= Solution
The <Möbius group> consists of all <Möbius transformations> of the <Riemann sphere> $\mathbb C_\infty=\mathbb C\cup\{\infty\}$, with operation <function composition>. A nonzero-determinant <matrix> $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ represents the map $z\mapsto(az+b)/(cz+d)$. If $c\ne0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.
Define $\phi(M)$ by this action for $M\in SL_2(\mathbb C)$. Substitution of fractional-linear expressions shows $\phi(MN)=\phi(M)\circ\phi(N)$, so $\phi$ is a <group homomorphism>. Any invertible matrix can be rescaled into the <special linear group>: choose $\lambda\in\mathbb C^\times$ with $\lambda^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. \b[Thus]
$$
\boxed{\ker\phi=\{I,-I\},\qquad\mathcal M\cong SL_2(\mathbb C)/\{\pm I\}.}
$$
For the fixed-point claim, a nonidentity <Möbius transformation> has at least one <fixed point of a Möbius transformation>: when $c\ne0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a <Möbius transformation>. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so \b[a nonidentity finite-order <Möbius transformation> has exactly two fixed points].
Now let $K$ be a <finite abelian subgroup of the Möbius group>. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a <root of unity> of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A <Möbius transformation> preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force $\zeta^2=1$, impossible. Thus all elements of $K$ are scalings. A <finite subgroup of a field multiplicative group is cyclic>; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a <cyclic group>. Hence $K$ is cyclic.
Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto-z$. The action of $K$ on $\{0,\infty\}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\leq4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. \b[The classification is]
$$
\boxed{K\text{ is cyclic, or }K\cong C_2\times C_2.}
$$
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