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Questions about the branch of algebra that deals with groups.
2
votes
On direct product of capable groups
The converse is not true in general. Beyl, Felgner and Schmid presented a condition in which the capability of a direct product of finitely many of groups implies the capability of each of the factor …
3
votes
Accepted
Presentation of $\mathrm{Sp}(4, \mathbb{Z})$
Presentations for the Siegel modular groups $Sp(2g,\mathbb{Z})$ are closely related to presentations of the surface mapping class group.In general, the braid group $B_{2g+2}$
acts on the free group $F …
4
votes
Generalizations of Fitting subgroup
A good reference is the book "Finite group Theory" of M. Aschbacher.
Chapter 11 is devoted to the generalized Fitting subgroup $F^*(G)$ of $G$,
and is quite detailed. The second one, $\tilde{F}(G)$ ha …
2
votes
Accessible problems on classical groups over complex or real numbers.
Classical groups are really a classical topic and have been studied a lot. It is therefore not easy to find some interesting topics which are still accessible. Certainly there are some aspects of unit …
1
vote
frobenius group
Frobenius groups have trivial center. This is helpful for non-examples, e.g., to see that
$p$-groups, or the quaternion group is not Frobenius (and all nilpotent groups).
Further examples are: non-abe …
22
votes
3
answers
3k
views
Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ?
All free groups of finite or infinite countable rank are subgroups of the free non-abelian group $F_2$, which is linear. However, a free group of infinite uncountable rank will not be a subgroup of $F …
5
votes
1
answer
324
views
Are all dinilpotent groups solvable, i.e., groups G=AB with nilpotent subgroups A, B ?
The Wielandt-Kegel theorem states that all finite dinilpotent groups G=AB are solvable.
This results extends to many infinite groups, e.g., to finitely-generated linear groups.
In general however, I t …
3
votes
Bound on the index of an abelian subgroup in discrete subgroup of the euclidean group?
Every discrete group of Euclidean isometries acts properly discontinuously and cocompactly on some subspace of $\mathbb{R}^n$ (a result of Bieberbach). If we are in the crystallographic case, then the …
4
votes
$nse$ for which simple group was determined?
I found explicit results for $nse(G)$ for many finite simple groups $G$. Indeed, there is a large literature on the characterisation of finite simple groups $G$ by their order $\mid G\mid$ and the set …
2
votes
Knot group epimorphism from a prime knot
Yes, in general $L$ need not be a prime knot, if $K$ is a prime knot, given a knot epimorphism $f\colon \pi K \to \pi L$, see section $4$ of D.S. Silver and W. Whitten, Knot group epimorphisms II (whi …
11
votes
Accepted
Groups as Union of Proper Subgroups: References
The mentioned result of Cohn has been further extended. Let us write $σ(G) = n$ whenever $G$ is the union of $n$ proper subgroups, but is not the union of any smaller number of proper sub- groups. Thu …
5
votes
Can group solvability be detected from identities among the generators?
For finite groups, solvability can be detected from Engel-like identities. This was not really the question, but
it is very interesting in this context, I think.
The proof is surprisingly complicated …
7
votes
How to compute the Alexander polynomial of general torus knot
This is an exercise in many topology books. Here is a reference with a complete
proof: Look up Example 9.15 in the book "Knots" by G. Burde and H. Zieschang.
The Jacobian of the presentation $G(T_{p,q …
9
votes
Which finite p-groups occur as commutators of finite p-groups?
There are some results for special cases. Burnside has proved in $1912$ that,
if $G$ is a non-metabelian $p$-group, then the centre of the derived group of $G$ cannot be
cyclic. In particular, a non- …
8
votes
Accepted
orders of maximal abelian subgroups
For a finite group $G$, denote the maximum of the orders of its Abelian subgroups by $a(G)$. Then we have, for $G=F_4(q)$ and $q$ even,
$$
q^{11}\le a(G)\le q^{17},
$$
and for $G=F_4(q)$, $q$ odd,
$$
…