= Solution
A <group homomorphism> preserves multiplication: $\Phi(gh)=\Phi(g)\Phi(h)$. Its <group kernel> $K=\{g:\Phi(g)=e_H\}$ is a subgroup, and $\Phi(gkg^{-1})=e_H$ for $k\in K$. Thus $gKg^{-1}=K$, proving it is a <normal subgroup>. The map $gK\mapsto\Phi(g)$ is well-defined because two representatives of a coset differ by an element of $K$. It is a homomorphism, has trivial kernel, and is surjective when $\Phi$ is. This proves the <first isomorphism theorem>, $G/K\cong H$. This proof does not use finiteness and therefore also applies below.
The identity <matrix> belongs to $SL_2(\mathbb C)$. The <determinant> is multiplicative, giving closure, and <matrix> multiplication is associative. The inverse of a determinant-one <matrix> $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, also of determinant one. Thus this <special linear group> is a group.
The fractional linear maps must be interpreted on the <Riemann sphere>, including infinity: a finite pole is sent to infinity and infinity is sent to $a/c$ when $c\ne0$. As literal everywhere-defined maps of the complex plane they would not form the asserted group. Substitution into the two rational expressions gives $f_A\circ f_B=f_{AB}$. Matrix identities and inverses therefore give the identity and inverse transformations, proving that the <Möbius group> is a group and that $\Phi:A\mapsto f_A$ is a surjective homomorphism.
For $f_A$ to be the identity, $(az+b)/(cz+d)=z$ on all finite points where defined. Hence $cz^2+(d-a)z-b$ is the zero polynomial, so $b=c=0$, $d=a$, and $a^2=1$ from the determinant. Therefore
$$
\boxed{\ker\Phi=\{I,-I\},\qquad\mathcal M\cong SL_2(\mathbb C)/\{\pm I\}=PSL_2(\mathbb C).}
$$
The kernel is central and hence normal. In particular $\Phi$ is \b[not an <isomorphism>], since $A$ and $-A$ represent the same transformation. The quotient is the <projective special linear group> in this complex two-dimensional case.
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