= Solution
A <Möbius transformation> is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equation
$$
\alpha|z|^2+\beta z+\overline\beta\,\overline z+\gamma=0,
$$
where $\alpha,\gamma\in\mathbb R$; the case $\alpha=0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w|^2$ to obtain
$$
\gamma|w|^2+\overline\beta w+\beta\overline w+\alpha=0,
$$
which is again such an equation. The generators, and hence every Möbius transformation, therefore preserve <Generalized circle under a Möbius transformation>.
Now let
$$
f(z)=\frac{az+b}{cz+d}.
$$
If $c\ne0$, its finite pole $-d/c$ lies on some Euclidean circle, and that circle maps to a line because its image contains infinity. Thus $f$ takes every circle to a circle exactly when $c=0$, so the <subgroup> in question is the <Affine subgroup of the Möbius group>, consisting of $z\mapsto az+b$ with $a\ne0$.
It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne0$, then $T_b$ is affine but
$$
IT_bI(z)=\frac{z}{1+bz}
$$
has a finite pole and is not affine. Hence conjugation by $I$ does not preserve the <subgroup>.
Back to article page
