Example 2

Let $\mathfrak{g}=\mathfrak{g}\mathfrak{l}(n, k)$, and let $\mathfrak{h}$ be the set of diagonal matrices belonging to $\mathfrak{g}$. We show that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$. First, $\mathfrak{h}$ is commutative, hence nilpotent. Let $(E_{ij})$ be the canonical basis of $\mathfrak{g}\mathfrak{l}(n, k)$, and let $x=\sum\mu_{ij}E_{ij}$ be an element of the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$. If $i\not=j$, formulas (5) of Chap. I, §1, no. 2 show that the coefficient of $E_{ij}$ in $[E_{ii}, x]$ is $\mu_{ij}$. Since $E_{ii}\in \mathfrak{h}$, $[E_{ii}, x]\in \mathfrak{h}$, and the coefficient in question is zero. Thus $\mu_{ij}= 0$ for $i\not=j$, so $x\in \mathfrak{h}$, which shows that $\mathfrak{h}$ is indeed a Cartan subalgebra of $\mathfrak{g}$.