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comment Bound on the index of an abelian subgroup in discrete subgroup of the euclidean group?
a discrete group $\Gamma$ of isometries of $\mathbf{R}^n$ indees acts properly cocompactly by isometries on $\mathbf{R}^k$ for some $k\le n$ (namely any minimal nonempty invariant affine subspace of $\mathbf{R}^n$ for the given action, but you have to take into account that the new action has a finite kernel, possibly nontrivial, and precisely equal to the maximal finite normal subgroup in $\Gamma$.
Jul
19
comment Generating finite groups
My example gives an example with the free profinite group on 2 generators $PF_2$: just take a continuous homomorphism from $PF_2$ onto $S^3$, define the new $H$ as the pull-back of $H\subset S^3$, and $x_1,x_2$ any lifts of the old $x_1,x_2$.
Jul
19
revised Generating finite groups
added 89 characters in body
Jul
19
answered Generating finite groups
Jul
13
comment On groups satisfying a law
This subject has a long history. If you google keywords such as "varieties of groups" you will find a lot.
Jul
12
comment Is any finitely generated nilpotent pro-$p$ group necessarily the pro-$p$ completion of some finitely generated nilpotent group?
When you have a f.dim. nilpotent Lie algebra $\mathfrak{g}$ over $\mathbb{Q}_p$, the BCH formula provides a $p$-adic Lie group with Lie algebra $\mathfrak{g}$, and the clopen subgroups also admit $\mathfrak{g}$ as Lie algebra. Moreover taking Lie algebras is functorial for $p$-adic Lie groups and continuous homomorphisms, hence topological group isomorphisms induce Lie algebra isomorphisms.
Jul
12
comment Is any finitely generated nilpotent pro-$p$ group necessarily the pro-$p$ completion of some finitely generated nilpotent group?
because the Lie algebra of $K_c$ is $\mathfrak{g}_c(\mathbb{Q}_p)$.
Jul
12
comment Is any finitely generated nilpotent pro-$p$ group necessarily the pro-$p$ completion of some finitely generated nilpotent group?
In my email I gave you an argument to encompass $p=2,3$: you have the $G_c(\mathbb{Q}_p)$ that are (finite-to-one-wise) non-locally-isomorphic; then if you just pick a compact open open subgroup $K_c$ inside, they are also (finite-to-one-wise) non-isomorphic.
Jul
8
comment solvable word problem without algorithm
@StevenStadnicki I'm aware that these kind of results exist, but they ought to be stated in a proper way, which is, I'm afraid, not the case. What's the input? A "f.g. group with solvable word problem" is not an input. Should it be understood that the input is a recursive presentation and the question is whether, when the resulting group has solvable word problem, the output provides a algorithm? Possibly there are other interpretations of the question, and for this reason I don't consider it's asked in a proper way.
Jul
8
comment Union of conjugates of a closed subgroup of a compact group
@Benjamin: the assertion is that the condition holds for finite groups, which, as you mention, is a tautology.
Jul
8
comment solvable word problem without algorithm
2) should be formulated in a way to make sense, what's the input? Right now it sounds like "given a non-empty subset of positive integers, can we exhibit an element in this set?"
Jul
7
comment What are the possible finite non-solvable quotients of one relator groups?
Moshe: I think it's only true for Baumslag-Solitar $BS(m,n)$ with $m,n$ coprime.
Jul
7
comment The set of (property) elements of a locally compact group is closed
I think it follows from a result of Trofimov that the topological FC-center, namely those $g$ whose conjugacy class has a compact closure, is closed if $G$ is in addition compactly generated; this is not true for general locally compact groups.
Jul
6
comment A sequence of subsets of an infinite group
OK, but once we are reduced to finitely generated quasi-finite groups, this discussion is of little relevance.
Jul
6
comment A sequence of subsets of an infinite group
I'm not sure what you mean by "no nondiscrete metric". What do you require about the metric?
Jul
6
answered A sequence of subsets of an infinite group
Jul
6
comment A sequence of subsets of an infinite group
@Misha I'm not sure of any link with approximate groups, which involve a parameter and finite subsets. On the other hand there should be a link with non-topologizable groups.
Jun
30
comment Compact Lie groups with only 3 dimensional cohomology generators
I don't see the point in "but also $SO(4)$". Either you say it's a quotient of a finite direct product of $SU(2)$'s by a central subgroup, or if you really want a full classification, there are infinitely many possibilities which are not nice to enumerate (for $n$ factors, they are indexed by the quotient of the Grassmannian of $(Z/2Z)_2^n$ by the symmetric group $S_n$)... If we want only those that are not direct products, we should remove those element in the Grassmannian that split as products according to the direct product decomposition of $(Z/2Z)^n$.
Jun
21
comment examples of polyclic groups
1) A polycyclic group is is a solvable unimodular Lie group, and is a lattice in itself. Do you mean connected or with finitely many components? 2) I can't guess the meaning of the sentence including "I will just say". 3) $SL_n(\mathbf{Z})$ is not polycyclic.
Jun
5
comment Lie groups vs Lie monoids
One example of Lie monoid is that of invertible real matrices with positive entries. It's open in $GL_n$ so should have the same Lie algebra, and the exponential map cannot be defined everywhere.