Proving those elliptic matrices in $\operatorname{SL}_2(ℤ)$ are not conjugate Set $\mathbf{\Gamma} = \operatorname{SL}_2(ℤ)$, let $\mathbf{H}$ denote the upper half plane. and let
$$\Gamma_0 (N) = \left\{ \left[\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right] ∈ \mathbf{\Gamma};\; c \equiv 0 \mod N\right\}$$
As known, $\mathbf{\Gamma}$ acts on $\mathbf{H}$ by Möbius transformations , i.e. $\left[\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right]z = \frac{az+b}{cz+d}$.
I know:


*

*that for any $z ∈ \mathbf{H}$, the isotropy $\mathbf{\Gamma}_z$ is cyclic of order $6$ or $4$,

*that any element in $\mathbf{\Gamma}$ of order $6$ or $4$ is conjugate in $\mathbf{\Gamma}$ to $\left[\begin{smallmatrix} 0 & -1 \\ 1 & 1 \end{smallmatrix}\right]^{\pm 1}$ or $\left[\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right]^{\pm 1}$ respectively,

*that any two different elements in $\mathbf{\Gamma}_z$ for $z ∈ \mathbf{H}$ are not conjugate in $\mathbf{\Gamma}$, and

*that any element $\gamma ∈ \mathbf{\Gamma} $of order $6$ or $4$ stabilizes exactly one element $z ∈ \mathbf{H}$, and also $\overline{z} ∈ -\mathbf{H}$.


Using this information, how can I prove that any $\gamma ∈ \Gamma_0(N)_z$ of order $6$ or $4$ cannot be conjugate to $\gamma' = \left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right]\, \gamma \,\left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right]$ by some element of $\Gamma_0(N)$? Please give me only hints if possible.
I know that $\gamma '$ stabilizes $-z$ in $ℂ$ and therefore it stabilizes $\overline{-z}$ as well. I tried several approaches which lead nowhere.
 A: First, $N$ is a red herring. What one actually proves is the more general statement that any $\gamma \in \Gamma$ of order 6 or 4 (or 3 for that matter) cannot be conjugate to $\gamma'$ by some element of $\Gamma$. 
For me it helps to take a geometric perspective on this problem and not get caught up in matrices. $\mathbf{H}$ is the hyperbolic plane, and its orientation preserving isometry group is $SL(2,\mathbb{R})$. The reason the third bullet point is true is because if $\gamma$ is a finite order isometry of $\mathbf{H}$, or for that matter any elliptic isometry, and if $z \in \mathbf{H}$ is its fixed point, then $\gamma$ rotates around $z$ by some angle $\alpha$, different elements of the isotropy group of $z$ have different rotation angles, and this rotation angle $\alpha$ is invariant under conjugacy by any orientation preserving isometry (element of $SL(2,\mathbb{R})$).
Next, the rotation angles of $\gamma$ about $z$ and about $\bar z$ are of opposite sign and therefore are unequal. This is where the proof uses order 4, or 6 (or 3, but order 2 does not work).
Next, the rotation angles of $\gamma$ about $z$ and of $\gamma'$ about $-z$ are equal.
Combining these, the rotation angles of $\gamma$ about $z$ and of $\gamma'$ about $-\bar z$ are unequal, so they are not conjugate in $\Gamma$.  
If you like you can translate this geometric proof into more analytic language, using complex derivatives $D_z\gamma$. Thinking of the rotation angle $\alpha$ as an element of $\mathbb{R}/2\pi\mathbb{Z}$, if $z$ is the fixed point of the elliptic isometry $\gamma$ then $D_z\gamma = e^{-i\alpha}$, and then you can check using the chain rule that the derivative of any elliptic isometry at its fixed point is invariant under conjugacy by an element of $SL(2,\mathbb{R})$.
What is proved here shows that even $\Gamma$ itself (and the numbers 4 and 6) are red herrings: for any non-identity elliptic isometry $\gamma \in SL(2,\mathbb{R})$ fixing a point $z \in \mathbf{H}$, as long as the rotation angle of $\gamma$ is not equal to $\pi$ the isometries $\gamma$ and $\gamma'$ are not conjugate to each other by any element of the orientation preserving isometry group $SL(2,\mathbb{R})$ of $\mathbf{H}$.
A: I think, I got it. Please someone check, if this is correct:
Let $\gamma = \left[\begin{smallmatrix}a & b \\c & d\end{smallmatrix}\right]∈ \Gamma_0(N)$ and $\alpha = \left[\begin{smallmatrix}p & q \\r & t\end{smallmatrix}\right]∈ \Gamma_0(N)$ such that $\alpha \gamma \alpha^{-1} = \gamma'$ and $\gamma ≠ \pm 1$.
If $\gamma$ stabilizes $z ∈ \mathbf{H}$, then $\gamma'$ stabilizes $\overline{-z} ∈ \mathbf{H}$. Therefore, since $\gamma'$ stabilizes $\alpha z$ as well and since there can only be one point in the upper half plane stabilized by some non-trivial action of the modular group, we have $\alpha z = \overline{-z}$, so $\Im (\alpha z) = \Im (\overline{-z}) = \Im (z)$.
Now, the following conclusion is wrong (see my comment):

This is only possible if $r = 0$ or $\Im (z) = 1$. (You can show that $\Im(\alpha z) ≤ 1/\Im (z)$ for $r ≠ 0$.)

$\Im (z) = 1$ cannot happen: Else, by some translation $\left[\begin{smallmatrix}1 & j \\0 & 1\end{smallmatrix}\right] ∈ \Gamma_0(N)$ the point $z$ would be equivalent to an elliptic point in the fundamental domain of $\mathbf{\Gamma}$ of imaginary part $1$, which can only be $\mathrm{i}$, which is not elliptic with respect to $\Gamma_0(N)$, for $\left[\begin{smallmatrix}0 & -1 \\1 & 0\end{smallmatrix}\right] \notin \Gamma_0(N)$.
So $c = 0$ and $\alpha = \pm \left[\begin{smallmatrix}1 & p \\0 & 1\end{smallmatrix}\right]$. Now, it’s easy to check that: $\gamma' = \left[\begin{smallmatrix}a & -b \\-c & d\end{smallmatrix}\right]$ and conjugating $\gamma$ by $\alpha$ doesn’t change the entry $c$ in the lower left corner. So $c = -c = 0$, implying that $\gamma$ has infinite order.
Therefore such a $\gamma$ cannot have order $4$ or $6$.
