Questions tagged [finite-groups]

Questions on group theory which concern finite groups.

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Centraliser of a finite group

Let $G=\operatorname{Sp}(8,K)$ be a symplectic algebraic group over an algebraically closed field of characteristic not $2$. We have a vector space decomposition $V_8=V_2\otimes V_4$ where the $2$-...
user488802's user avatar
14 votes
1 answer
571 views

Is applying Feit–Thompson’s theorem for the nonexistence of a simple group of order $1004913$ really a circular argument?

In p.212 of Dummit–Foote’s Abstract Algebra, 3rd Edition, an analysis of a hypothetical simple group $G$ of order $1004913 = 3^3 \cdot 7 \cdot 13 \cdot 409$ is carried out. The authors write: We ...
Kazune Takahashi's user avatar
8 votes
2 answers
423 views

When are two semidirect products of two cyclic groups isomorphic

(I have posted this question in Math Stack Exchange, only to have received no answer.) It is well known that a semidirect product of two cyclic groups $C_m$ and $C_n$ has the form $$ C_m \rtimes_k C_n ...
Jianing Song's user avatar
9 votes
1 answer
245 views

A question related to Jordan's theorem on subgroups of $\mathrm{GL}_n(\mathbb{C})$

$\newcommand{\C}{\mathbb{C}}$ $\newcommand{\mr}{\mathrm}$ For any positive integer $n$, let $f(n)$ be the minimal integer with the following property: For any finite subgroup $G < \mr{GL}_n(\C)$ ...
naf's user avatar
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1 vote
0 answers
200 views

Generalized conjugacy classes in (topological) groups

Let $G$ be a topological group. We define an equivalence relation on $G$ as follows: For $a,b\in G$ we set $a\sim b$ if the following two maps are topologically conjugate: $$x\mapsto ax,\qquad x\...
Ali Taghavi's user avatar
1 vote
0 answers
108 views

What is $H^*(\mathbb{CP}^{2^N-1}/\Sigma_n;\mathbb{Z})$ when $N=\binom{n}{2}$?

$H^*((S^3)^N/\Sigma_n;\mathbb{Q})$ is computed here. It makes a little more sense to compute $H^*((S^2)^N/\Sigma_n;\mathbb{Q})$ given that global phase is irrelevant. The proof is exactly the same. ...
Jackson Walters's user avatar
1 vote
0 answers
97 views

Reduction mod 2 for orthogonal groups

Setting Let $k$ be a real quadratic field, $\mathbb Z_k$ its ring of integers. Let $n$ be an even integer $A$ a symmetric $n$-by-$n$ matrix with coefficients in $\mathbb Z_k$. Let $L$ be the lattice $\...
Jean Raimbault's user avatar
2 votes
0 answers
109 views

Subgroups of a finite group whose conjugates intersect to conjugates of a specified subgroup

I have encountered a mysterious condition on finite groups in my research, and would like help understanding it better. Let $G$ be a finite group, and let $H\leq K\leq G$ be a chain of subgroup ...
Chase's user avatar
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172 views

Subgroups of the symmetric group and binary relations

Motivation The following came up in my work recently. (NB this is the motivation, not the question I'm asking. You can skip to the actual question below, which is self-contained, but not self-...
Z. A. K.'s user avatar
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3 votes
1 answer
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Finite subgroup of $\operatorname{Sp}(2n,K)$

Let $G$ be the algebraic group $\operatorname{Sp}(2n, K)$ where $K$ is an algebraically closed field of characteristic not $2$. There is a quaternion subgroup $Q$ such that $Q/Z(G)$ is elementary ...
user488802's user avatar
2 votes
0 answers
59 views

Diameters of permutation groups with transitive generators

Suppose that for a permutation puzzle on $n$ elements we have a subroutine to place any $m>3$ elements in arbitrary positions (possibly scrambling the rest). Can we solve the puzzle (if it is ...
Dmytro Taranovsky's user avatar
4 votes
0 answers
163 views

A different approach to proving a property of finite solvable groups

Edit: I'd be happy to hear any vague thoughts you might have, however far they may be from a complete solution! I asked this on math.stackexchange a couple of days ago, but it didn't attract any ...
semisimpleton's user avatar
6 votes
1 answer
196 views

