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Let $n$ be either a product of two distinct primes or a power of a prime. Then is it true that normal subgroups of $SL(2, \mathbb Z/n \mathbb Z)$ remains normal in $GL(2, \mathbb Z/n\mathbb Z)$ ?

The $n$=prime case was dealt with here For prime $p$, normal subgroups of $SL(2, \mathbb Z/p\mathbb Z)$ remains normal in $GL(2, \mathbb Z/p\mathbb Z)$?

  • I suspect you want to show that $SL(2,\mathbb{Z}/n\mathbb{Z})$ is a characteristic subgroup of $GL(2,\mathbb{Z}/n\mathbb{Z})$ from which your question will follow, but I can't think how to prove it (or even if it's true generally). – Matt B Feb 04 '18 at 16:59
  • This is certainly true in arbitrary dimension when $n \geq 3$ is prime as it is the commutator/derived subgroup which is always characteristic. – Matt B Feb 04 '18 at 17:02
  • @Matt B: thanks for your interest. Without even bothering about generality, it would be very helpful just to settle the question I ask .. –  Feb 04 '18 at 17:55
  • ${\rm SL}(2,Z/nZ)$ is the direct product of the groups ${\rm SL}(2,Z/qZ)$, where the $q$ are the prime power factors of $n$. And, if $q = p^k$ for prime $p$, then ${\rm SL}(2,Z/qZ)$ has a normal $p$-subgroup with quotient ${\rm SL}(2,p)$. I think all of the normal subgroups of ${\rm SL}(2,Z/nZ)$ are characteristic. – Derek Holt Feb 04 '18 at 21:23
  • @Derek Holt : characteristic in where ..? –  Feb 05 '18 at 06:45
  • What do you mean? The sentence "all normal subgroups of the group $G$ are characteristic" is clear and unambiguous. BTW, I voted to close this question for lack of context. – Derek Holt Feb 05 '18 at 10:03

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