}$. I don't see why this is true. I am new to Lie algebras and am probably missing something obvious. I would appreciate if someone could show me why. Thanks!

Answer 1

If $x$ and $y$ are arbitrary elements of $\mathfrak{g}$, we can denote by $\bar{x}$ and $\bar{y}$ their images in $\mathfrak{g}/i$. Since the quotient is abelian, we have $[\bar{x},\bar{y}] = 0$, which means that $[x,y]\in i$, and hence that $[\mathfrak{g},\mathfrak{g}]\subseteq i$.