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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 ...
LSpice's user avatar
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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 ...
Alexander Chervov's user avatar
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 ...
Will Sawin's user avatar
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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 ...
Geoff Robinson's user avatar
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 ...
KhashF's user avatar
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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 ...
Geoff Robinson's user avatar
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 $\...
Sean Eberhard's user avatar
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 $q=p^m$. ...
Joseph Van Name's user avatar
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 ...
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 ...
Dave Benson's user avatar
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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 ...
Dave Benson's user avatar
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