= Solution
{c}
Use the general definition of a <Cartan subalgebra>: a <nilpotent Lie algebra> that is a <Lie subalgebra> and equals its <normalizer of a Lie subalgebra>. In a <semisimple Lie algebra>, the stated structural fact says every <Cartan subalgebra> $\mathfrak h$ is a <toral subalgebra> and an <abelian Lie algebra>. Part (i) therefore gives
$$
C_{\mathfrak g}(\mathfrak h)=N_{\mathfrak g}(\mathfrak h)=\mathfrak h.
$$
Conversely, suppose $\mathfrak h$ is a <toral subalgebra> and an <abelian Lie algebra> with $C_{\mathfrak g}(\mathfrak h)=\mathfrak h$. An <abelian Lie algebra> is a <nilpotent Lie algebra>, because its <lower central series of a Lie algebra> vanishes at the second term. Part (i) gives $N_{\mathfrak g}(\mathfrak h)=C_{\mathfrak g}(\mathfrak h)=\mathfrak h$. Thus $\mathfrak h$ satisfies the defining conditions of a <Cartan subalgebra>, proving
$$
\boxed{\mathfrak h\text{ is Cartan}\iff\begin{gathered}\mathfrak h\text{ is toral and abelian},\\C_{\mathfrak g}(\mathfrak h)=\mathfrak h.\end{gathered}}
$$
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