 Solution

Because $F_1$ is projective twistor space, the printed column represents the line $p=[e_1]$, not a distinguished nonzero vector. Its stabilizer subgroup consists of matrices whose first column is a nonzero multiple of $e_1$:The block form is necessary to preserve the line, and conversely every such determinant-one matrix preserves it. By transitivity, the orbit map $g\mapsto g[e_1]$ is onto $F_1$. Its two values agree precisely when $g^{-1}{g}^{\prime }\in H_p$, equivalently when $g{H}_p=g^{\prime }{H}_p$. This gives the homogeneous space identificationThe dimensions also agree: ${\dim }_{\mathbb{C}}SL(4)=15$, while the free $B$ and $b$ parameters give ${\dim }_{\mathbb{C}}{H}_p=9+3=12$; the quotient has dimension three.

This coset space is not a quotient group, because $H_p$ is not a normal subgroup. For example $h=I+E_{12}\in H_p$. Let $g$ have upper-left block $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and identity on the other two coordinates. Then $g\in SL(4,\mathbb{C})$ but $gh{g}^{-1}=I-E_{21}$ sends $e_1$ to $e_1-e_2$, so it is not in $H_p$. Thus multiplying coset representatives does not define a well-defined group operation.

If instead one required the literal vector $e_1$ to be fixed, one would have $a=1$ and $B\in SL(3,\mathbb{C})$; its quotient would be ${\mathbb{C}}^4\setminus \{0\}$, not $F_1$. The projective reading is required by the printed target quotient and the flag manifold definition.