On a generalization of Schur-Zassenhaus

Disclaimer: I'm not a group theorist, I arrived at the following question from algebraic geometry. The first half of the Schur-Zassenhaus theorem states that, if $N$ is a normal subgroup of a finite ...
Giulio Bresciani's user avatar
0 votes
0 answers
99 views

$G\cdot H$ with $G,H$ non-Abelian finite simple

Can a non-split extension of one non-Abelian finite simple group by another exist?
Daniel Sebald's user avatar
1 vote
0 answers
79 views

Central-by-cyclic

This is a following-up question of this. Lemma 2.4 from Robert Griess' paper "Elementary abelian $p$-subgroups of algebraic groups" states: (i) Let $T$ be a finite $p$-group whose Frattini ...
user488802's user avatar
1 vote
1 answer
138 views

$|C(E):C(E)\cap C(Z(U))|=1$ or $p$

Lemma 2.4 from Robert Griess' paper "Elementary abelian $p$-subgroups of algebraic groups" states: (i) Let $T$ be a finite $p$-group whose Frattini subgroup is cyclic and central. Then $T'$ ...
user488802's user avatar
0 votes
0 answers
42 views

A question on width vs covering of the subgroup generated by a conjugacy class in a finite group

Let $G$ be a finite group and $C$ be a conjugacy class of $G$. It is clear that there exists $k\in \mathbb{N}$, such that $1\cup C\cup C^2 \cup \cdots \cup C^k=\langle C \rangle$. Note that $\langle C ...
Riju's user avatar
  • 420
5 votes
1 answer
239 views

Extension of base field for modules of groups and cohomology [duplicate]

Let $G$ be a group and let $K/k$ be a field extension. Suppose that $V$ is a $kG$-module, and let $V_K = K \otimes_k V$ be the $KG$-module given by changing the base field. Is it true that $H^n(G,V_K) ...
testaccount's user avatar
35 votes
2 answers
3k views

Why does the monster group exist?

Recently, I was watching an interview that John Conway did with Numberphile. By the end of the video, Brady Haran asked John: If you were to come back a hundred years after your death, what problem ...
Leibniz's Alien's user avatar
1 vote
0 answers
94 views

Finite groups of prime power order containing an abelian maximal subgroup

Let $G$ be a finite $p$-group containing an abelian maximal subgroup. Then it is a well-known result that $|G:Z(G)|=p|G'|$. If in addition $G$ is of nilpotent class 2, then $|G:Z(G)|\leq p^{r+1}$, ...
Hamid Shahverdi's user avatar
4 votes
1 answer
158 views

Mackey coset decomposition formula

I have a question about following argument I found in these notes on Mackey functors: (2.1) LEMMA. (page 6) Let $G$ be a finite group and $J$ any subgroup. Whenever $H$ and $K$ are subgroups of $J$, ...
user267839's user avatar
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114 views

normalizer info for subgroups

In [1], Griess classified the maximal nontoral elementary abelian subgroups of algebraic groups. For the exceptional types, normalizer info was also given. Is there any work out there providing ...
user488802's user avatar
2 votes
0 answers
57 views

Are the integer points of a simple linear algebraic group 2-generated?

Set Up: Let $ K $ be a totally real number field. Let $ \mathcal{O}_K $ be the ring of integers of $ K $. Let $ G $ be a simple linear algebraic group. Suppose that $ G(\mathbb{R}) $ is a compact Lie ...
Ian Gershon Teixeira's user avatar
4 votes
0 answers
143 views

New characters from old

(All groups in the following discussion are assumed to be finite.) Character induction is an operation that produces a character of a group given a character of a subgroup. I'm aware that there are ...
semisimpleton's user avatar
8 votes
1 answer
372 views

Trivial group cohomology induces trivial cohomology of subgroups

From the answer to another question I asked (Projective representations of a finite abelian group) and from the structure theorem of finite abelian groups it follows that if $A$ is a finite abelian ...
Andrea Antinucci's user avatar
4 votes
1 answer
152 views

Projective representations of a finite abelian group

Projective representations of a group $G$ are classified by the second group cohomology $H^2(G,U(1))$. If $G$ is finite and abelian, it is isomorphic to the direct product of cyclic groups $$ G\cong ...
Andrea Antinucci's user avatar
4 votes
1 answer
178 views

