# Tag Info

4

No. Indeed, every Coxeter group with no $\infty$ edge has Serre's Property FA and hence cannot be written as a nontrivial amalgam. Hence if it's not spherical or affine, it does not belong to the class you're defining. This notably applies to many Coxeter groups on 3 Coxeter generators. (A group generated by a finite subset $S$ such that every element of the ...

4

No, it's not possible because the isometry group of the Euclidean plane is amenable (as discrete group, since it's solvable), so every $G$-set admits an invariant mean defined on all subsets. If $S$ is $G$-invariant, this applies to $S$ as $G$-set and an immediate contradiction follows.

4

Let $\tilde Y$ denote the universal cover of $Y$. Then $\tilde{X'}$ is formed from $\tilde X$ by gluing a copy of $S^2$ to each preimage of the basepoint of $X$. Then by the Hurewicz isomorphism, we have $$\pi_2(X') = \pi_2(\tilde{X'}) = H_2(\tilde X) \oplus \bigoplus_{g \in \pi_1(X)} \mathbb Z = \pi_2(X) \oplus \mathbb Z[\pi_1(X)],$$ since the classes of ...

3

There is a theoretical answer (as opposed to an algorithmic answer) found in Björner and Brenti's "Combinatorics of Coxeter groups", Section 1.5. (They seem to credit it to Matsumoto.) Their Theorem 1.5.1: Suppose $W$ is a group generated by a subset $S$ consisting of elements of order $2$. Then TFAE: $(W,S)$ is a Coxeter system (i.e. $S$ ...

2

Since a theorem of Borel (which you quote in the comment) tells you that this regular map is dominant for every $w$, its differential is surjective on a Zariski-dense open subset $U_w$. Hence there intersection (over all $w\neq 1$) of these subsets $U_w$ is a $\mathrm{G}_\delta$-dense subset of full measure.

2

I don't think this is what the questioner means, so this is not really an answer. But it's worth mentioning and it's too long for a comment. If we know that $t_1,\ldots,t_n$ are transpositions, then $G$ is a "reflection subgroup" of $S_m$ (a subgroup generated by reflections). Then a theorem of Deodhar ("A note on subgroups generated by ...

1

news to me: this has a quick answer for indefinite case. Lecture notes at Chicago, https://math.uchicago.edu/~dannyc/courses/4manifolds_2018/4_manifolds_notes.pdf He refers to Serre's little book for part of it, chapter V is pages 48-58. The definite case has bounds involved, there is a command in magma https://magma.maths.usyd.edu.au/magma/overview/2/17/9/ ...

1

Another answer from a combinatorial point of view: the set of all transpositions is a generating set for $S_n$. Obviously this is a huge overkill. But a nice thing about including all transpositions is now your generating set is closed under conjugation. In fact, this generating set is used to define the so-called absolute order or reflection order (which ...

1

You can take any tree $T$ on $[n]$ and the set of transpositions $(i,j)$ for $\{i,j\}$ an edge of $T$ will give a (minimal) generating set for $S_n$. This obviously includes the "usual" generating set of adjacent transposition, as well as the "star transpositions" mentioned by Igor Pak, but also many more examples. In general these ...

1

Contrary to OP's claim, this is false even for finitely generated abelian groups, including $\mathbf{Z}^2$. Indeed, fix a dense group embedding $i:\mathbf{Z}^2\to\mathbf{Z}_p$ (namely $i(n,m)=n+mt$ for some fixed irrational $t\in\mathbf{Z}_p$). Define $G_n=i^{-1}(p^n\mathbf{Z}_p)$. Then $(G_n)$ is a decreasing sequence of subgroups with intersection reduced ...

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