I am trying to understand the Corollary 2.3.4 of "A first course in Modular forms", here the authors claim the following statement:
The elliptic points for $SL_2(\mathbb{Z})$ are $SL_2(\mathbb{Z})i$ and $SL_2(\mathbb{Z})u_3$. The modular curve $Y(1)$ has just these two elliptic points and the isotropy groups are finite cyclic.
Now, by Proposition 2.3.3 we know that for any $\tau\in{H}$, if $\gamma\in SL_2(Z)$ makes $\gamma(\tau)=\tau$, the $\gamma$ has order $1,2,3,4,6$, and the only nontrivial options are $3,4,6$, and since then $\gamma$ will conjugate to some matrix.
Now I am confused why this can conclude only two elliptic points and with finite cyclic isotropy groups?
Assume $\gamma$ fixes some $\tau \in \mathbb{H}$ and has order $4$. Then we can write $\gamma=M\sigma M^{-1}$ where $M \in SL_2(\mathbb{Z})$ and $\sigma=\begin{bmatrix}0 & -1\\1&0\end{bmatrix}$. Thus $\sigma$ fixes $M\tau$, and thus $M\tau=i$ so $SL_2(\mathbb{Z})\tau=SL_2(\mathbb{Z})i$.
Moreover, you can check that the isotropy group $G_i$ of $i=M\tau$ can be written as $MG_{\tau}M^{-1}$ where $G{\tau}$ is the isotropy group of $\tau$, so that $G_{\tau}$ and $G_i$ are isomorphic. But it’s a straightforward calculation to see that $G_i=\langle \sigma \rangle$ is finite cyclic with order $4$.
The same principle applies when $\gamma$ has order $3$ or $6$.