= Solution
A <Möbius transformation> is $M_A(z)=(az+b)/(cz+d)$ with $\Delta=ad-bc\ne0$. On the <Riemann sphere>, a zero denominator represents infinity, and $M_A(\infty)=a/c$ when $c\ne0$; when $c=0$, infinity is fixed. The numerator and denominator cannot vanish together because $\Delta\ne0$.
The <matrix> $A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ acts on homogeneous coordinates $[z:1]$, so composition satisfies $M_A\circ M_B=M_{AB}$. The product is invertible since $\det(AB)=\det A\det B\ne0$, proving closure. Composition is associative because these are maps of the sphere. The identity is $z\mapsto z$, and
$$
M_A^{-1}(z)=\frac{dz-b}{-cz+a}
$$
is again a <Möbius transformation>. Thus the transformations form a <group>. Nonzero scalar multiples of a <matrix> give the same transformation, but the identity and inverse assertions concern the transformations themselves.
Write $T_q(z)=z+q$, $S_k(z)=kz$ with $k\ne0$, and $H(z)=1/z$. If $c=0$, then $a,d\ne0$ and $M_A=T_{b/d}\circ S_{a/d}$. If $c\ne0$, division gives
$$
\frac{az+b}{cz+d}=\frac ac-\frac{\Delta/c^2}{z+d/c}.
$$
Hence the <Möbius translation-scaling-inversion factorization> is
$$
\boxed{M_A=T_{a/c}\circ S_{-\Delta/c^2}\circ H\circ T_{d/c}\quad(c\ne0).}
$$
The scaling coefficient is nonzero. At $z=-d/c$ the reciprocal sends the translated zero to infinity; at infinity it sends infinity to zero. Thus the factorization holds on the entire <Riemann sphere>, including both exceptional points, and proves the stated generating property.
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