Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras. Therefore, the generators are not a subset of the original's group generators.

That is not quite right. A special subalgebra is one such that their step operators do not form a subset of the algebra step operators.That is what is meant by a root system not being a subset of another root system. It does not imply that the subalgebra generators are not a subset of the algebra generators.

Example: Consider the three dimensional representation of $\mathfrak g=\mathfrak{su}(3)$, whose generators $T_a=\lambda_a/2$, where $\lambda_a$ are the Gell-Mann matrices. The step operators are

\begin{align*}

E_{\pm\alpha_1}=T_1\pm iT_2,\\

E_{\pm\alpha_2}=T_6\pm iT_7,\\

E_{\pm(\alpha_1+\alpha_2)}=T_4\pm iT_5,

\end{align*}

**Regular embedding:** The generators $T_1$, $T_2$ and $T_3$ form a subalgebra $\mathfrak{su}(2)$. This subalgebra step operators are $t_\pm=T_1\pm iT_2=E_{\pm\alpha_1}$. The subalgebra step operators form a subset of the algebra step operators.

**Special embedding:** Another $\mathfrak{su}(2)$ subalgebra is given by $T_2$, $T_5$ and $T_7$. In this case the step operators are $t_\pm=T_5\pm iT_7$ which cannot be written as a subset of the step operators of $\mathfrak{su}(3)$.

Notice the branching rules for the first embedding is $3=2\oplus 1$ while for the second is $3=3$.