# Let $M_{\alpha}, \alpha \in \mathbb{C}$ be the subgroup of the Möbius transformation mapping $\alpha$ to itself. Calculate $M_i$

Let $M_{\alpha}, \alpha \in \mathbb{C}$, be the subgroup of $M$ mapping $\alpha$ to itself, that is, the stabilizer of $\alpha$. Given that

$$M_0 = \left \{w = \frac{z}{cz + d}, d \neq 0 \right \}$$

compute the subgroup $M_i, i = \sqrt{-1}$.

$M$ is the Möbius transformation on the extended complex plane.

What I have said is that in order to compute this subgroup, I want some transformation, $L$, which will send map $z \mapsto z + i$. This subgroup can be worked out using the composition

$$M_i = L^{-1} \circ M_0 \circ L$$

So I know that $L = \frac{(z + i)}{c(z + i) + d}$, but I'm not sure how I get $L^{-1}$. I'm guessing from here, it'll be fairly simple to work out the composition.

EDIT: Also, does order matter? I.e is

$$M_i = L^{-1} \circ M_0 \circ L = L \circ M_0 \circ L^{-1}$$

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If $w = z+i$, then $z=w-i$. This should help you to compute $L^{-1}$. –  mrf Jan 8 '13 at 11:44
@mrf So would I say my $L^{-1}$ maps $z \mapsto z - i$? –  Kaish Jan 8 '13 at 11:46
Yes.${}{}{}{}{}$ –  mrf Jan 8 '13 at 11:49

Yes: you want to move $i$ to $0$, then apply a map that stabilizes $0$, then move $0$ back to $i$ -- in this order. The simplest maps to use are translations by $\pm i$ (suggested by mrf). Thus, $$M_i = \left \{w = i+\frac{z-i}{c(z-i) + d}, d \neq 0 \right \}$$ which you can simplify if desired.