New answers tagged finite-groups
9
votes
Accepted
A pair of non-conjugate subgroups: a simple proof
I think that the elements $g = \dfrac1{\sqrt2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^{\oplus3}$ and $h = \dfrac1 2\begin{pmatrix} 1 & 1 & 1 & 0 & 1 & 0 \\ -1 & 1 ...
1
vote
Diameter of the "Masterball-puzzle" permutation groups by a kind of Cartier-Foata enumeration?
Some partial information.
Cartier-Foata - Koszul duality
The classical Cartier-Foata counting formula can be seen as particular case of the Koszul duality. Consider algebra generated by several ...
4
votes
Accepted
Irreducible subspaces in the space of functions on Grassmannian acted by $\mathrm{GL}_n(\mathbb{F}_q)$
For $j\leq i$, functions on $\operatorname{Gr}_{j,n}(\mathbb F_q)$ map to functions on $\operatorname{Gr}_{i,n}(\mathbb F_q)$ by sending the delta function on a $j$-dimensional subspace to the ...
3
votes
Accepted
Finite-maximal subgroups of orthogonal groups
The group $(\mathbb{Z}/2\mathbb{Z}) \wr S_{n}$ is finite maximal for $n$ sufficiently large (probably $n>72$ will do, by a theorem of M.Collins on a sharp form of Jordan's theorem on finite complex ...
3
votes
Order of abelian subgroup of the automorphism group of an abelian group
Claim) Suppose $A$ is a finite non-trivial Abelian group. Let $B$ be an Abelian subgroup of ${\rm{Aut}}(A)$. Then, $|B|\leq |A|-1$ if $|A|$ and $|B|$ are coprime.
An elementary proof by induction that ...
3
votes
Number of conjugacy classes of pairs of commuting elements
I think it is false in general that $r_{G} \geq p^{\frac{3}{2}},$ where $p$ is the largest prime divisor of the order of $G$.
If we take a Frobenius group $G$ of order $pq,$ where $p,q$ are primes ...
8
votes
Accepted
Order of abelian subgroup of the automorphism group of an abelian group
As Nick Gill points out, one can certainly have $|B| > |A|$ if you do not assume coprimality. If you assume coprimality then yes $|B| < |A|$.
If $A_p$ is the Sylow $p$-subgroup of $A$ then $\...
2
votes
How small can maximal subgroups be?
I claim that there are finite groups whose order has arbitrarily many repeating prime factors but which contain a cyclic maximal subgroup of prime order.
Let $F$ be a finite field of order $q=p^m$. ...
1
vote
Collecting proofs that finite multiplicative subgroups of fields are cyclic
There is a quite elementary proof that uses only the fact that a polynomial of degree $n$ has at most $n$ roots in a field.
So, let $F$ be a field and $G$ be a finite subgroup of the multiplicative ...
Community wiki
18
votes
Accepted
How small can maximal subgroups be?
$A_5$ is a maximal subgroup of $\operatorname{\rm PSL}(2,p)$ when $p$ is $\pm 1$ mod $10$. Combining this with Dirichlet's theorem for primes congruent to one modulo $10$ times a highly composite ...
6
votes
Finite groups and noncommutative algebraic geometry
For question 1, I'm not sure if this is the sort of thing you're looking for, but in work I've been doing with Radha Kessar and Markus Linckelmann, non-principal blocks of a certain family of finite ...
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