Quadratic refinements of a bilinear form on finite abelian groups

$\DeclareMathOperator\Hom{Hom}$Let $A$ be a finite abelian group and $\text{Sym}(A)$ the (abelian) group of symmetric bilinear forms over $A$ valued in $\mathbb{R}/\mathbb{Z}$. A quadratic function on ...
Andrea Antinucci's user avatar
9 votes
0 answers
161 views

Cyclic numbers of the form $2^n + 1$

A cyclic number (or cyclic order) is a number $m$ such that the only group of order $m$ is the cyclic group $\mathbb{Z}/m\mathbb{Z}$. The set of cyclic numbers admits a couple of cute number-theoretic ...
Z. A. K.'s user avatar
  • 281
24 votes
0 answers
689 views

Revising the proof of CFSG

This is an oft-quoted excerpt from John Thompson's article "Finite Non-Solvable Groups": “... the classification of finite simple groups is an exercise in taxonomy. This is obvious to the ...
semisimpleton's user avatar
4 votes
1 answer
222 views

Condition on $q$ for inclusion $p^{1+2r}\cdot\operatorname{Sp}(2r,p)\leqslant \operatorname{GU}(p^r,q)$

Let $p$ be an odd prime. What's the condition on $q$ for $$ p^{1+2r}\cdot\operatorname{Sp}(2r,p)\leqslant \operatorname{GU}(p^r,q)\;? $$ I did some computation and seemed that $q\equiv -1$(mod $p$) ...
user488802's user avatar
4 votes
1 answer
118 views

CFSG-free proof for classifying simple $K_3$-group

Let $G$ be a finite nonabelian simple group. We call $G$ a $K_3$-group if $|G|=p^aq^br^c$ where $p,q,r$ are distinct primes and $a,b,c$ are positive integers. My question is: Is there a CFSG-free ...
user44312's user avatar
  • 385
2 votes
1 answer
160 views

$\mathrm{PSL}_3(4)$ inside the Monster group

Which quasisimple groups with central quotient $G\cong\mathrm{PSL}_3(4)$ are isomorphic to subgroups of the Monster sporadic group? So far I know that $G$ itself is not and that $2\cdot G$, $2^2\cdot ...
Daniel Sebald's user avatar
2 votes
1 answer
182 views

Sparsity of q in groups PSL(2,q) that are K_4-simple

One of the problems that has come up during my research concerns $K_4$-simple groups (simple groups with $4$ prime divisors). The only (potentially) infinite family of groups satisfying this condition ...
abiteofdata's user avatar
6 votes
0 answers
150 views

Singularity category of a hypersurface associated to $M_{11}$

For reasons to do with classifying spaces of finite groups, I have the following algebra. Let $k$ be a field of characteristic two, and let $R = k[x,y,z]/(x^2 y + z^2)$, as a graded $k$-algebra with $|...
Dave Benson's user avatar
  • 8,439
4 votes
1 answer
104 views

$\mathbb{Z}$-forms of rational representation of finite group

Let $G$ be a finite group and let $\rho\colon G\to \mathrm{GL}_n(\mathbb{Q})$ be a representation of $G$. How does one go about classifying the $\mathbb{Z}$-forms of $\rho$? For example: here it is ...
Sam Hughes's user avatar
0 votes
0 answers
61 views

Is a Lagrangian subgroup of a metric group isomorphic to its quotient?

A metric group is a finite abelian group $G$ with a quadratic function $$q:G\rightarrow \mathbb R/\mathbb Z\;,$$ that is, $$M(a,b):= q(a+b)-q(a)-q(b)$$ is bilinear in $a$ and $b$ [edit: and non-...
Andi Bauer's user avatar
  • 2,799
7 votes
2 answers
572 views

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

I’m wondering what is known about the cohomology of $\operatorname{GL}_3(\mathbb{Z}_2)$, the general linear group of $3\times3$ matrices over the finite field $\mathbb{Z}_2$. There are results in ...
Noah B's user avatar
  • 379
5 votes
2 answers
212 views

