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Questions about the branch of algebra that deals with groups.

41 votes
1 answer
2k views

Known and fixed gaps in the proof of the CFSG

As the "second-generation" proof of the Classification of Finite Simple Groups is being written up in the volumes by Gorenstein, Lyons, Aschbacher, Smith, Solomon, and others (see e.g. this question) …
Carl-Fredrik Nyberg Brodda's user avatar
24 votes

Recognizing free groups

As indicated in the comments, it's undecidable in general to take as input a finite presentation of a group and try to output whether or not this group is free or not. This is a direct consequence of …
Carl-Fredrik Nyberg Brodda's user avatar
20 votes
1 answer
977 views

Proof of CFSG assuming every simple group is two-generated

It is well-known that one of the corollaries of the classification of finite simple groups (CFSG) is that every finite simple group can be generated by two elements. In a comment on an answer to an ol …
Carl-Fredrik Nyberg Brodda's user avatar
19 votes
Accepted

Number of finite index subgroups in a free abelian group

Yes. This is given by OEIS sequence A160870. The number of subgroups of index $n$ in $\mathbf{Z}^m$ is there denoted $T(n,m)$. There is a recursive formula in terms of the divisors of $n$ given at thi …
Carl-Fredrik Nyberg Brodda's user avatar
18 votes

Results from abstract algebra which look wrong (but are true)

In combinatorial group theory, loosely speaking almost any problem one can imagine, in full generality, turns out to be undecidable. This includes the word problem, the isomorphism problem, the trivia …
18 votes
1 answer
750 views

Is solvability semi-decidable?

Let $G = \langle A \mid R \rangle$ be a finitely presented group, given by a finite presentation. If $G$ is abelian, then we can verify this fact: simply verify the fact that $[a, b] = 1$ for all gene …
Carl-Fredrik Nyberg Brodda's user avatar
13 votes

Conceptual reason why the sign of a permutation is well-defined?

Obviously, there are lots of answers already, but I thought I'd give the proof of (5) as given by Jordan already in 1870 in his Traité des substitutions -- this has the benefit of being quite clear to …
Carl-Fredrik Nyberg Brodda's user avatar
12 votes
1 answer
406 views

Commutator problem vs conjugacy/word problem

For a finitely presented group $G$, generated by a finite set $A$, the commutator problem is the decision problem: given a word $w$ over the alphabet $A \cup A^{-1}$, can one decide if $w$ is a commut …
Carl-Fredrik Nyberg Brodda's user avatar
11 votes
Accepted

Reference request: Recent progress on the conjugacy problem for torsion-free one-relator gro...

As mentioned in the comments, this is still considered an open problem. I thought I'd flesh out a few aspects. A solution was claimed in 1992 by Juhasz, but it seems to have failed to convince experts …
Carl-Fredrik Nyberg Brodda's user avatar
11 votes
Accepted

Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$

As you mention in your update, you have a general answer, but if you want a concrete answer for the low-dimensional integral cohomology of $G = \operatorname{GL}(3,2)$ (or any other finite group!), yo …
Carl-Fredrik Nyberg Brodda's user avatar
10 votes

Analogous results in geometric group theory and Riemannian geometry?

Here is a very classical example. As stated in the comments, Gromov was an early proponent of importing ideas from geometry to group theory, but already thirty years earlier there was work in this dir …
Carl-Fredrik Nyberg Brodda's user avatar
9 votes
1 answer
225 views

Yang-Mills algebra and lower central series of surface groups

Here is a connection that I recently noticed, but I haven't quite been able to make sense of. It might follow from well-known facts; apologies, if so. This is quite far from my area. First, in "Yang-M …
Carl-Fredrik Nyberg Brodda's user avatar
9 votes
Accepted

Is the isomorphism problem solvable for torsion-free groups?

Novikov's centrally-symmetric group $\mathfrak{A}_P$ is a torsion-free group with undecidable word problem, constructed in [1]. Novikov did not prove it is torsion-free but, as Adian points out in [Ad …
Carl-Fredrik Nyberg Brodda's user avatar
9 votes
3 answers
490 views

Residually solvable Bianchi groups

Let $d$ be a square-free positive integer, and let $\mathcal{O}_d$ be the ring of integers of the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$. Consider the Bianchi group $\Gamma_d = \oper …
Carl-Fredrik Nyberg Brodda's user avatar
8 votes
0 answers
121 views

The conjugacy problem for two-relator groups

Is the conjugacy problem for two-relator groups known to be undecidable? The word problem for two-relator groups is a famous open problem (appearing e.g. as Question 9.29 in the Kourovka notebook), an …
Carl-Fredrik Nyberg Brodda's user avatar

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