Search Results

Here are two examples. I describe it as Lie algebras (over any field $K$). (1) The 7-dimensional, 3-step nilpotent Lie algebra with basis $(X_1,\dots,X_7)$ and nonzero brackets $$ [X_1,X_2]=X_4,[X_1 …

If this is your question, yes it's true that for every nilpotent compact Lie group $G$, $G_0$ is central. Indeed as you already noticed, $G_0$ is abelian, so the action by conjugation on $G_0$ facto …

Here's an alternative proof (at least, an alternative wording), making $H$ very explicit. For $r$ the rank, let $G\to\mathbf{Z}^r$ be a surjective homomorphism (e.g., take the abelianization homomorp …

Consider the subgroup of $\mathrm{GL}_4(\mathbf{Z})$ generated by the matrices $u=\begin{pmatrix}I_2 & 0\\0 & -I_2\end{pmatrix}$ and $v=\begin{pmatrix}0 & A\\I_2 & 0\end{pmatrix}$, where $A$ is a matr …

There is no such group that is $\mathbf{C}$-irreducible. Indeed, let $G$ be its Zariski closure. Let $U$ be its unipotent radical: by irreducibility, $U$ is trivial [this very point applies even assum …

There exists such example (thus also disproving Conjecture 4.1.28 here). Indeed there exists a nonzero nilpotent Lie algebra $\mathfrak{g}$ with rational coefficients whose automorphism group $A$ is …

It's true, and due to Ph. Hall, when $G$ is virtually nilpotent, and more generally (Roseblade) when $G$ is virtually polycyclic. When $G=\mathbf{Z}\wr\mathbf{Z}$ there exists an infinite simple $\mat …

No. For a scalar (= unital associative commutative) ring $R$, consider $V=V(R)=R[X]$, the polynomial ring, and $q$ the operator $X^n\mapsto X^{n-1}$, $X\mapsto 0$. Let $V_n=V_n(R)$ be the $R$-submodul …

Let $G$ be any finite $p$-group, whose center and derived subgroup both have order $p$. Then every proper quotient of $G$ is abelian. In particular, if $G$ is not generated by 2 elements, it answers t …

No. Let $U_n(R)$ be group upper triangular $n\times n$ matrices with identity diagonal over the ring $R$. The groups $U_3(\mathbf{Z}[\sqrt{d}])$ are pairwise non-abstractly-commensurable, when $d\ge 2 …

No. Call the minimal cardinal of a subset $S$ such that $H\cup S$ generates $G$, the (generating) corank of $H$ in $G$. I claim that There exists solvable (actually metabelian) finite groups for …

Let me call "left cobounded" and "right cobounded" subsets of a group which you call "right syndetic" and "left syndetic" respectively. That is, $X\subset G$ is left cobounded if there exists $F$ fini …

Here's a sketch of proof (which intersects yours): Let $\Gamma$ be QI to NIL. As you say, by Gromov's theorem, $\Gamma$ is virtually nilpotent; let some finite index subgroup be a lattice in some sim …

Yes, in a simply connected nilpotent Lie group with a $\mathbf{Q}$-form (i.e. with a lattice), there exist lattices approximating the whole group. In particular, if the group has positive dimension, t …