For each $n \geq 2$, is $\Gamma(2^{n})$ the unique normal subgroup of $\Gamma(2)$ with quotient isomorphic to $\Gamma(2) / \Gamma(2^{n})$ (here we are talking about principal congruence subgroups of $\mathrm{SL}_{2}(\mathbb{Z})$)? I can show this using elementary group theory for up to $n = 3$ and suspect there's a group theoretic argument that would prove it for all $n$.
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$\begingroup$ Isn't $\Gamma(2)$ (in $PSL_2(\mathbb{Z})$) a free group? $\endgroup$– HJRWCommented Mar 14, 2015 at 22:51
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$\begingroup$ Yes, and $\Gamma(2) \lhd \mathrm{SL}_{2}(\mathbb{Z})$ is the direct product of a free group and $\{\pm 1\}$. $\endgroup$– Jeff YeltonCommented Mar 15, 2015 at 1:18
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$\begingroup$ So then I don't see how what you want can possibly be true. Any 2-generated finite group $Q$ will have many epimorphisms $\Gamma(2)\to Q$ (one for each generating pair), and the kernels will be different unless they differ by an automorphism of $Q$. (Unless the groups $\Gamma(2)/\Gamma(2^n)$ have the very special property that every generating pair differs by an automorphism.) $\endgroup$– HJRWCommented Mar 15, 2015 at 10:19
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$\begingroup$ It can be proven true for $\Gamma(8)$, using a nice presentation of $\Gamma(2) / \Gamma(8)$ from which it's apparent that any pair of generators satisfies these relations. I suspect that with some effort, I would find that it's true for $\Gamma(16)$ etc. in the same way, but I hoped there was some variant of that argument that would work for all $n$ (or that maybe it would be known by some other argument). $\endgroup$– Jeff YeltonCommented Mar 15, 2015 at 14:10
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1$\begingroup$ @HJRW I wonder what the structure of the quotient is (it is of order $2^{3 (n-1)},$ so is nilpotent, but what else. Maybe I will ask this as a separate question. $\endgroup$– Igor RivinCommented Mar 15, 2015 at 18:06
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