We know that principal congruence subgroups are characteristic in $SL(n,\mathbb Z)$. Suppose $\Gamma$ is a finite index subgroup of $SL(n,\mathbb Z)$ and $\Gamma_m$ is a principal congruence subgroup of level m contained in $\Gamma$. Will it be characteristic in $\Gamma$?
This is false. Consider the case $n=2$, and let $p$ be a prime. Let $A=\left[\begin{array}{cc}0 & 1 \\\ p & 0\end{array}\right] $ be an AtkinLehner involution (considered as an element of $PGL_2(\mathbb{Q})$), and consider the subgroup $\Gamma_0(p) = \{ \left[\begin{array}{cc}a & b \\\ c & d\end{array}\right]\in SL_2(\mathbb{Z})  c\equiv 0(\mod p) \}$. Then one may check that for $B\in \Gamma_0(p)$, $A^{1} B A = \left[\begin{array}{cc}d & c/p \\\ pb & a\end{array}\right] \in \Gamma_0(p)$, so $A\in Aut(\Gamma_0(p))$. Also, the principal congruence subgroup $\Gamma(p) \leq \Gamma_0(p)$. However, consider the matrix $C=\left[\begin{array}{cc}1 & 0 \\\ p & 1\end{array}\right] \in \Gamma(p)$. Then one has $A^{1} C A = \left[\begin{array}{cc}1 & 1 \\\ 0 & 1\end{array}\right] \notin \Gamma(p)$. Thus, the subgroup $\Gamma(p)\leq \Gamma_0(p)$ is not characteristic. This extends to all $n$, taking the appropriate congruence subgroup by extending trivially $\mod p$.

$\begingroup$ Is it standard to use "AtkinLehner involution" to refer to the matrix and not just its inner automorphism (or PGL class)? I don't have any alternative terminology to suggest, but it really bothered me when I read the assertion that the matrix is an involution. $\endgroup$ – Ben Wieland Sep 6 '11 at 20:36

$\begingroup$ @ Ben: I'm referring to the element of PGL. I realize it's usually normalized to have determinant 1, but I wanted to make it clear that it lies in $PGL_2 Q$. $\endgroup$ – Ian Agol Sep 6 '11 at 20:56

$\begingroup$ I would refer to A as an AtkinLehner involution but would probably apologize for doing so if Ben got in my face about it. $\endgroup$ – JSE Sep 6 '11 at 21:58

$\begingroup$ @Ben: I apologize. I guess I can't ask A.O.L. if he would mind. $\endgroup$ – Ian Agol Sep 7 '11 at 0:48

$\begingroup$ @Agol: we can simply take $A \in SL(2,\mathbb Q)$. Thanks for the answer. $\endgroup$ – Poove Sep 8 '11 at 20:55
As noted by in the answer to this question:
Automorphisms of $SL_n(\mathbb{Z})$
by @Guntram, Margulis super rigidity implies that an authomorphism of a lattice (which a finite index subgroup of $SL(n, \mathbb{Z})$ is) extends to an automorphism of $SL(n, \mathbb{Q}).$ So, the answer to your previous question
Principal congruence subgroups of $SL(n, \mathbb{Z})$
works just as well here.

$\begingroup$ That is for $n\ge 3$. For $n=1$ the result is also true, but for $n=2$ a finite index subgroup may not contain a congruence subgroup at all. $\endgroup$ – user6976 Sep 6 '11 at 14:15

$\begingroup$ @Mark: yes, I was assuming higher rank (since, as you say, for $n=2$ the situation is completely different...) $\endgroup$ – Igor Rivin Sep 6 '11 at 14:27

1$\begingroup$ $\Gamma_m$ isn't characteristic in $SL_n(Q)$. It isn't even normal. Also, superrigidity only says that an automorphism of a lattice virtually extends to an automorphism of the ambient group. $\endgroup$ – Ben Wieland Sep 6 '11 at 18:41

1$\begingroup$ I never said $\Gamma_m$ was characteristic in $SL_n(\mathbb{Q}).$ $\endgroup$ – Igor Rivin Sep 6 '11 at 19:58

$\begingroup$ @Igor Rivin: Any automorphism of $\Gamma$ extends(super rigidity) to an automorphism of $SL(n,\mathbb R)$, for n>2. But it not necessarily restricts as an automorphism to $SL(n,\mathbb Z)$, instead it restricts to the normalizer of $\Gamma$ in $SL(n,\mathbb R)$, Which is again a proper finite index subgroup of $SL(n,\mathbb Z)$. So mathoverflow.net/questions/62446/… doesn't say $\Gamma_m$ is characteristic in $\Gamma$. $\endgroup$ – Poove Sep 8 '11 at 21:09