For Lie algebra $\mathfrak{g}=\{e1,e2,e3,e4\}$ with commutations $[e1, e3]=\,e1, [e2, e3]=\,\alpha\,e2$, I have calculated normalizer for sub-algebra $\mathfrak{q}=\,\{e1+e2\}$ as
$\text{Nor}_{\mathfrak{g}}\left(\mathfrak{q}\right)=\,\{e1, e2, e4\}$.
I wonder what could be the quotient algebra $\frac{\text{Nor}_{\mathfrak{g}}\left(\mathfrak{q}\right)}{\mathfrak{q}}$ ?
I can explain how I use to calculate quotient algebra, for example, for sub-algebra $\mathfrak{p}=\{e1\}$, the normalizer is $\text{Nor}_{\mathfrak{g}}\left(\mathfrak{p}\right)=\,\{e1, e2, e3, e4\}$ and for quotient algebra $\frac{\text{Nor}_{\mathfrak{g}}\left(\mathfrak{p}\right)}{\mathfrak{p}}$, I simply remove $e1$ from $\text{Nor}_{\mathfrak{g}}\left(\mathfrak{p}\right)$ and obtain quotient algebra for normalizer as
$\frac{\text{Nor}_{\mathfrak{g}}\left(\mathfrak{p}\right)}{\mathfrak{p}}=\,\{e2, e3, e4\}$
But in above case I don't know how to proceed. This problem seems to be too simple for experts in Lie algebra, but as I belong to applied mathematics so I feel handicapped here, Please help !!
PS. I have calculated these normalizers using Lie Algebra package in Maple.
Asumming that $[e_1, e_2] = [e_1, e_4] = [e_2, e_4] = [e_3, e_4] = 0$ we have $$ \operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) = \begin{cases} \langle e_1, e_2, e_4 \rangle & \text{if $\alpha \neq 1$}, \\ \langle e_1, e_2, e_3, e_4 \rangle & \text{if $\alpha = 1$}, \end{cases} $$ where $\langle - \rangle$ denotes the linear span. (The computation can be found below.)
We first consider the case $\alpha \neq 1$. Then $$ \operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) = \langle e_1, e_2, e_4 \rangle = \langle e_1 + e_2, e_2, e_4 \rangle $$ with the three generators being a basis. Therefore $$ \operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) / \mathfrak{q} = \langle \overline{e_2}, \overline{e_4} \rangle, $$ where $\overline{x}$ denotes the residue class of $x \in \operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$, and the two generators being a basis. Because $e_4$ is central in $\mathfrak{g}$ we find that $\overline{e_4}$ is central in $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) / \mathfrak{q}$. So $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) / \mathfrak{q}$ is the two-dimensional abelian Lie algebra.
Consider now the case $\alpha = 1$, i.e. $[e_2, e_3] = e_2$. Then $$ \operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) = \langle e_1, e_2, e_3, e_4 \rangle = \langle e_1 + e_2, e_2, e_3, e_4 \rangle $$ with the four generators being a basis. Therefore $$ \operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) / \mathfrak{q} = \langle \overline{e_2}, \overline{e_3}, \overline{e_4} \rangle $$ with the three generators being a basis. As before we find that $\overline{e_4}$ is central, but in this case we also have the addition relation $$ [\overline{e_2}, \overline{e_3}] = \overline{e_2}. $$ So $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) / \mathfrak{q}$ is the direct sum of the two-dimensional non-abelian Lie algebra $\langle \overline{e_2}, \overline{e_3} \rangle$ and the one-dimensional (abelian) Lie algebra $\langle \overline{e_4} \rangle$.
Edit: I computed the normalizer $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$ as follows: We have $$ [e_1, e_1 + e_2] = [e_2, e_1 + e_2] = [e_4, e_1 + e_2] = 0 $$ and $$ [e_3, e_1 + e_2] = - [e_1 + e_2, e_3] = - [e_1, e_3] - [e_2, e_3] = - e_1 - \alpha e_2. $$ So we have $e_1, e_2, e_4 \in \operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$, independent of $\alpha$.
If $\alpha = 1$ then $$ [e_3, e_1 + e_2] = - e_1 - \alpha e_2 = - e_1 - e_2 = -(e_1 + e_2) \in \langle e_1 + e_2 \rangle = \mathfrak{q}, $$ so in this case we also have $e_3 \in \operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$ and thus $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) = \langle e_1, e_2, e_3, e_4 \rangle = \mathfrak{g}$.
If $\alpha \neq 1$ then $-e_1 - \alpha e_2$ is no scalar multiple of $e_1 + e_2$, and thus $e_3 \notin \operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$. In particular $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q}) \neq \mathfrak{q}$, which is why $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$ is at most three-dimensional. Because $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$ already contains the three linear independent elements $e_1$, $e_2$ and $e_4$ we also know that $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$ is at least three dimensional. Hence $\operatorname{Nor}_\mathfrak{g}(\mathfrak{q})$ must be three-dimensional and thus equals its three-dimensional subspace $\langle e_1, e_2, e_4 \rangle$.