Unimodality of sequence of number of subgroups in $p$-groups

It's easy to know that the sequence of number of subgroups is unimodal for elementary abelian $p$-groups. I want to know if the result is true for any $p$-group. More, precisely, let $G$ be a finite $...
gdre's user avatar
  • 51
3 votes
1 answer
170 views

normalizer quotient is $\operatorname{GL}_2(p)$

Let $p$ be a prime and let $w$ be a primitive $p$-th root of unity in $\mathbb{C}$. There is an element $e$ of order $p$ in $G=\operatorname{PGL}_n(\mathbb{C})$ where $n=pk$ and $$e=\left[\left(\begin{...
user488802's user avatar
2 votes
0 answers
74 views

Reference request: structure of group of units of finite group ring

Let $G$ be a finite group, let $F$ be a finite field and let $F[G]$ be the group algebra of $G$ over $F$. What is known about the structure of the group of units $F[G]^\times$? Of course, it must ...
semisimpleton's user avatar
5 votes
1 answer
199 views

Product of all conjugacy classes

Related to this post of my coauthor Sebastien, I should also mention that one can also prove the following dual result: For any finite group G, the following identity holds: $$ \left(\prod_{j=0}^m \...
Sebastian Burciu's user avatar
0 votes
0 answers
109 views

Comparing the perfect groups of order 1344

Take two nonisomorphic perfect groups of order 1344 and label the elements of each with the numbers 1 through 1344, then superimpose their respective Cayley tables (for simplicity’s sake, the nth row ...
Daniel Sebald's user avatar
5 votes
0 answers
187 views

Adjoint identity on finite nilpotent groups

Let $G$ be a finite nilpotent group. Consider the following (adjoint) identity from [BuPa, Theorem 9.6]: $$\left(\prod_{\chi \in \mathrm{Irr}(G)} \frac{\chi}{\chi(1)} \right)^2 =\frac{|Z(G)|}{|G|}\...
Sebastien Palcoux's user avatar
3 votes
1 answer
289 views

A generalisation of induced representations

Let $G$ be a finite group, and $H\subseteq G$ a subgroup. Let $F$ be a field. Let $W$ be a finite-dimensional $F[H]$-module. Let $T$ be a left transversal of $H$ in $G$. Then we can define: $W^G=\sum_{...
semisimpleton's user avatar
1 vote
1 answer
117 views

Example of a group algebra with commutative Jacobson radical

I am searching a simple example of a finite group $G$, so that the Jacobson radical $J(FG)$, of group algebra $FG$ is commutative, where $F$ is a finite field. I know example for that if $G$ is any ...
neelkanth's user avatar
  • 141
12 votes
1 answer
415 views

abelian quotients of permutation groups

Let $G$ be a subgroup of the permutation group $S_n$, and let $H$ be a normal subgroup of $G$ such that the quotient group $G/H$ is abelian. What is the best known upper estimate for the cardinality $...
Yuri Bilu's user avatar
  • 1,036
9 votes
0 answers
99 views

Is there a strictly coassociative resolution of polynomial growth, for a finite group?

Let $G$ be a finite group and $k$ a field of characteristic $p$. It is well known, thanks to the work of Quillen, that the trivial $kG$-module $k$ has a projective resolution of polynomial growth. To ...
Dave Benson's user avatar
  • 8,439
0 votes
1 answer
187 views

Are these two natural cohomology classes of a manifold constructed from a 1-cochain and a group extension equal?

Let $X$ be a manifold, $G$ and $A$ finite abelian groups and $\epsilon \in H^2(G,A)$ a group cohomology class (for the moment I am assuming there is no action of $G$ on $A$). Given $\alpha \in H^1(X,G)...
Andrea Antinucci's user avatar
2 votes
2 answers
202 views

is the embedding $\mathrm{Sp}_{2m}(p)\leqslant S_{p^m-1}$ possible?

Is the following embedding possible? $\mathrm{Sp}_{2m}(p)\leqslant S_{p^m-1}$ where $S_{p^m-1}$ is a symmetric group and $p$ is prime. I see that when $p=3$ and $m=3$, the order of the former does ...
user488802's user avatar
1 vote
0 answers
107 views

Cohomology of the classifying space of a semidirect product, and some specific examples with cyclic groups

Let $G$ and $A$ be finite abelian groups and $\rho :G \rightarrow \text{Aut}(G)$ a representation of $G$. We can form the semidirect product $A\rtimes _{\rho} G$. Just to agree on the notation this is ...
Andrea Antinucci's user avatar

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