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I know we can use Margulis normal subgroup theorem to prove the normal subgroup of the finite index subgroup of $SL(3,\mathbb{Z})$ is either finite or finite index, could we prove this fact without Margulis normal subgroup theorem? Since Margulis normal subgroup theorem states the situation more general, but now we just focus on $SL(3,\mathbb{Z})$.

Antoine
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  • I remember reading somewhere that this was in fact known for SL(3, $\mathbb{Z}$) before Margulis' work. However, I can't remember where, and don't know the proof myself. – Owen Sizemore Mar 25 '14 at 13:46
  • @OwenSizemore In fact, I have read a proof for $SL(3,\mathbb{Z})$, it considers the g.c.d of the entries of $SL(3,\mathbb{Z})$, but it fails to prove the finite index subgroup of $SL(3,\mathbb{Z})$. – Antoine Mar 25 '14 at 14:05

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