I am trying to solve this problem from Koblitz "Introduction to Elliptic Curves and modular forms" page 144, where $\Gamma=\mathrm{SL}_2(\mathbb{Z}),$ and $T=\left(% \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}% \right).$ I showed (a) as follow :
Let $\beta\in(\alpha\Gamma'\alpha^{-1})_{\infty}$ then we can write $\beta=\alpha\gamma\alpha^{-1}$ with $\gamma\in\Gamma'$ so we have : \begin{eqnarray*} \beta.\infty=(\alpha\gamma\alpha^{-1}).\infty=\infty &\Leftrightarrow& (\alpha\gamma).\alpha^{-1}\infty=\infty \\ &\Leftrightarrow& (\alpha\gamma).s=\infty \\ &\Leftrightarrow& \alpha^{-1}\alpha\gamma.s=\alpha^{-1}.\infty\\ &\Leftrightarrow& \gamma.s=s \\ &\Leftrightarrow& \gamma\in\Gamma'_s \\ &\Leftrightarrow& \beta\in \alpha\Gamma'_s\alpha^{-1} \end{eqnarray*} finally $(\alpha\Gamma'\alpha^{-1})_{\infty}=\alpha\Gamma'_s\alpha^{-1}$
For (b) : we have $(\alpha\Gamma'\alpha^{-1})_{\infty}\subset \Gamma_\infty=\{\pm I\}.<T>$ and $[\Gamma_\infty :(\alpha\Gamma'\alpha^{-1})_{\infty}]<\infty$ so if we take $h=[\Gamma_\infty :(\alpha\Gamma'\alpha^{-1})_{\infty}]$ then we have : $(\alpha\Gamma'\alpha^{-1})_{\infty}=\{\pm I\}.<T^h>=\alpha\Gamma'_s\alpha^{-1}(*)$ I'm stuck here.
My question is : How I can conclude (IIa) and (IIb) from $(*)$ ?
Thanks!
You've made a slight error: $(*)$ implicitly assumes that $-I\in\Gamma_s$.
If $-I\in\Gamma'$, then $\{\pm I\}\subset(\alpha\Gamma'\alpha^{-1})_\infty\subset\Gamma_\infty=\{\pm I\}\cdot\langle T\rangle.$ As you've shown, it follows that $$\alpha\Gamma_s'\alpha^{-1} = \{\pm I\}\cdot\langle T^h\rangle = \{\pm T^{nh}\}_{n\in\mathbb Z}$$and hence that $$\Gamma_s = \pm\alpha^{-1}\{T^{nh}\}_{n\in\mathbb Z}\alpha.$$
Here $h$ is the minimal positive integer such that $T^h\in(\alpha\Gamma_s'\alpha^{-1})_\infty$, which exists as $[\Gamma_\infty:(\alpha\Gamma_s'\alpha^{-1})_\infty]<\infty$.
However, if $-I\notin \Gamma'$, then either $T^h$ or $(-T)^h$ is an element of $\alpha\Gamma_s'\alpha^{-1}$, but not both, so $(*)$ does not hold in this case. Either $$(\alpha\Gamma'\alpha^{-1})_\infty= \langle T^h\rangle = \{ T^{nh}\}_{n\in\mathbb Z}$$or$$(\alpha\Gamma'\alpha^{-1})_\infty=\langle (-T)^h\rangle = \{ (-T)^{nh}\}_{n\in\mathbb Z}.$$