19 votes
Accepted

Is every finite $d$-dimensional matrix group generated by $d$ elements?

The answer to the question is yes. In Theorem 1.2 of this paper. the authors (Colva Roney-Dougal and myself) prove that if $G \le {\rm GL}(n,F)$ with $F$ a field, where (a) $G$ is finite; (b) either $...
Derek Holt's user avatar
  • 36.4k
17 votes
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$\operatorname{Out}(F_n)$ is not linear for $n > 3$

There is an embedding $\text{Aut}(F_{n-1}) \hookrightarrow \text{Out}(F_n)$ for any $n \ge 2$, as follows. First one embeds $\text{Aut}(F_{n-1}) \hookrightarrow \text{Aut}(F_n)$ by extending any ...
Lee Mosher's user avatar
  • 15.3k
14 votes
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Transitive actions of finite subgroups of ${\rm GL}(n,\Bbb Z)$ on projective geometries

Probably final revision: I am indebted to Dave Witte-Morris, who added a reference to a refinement of Zsigmondy's Theorem by W. Feit, of which I was unaware, and pointed out that consequently, a ...
Geoff Robinson's user avatar
12 votes
Accepted

Sequence of epimorphisms of residually finite groups stabilizes

The answer is "no". The lamplighter group (which is infinitely presented) is a limit of a sequence of virtually free groups and surjective homomorphisms (see, for example, this question and ...
markvs's user avatar
  • 1,802
11 votes
Accepted

If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, must the fixed points of $1_\mathbb{Z}$ be a finitely generated group?

No. Fix $p\ge 2$. Take the group $$G=\{M(x,y,z;n):(x,y,z)\in\mathbf{Z}[1/p],n\in\mathbf{Z}\}$$where $$M(x,y,z;n)=\begin{pmatrix}1 & x & z \\ 0 & p^n & y\\ 0 & 0 & 1\end{pmatrix}...
YCor's user avatar
  • 60.1k
10 votes
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How can I detect the homology image of a unipotent group in the general linear group?

Suppose first that $F$ is a finite field of characteristic $p$. Then $U_n(F)$ is a Sylow $p$-subgroup of $GL_n(F)$, and so using the transfer in group homology one sees that the image of $f_k$ (for $k&...
Oscar Randal-Williams's user avatar
9 votes
Accepted

Double cosets of $U(n)\times U(n)$ in $U(2n)$

$K=U(n)\times U(n)$ is a symmetric subgroup of $G=U(2n)$. There is a discussion of $K$-double cosets in $G$ for any compact symmetric space in Helgason's book "Differential geometry, $\ldots$". See ...
Friedrich Knop's user avatar
9 votes

distribution of Young diagrams

For more details on this answer, together with two proofs of the following theorem, see arXiv:1705.07604 preprint ``External powers of tensor products as representations of general linear groups'' by ...
Piotr Śniady's user avatar
8 votes

Non-torsion part of the abelianisation of congruence subgroups

Suppose $r \geq 3$. Then one can show that $\Gamma=\mathrm{SL}(r,A)$ is a lattice in the group $G=\mathrm{SL}\big(r, {\mathbb F}_q(\!(1/t)\!)\big)$. The group $G$ has Kazhdan's property T and hence $\...
Venkataramana's user avatar
7 votes
Accepted

Generators of $SL(n,\mathbb F_2)$?

OK, I'll answer the intended question! ${\rm SL}(2,2) \cong S_3$ and the answer to the question is no in that case, so assume that $n>2$. Don Taylor wrote down explicit sets of two generators for ...
Derek Holt's user avatar
  • 36.4k
7 votes

Sequence of epimorphisms of residually finite groups stabilizes

In the same vein as dodd's answer, a counterexample can also be deduced from the second Houghton group $H_2$, which is defined as the group of bijections $L^{(0)} \to L^{(0)}$ that preserves adjacency ...
AGenevois's user avatar
  • 7,481
7 votes

Automorphisms of products of $GL_n(\mathbb{Z})$ 's

Here's a proof of finiteness. First, it's a general fact that for any centerless directly indecomposable groups $A_1,\dots,A_k$, any automorphism permutes the $A_i$. References are welcome; one (in ...
YCor's user avatar
  • 60.1k
7 votes

If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, must the fixed points of $1_\mathbb{Z}$ be a finitely generated group?

Here is another example, which I am fond of, because it played a role in a PhD thesis of a student that I supervised a long time ago. It is constructed as a central extension of $C_p \wr {\mathbb Z}$ ...
Derek Holt's user avatar
  • 36.4k
3 votes

Proving certain triangle groups are infinite

Here is an old paper that gives one answer to your question about "other explicit infinite families": G. A. Miller, Groups Defined by the Orders of Two Generators and the Order of their ...
Mikko Korhonen's user avatar
3 votes
Accepted

Non-torsion part of the abelianisation of congruence subgroups

In a more general setting, let $A$ be a commutative ring, $I, J$ its ideals, $n\geqslant3$, then $$[E(n,A,I),E(n,A,J)]\geqslant E(n,R,IJ),$$ where $E(n,A,I)$ is the normal closure in $E(n,A)$ of the ...
Andrei Smolensky's user avatar
2 votes
Accepted

Intransitive finite irreducible linear groups whose orbits are all large

Take an odd prime $p$, let $q=p^n$ and consider $V=\mathbb{F}_q$ as a vector space over $\mathbb{F}_p$. For $G_n=(\mathbb{F}_q^*)^2$ you get $c=1/2$.
Uri Bader's user avatar
  • 11.4k
2 votes
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Relation to the Bruhat cell

This is easily seen when working with matrices, so we'll work with matrices throughout. Thus, $B$ is identified with the set of upper triangular matrices, and every element $x\in \mathrm{SL}_n(\mathbb ...
kneidell's user avatar
  • 993
2 votes
Accepted

If $\Lambda \cap U$ is Zariski-dense in $U$, then $\Lambda$ contains $U(k\mathbb Z)$ for some $k ≥ 1$?

In this answer i will follow Yves' comment and add references. If $U = \mathbf{U}(R)$ with $\mathbf U$ an algebraic unipotent $\mathbb Q$-group then the two following facts hold : If $\Lambda \le U$ ...
Jean Raimbault's user avatar
2 votes
Accepted

Upper bounds for difference of entries between matrices and their inverses in $\mathsf{GL}_k(\mathbb Z)$

For $k=2$, the upper bound is zero. For $k>2$, there is no upper bound. E.g., let $$M=\pmatrix{1&1&1\cr9&10&11\cr-n&n&3n-1\cr}$$ Then $$M^{-1}=\pmatrix{10-19n&2n-1&...
Gerry Myerson's user avatar
1 vote
Accepted

Is the Singer cycle preserved by field automorphisms and graph automorphisms?

This is true by Proposition 4.3.6.(I) of Kleidman and Liebeck's book "The Subgroup Structure of the Finite Classical Groups", which says that, in all cases for the linear and unitary groups, there is ...
Derek Holt's user avatar
  • 36.4k
1 vote

classify antiholomorphic involutions of projective space

Coda (Lemma 11 from arXiv:2002.01355): Projective transformations of $\mathbb{C}P^{n}$ act on the solutions of the equation $L\bar L=I$, respectively $L\bar L=-I$, by transformations $L\mapsto U^{-1}...
Mikhail Skopenkov's user avatar

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