I am currently reviewing some examples of nilpotent groups. Why isn't $GL(2,\mathbb{Z})$ nilpotent?
2025-01-13 09:51:00.1736761860
$GL(2,\mathbb{Z})$ and nilpotency
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Recall that any nilpotent group is solvable. But $GL(2,\mathbb Z)$ is not solvable, because it possesses a free non-abelian subgroup, the so-called Sanov subgroup.