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I want to know whether the principal congruence subgroups of $SL(n, \mathbb{Z})$ are characteristic? please suggest me a reference.

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up vote 8 down vote accepted

Yes, they are caracteristic.

Let $\rho: GL(n, \mathbb{Z}) \to GL(n, \mathbb{Z}/m\mathbb{Z})$ be reduction mod $m$ and $\Gamma_n(m) := ker(\rho) \cap SL(n, \mathbb{Z})$ be a congruence sugroup.

According to the discussion in Automorphisms of $SL_n(\mathbb{Z})$ the automorphisms of $SL(n, \mathbb{Z})$ for $n > 2$ are generated by

  • $X \mapsto (X^T)^{-1}$
  • $X \mapsto AXA^{-1}, A \in GL(n, \mathbb{Z}).$

Obviously, $\Gamma_n(m)$ is invariant under these automorphisms and thus is characteristic.

In case $n = 2$ there is one more automorphisms of $SL(2, \mathbb{Z})$ that also leaves $\Gamma_n(m)$ invariant (for a description of this automorphism see the paper of Hua and Reiner mentioned in the link above).

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This is exactly the same answer as @Richard's – Igor Rivin Apr 20 '11 at 21:57
Yes it is. When I was starting to type there was no answer and after I saved mine, I saw Richard's. So I guess there was some overlapping in time. Nevertheless, I decided no to delete my answer since a) it confirms what Richard says he thinks and b) it has some more details about the generators (and the case $n = 2$) that might be interesting for readers who just overflow the question without looking into the mathoverflow link or the Hua-Reiner paper. – Ralph Apr 20 '11 at 22:13
you are right Ralph – Poove Apr 21 '11 at 5:56
@Ralph: agreed. – Igor Rivin Apr 25 '11 at 1:25
@Ralph: $\Gamma_n$ is characteristic. but what you said about the automorphism group of $SL(n,\mathbb Z)$ is true for n odd, because for n odd, all automorphisms of $SL(n,\mathbb Z)$ are induced by automorphisms of $GL(n,\mathbb Z)$(Hua-Reiner,theorem.3). Is it true if n even too? – Poove Apr 27 '11 at 9:30

I think the answer is yes.

See this question for a discussion of $\mathrm{Aut}(\mathrm{SL}(n,\mathbb{Z}))$. The generators given in the Hua-Reiner paper seem to preserve the principal congruence subgroups.

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Thanks a lot Richard. – Poove Apr 21 '11 at 5:55

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