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Draw a braid on $n$ strings sloppily, so that no three strings pass through the same point. Convince yourself that the parity of the number of crossings is the same in every other drawing of the sam …

No. Let $G=SL_n$, acting on its defining representation $V$, with $n\geq2$. Let $R=\mathbb{Z}[X_1,\dots,X_n]$ be the obvious $\mathbb{Z}$-form of the ring of polynomial functions on $V$. Let $p$ be a …

To get some clarity it may help to consider the example where $G=\{\pmatrix{1&t\cr 0&e^u}\mid t, u\in \mathbb R\}$. The adjoint action of its Lie algebra is not right for a unipotent group.

The normalizer in $SL_2(\Bbb R)$ is indeed $SL_2(\Bbb Z)$. [See the comments by Yves Cornulier for the normalizer in $GL_2(\Bbb R)$.] If $\pmatrix{a&b\\c&d}\in SL_2(\Bbb R)$ normalizes $SL_2(\Bbb Z) …

For any finite index subgroup $G$ there is a nonzero integer $m$ so that $G$ contains the elementary matrices $e_{ij}(m)$ that have ones on the diagonal, $m$ at the $(i,j)$ entry and zeroes elsewhere. …

The answer should be negative, because the $K_2$ of the reals is humongous. That is, there are nontrivial central extensions. (Please do not ask a question and then explain its negation.) Algebraic $K …

Rotate a quarter turn around the axis passing through the midpoints of two antipodal edges. That gives a different copy of the original icosahedron. A half turn preserves both icosahedra. So the state …

For $G=GL(2,\mathbb Z)$ there is no proper subgroup isomorphic to it. Consider the dihedral group $D$ of isometries of a regular 6-gon. There is only one conjugacy class in $G$ of subgroups isomorphic …

Indeed the "symmetrized square-free monomials" seem to generate. (Order lexicographically and look what the highest term in a product looks like. Now use that to concoct rewriting rules.) [Oops! T …