Derek Holt
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Registered User
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14h |
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Why are Schur multipliers of finite simple groups so small? I remember asking Michael Aschbacher a similar question (I think it was actually why are the outer automorphism groups of all finite simple groups solvable) many years ago, and he opined that questions like that were pointless. Many common properties of the finite simple groups are just consequences of the classification. |
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20h |
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Are residually finite, perfect groups residually alternating? I don't think that example is perfect. |
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20h |
answered | Are residually finite, perfect groups residually alternating? |
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2d |
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How to find quotients of infinite triangle groups or von Dyck groups? Finite quotients of $G(2,3,7)$ are known as Hurwitz groups and have been much studied. It is known, for example, that the alternating group $A_n$ is a Hurwitz group for all sufficiently large $n$. |
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2d |
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A catalog of faithful representations of finite groups? To answer your final query, all of the representations in the Atlas, apart from the trivial representation, are faithful for the group that they are listed under. You could also try this webpage (which I found with a google search): maths.manchester.ac.uk/~jm/wiki/… |
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May 15 |
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A catalog of faithful representations of finite groups? It's also worth observing that the smallest dimensional faithful representation is not always irreducible (for a noncyclic abelian group for example) so, even if you know the character table, then the computations are not always completely routine. |
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May 15 |
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A catalog of faithful representations of finite groups? Perhaps you need to specify more precisely exactly what information you are looking for, and for which which groups. |
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May 15 |
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A catalog of faithful representations of finite groups? But the Atlas of Finite Groups only has information about groups that are "close" to being nonabelian simple. If all of the "noteworthy finite groups" have this property, then fine, but it would be no use at all for the many other types of interesting finite groups. |
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May 13 |
awarded | ● Nice Answer |
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May 13 |
accepted | about the non-solvable group of order $120$ |
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May 13 |
answered | about the non-solvable group of order $120$ |
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May 11 |
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The role of the Automatic Groups in the history of Geometric Group Theory I am not aware of any specific examples at present. Many such examples in the past have turned out not to work, which may explain why people have given up! |
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May 11 |
answered | The role of the Automatic Groups in the history of Geometric Group Theory |
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May 9 |
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Generators of sections of free groups These numbers can certainly depend on the isomorphism type of $F/H$ (and not just on its order). |
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May 9 |
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Generators of sections of free groups The answer is yes, in the sense that there are algorithms to solve these problems, and it would not be particularly difficult to write programs in a language like GAP or Magma to do so. Is this what you are looking for, or is this more of a theoretical question? |
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May 7 |
awarded | ● Nice Answer |
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May 2 |
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Extensions with trivial induced outer action The assumptions are equivalent to $E = NC_E(N)$. |
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Apr 28 |
revised |
element of order n such that $\pi(n)=\pi(G)$, where $\pi(n)$ denote the prime divisors of $n$ Added group-theory tag |
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Apr 28 |
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element of order n such that $\pi(n)=\pi(G)$, where $\pi(n)$ denote the prime divisors of $n$ You can easily constuct non-nilpotent examples by taking a direct product of an arbitrary group with a cyclic group of the required order. But I would be surprised if there were any nonabelian simple groups with this property. |
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Apr 26 |
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Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators? @Alexey: Quantifiers! 1 says every geodesic, where 2 says there exists a geodesic. The subgroup $\langle xy \rangle$ of the free abelian group with free generators $x,y$ is qc using 2 but not using 1. |
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Apr 25 |
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Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators? Definition 3 is interesting when the combing is regular and part of an automatic structure of $G$. In that case, the combing words that lie in $H$ form a regular set, and it is often not difficult, starting from the automatic structure of $G$, to compute a finite state automaton to recognise membership of combing words $H$. Handling intersections of such subgroups can then be done by standard operations on finite state automata. |
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Apr 18 |
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Normal subgroups of finite index in free groups I am afraid that misread the definition of $H_{n,s}$ in your question. I took the definition to be $F_s/N$ where $N$ is the intersection of all normal subgroups of index exactly $n$ (rather than at most $n$, which is what you wrote). So Khalid Bou-Rabee's answer is more useful than mine for the question that you asked! |
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Apr 18 |
answered | Normal subgroups of finite index in free groups |
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Apr 18 |
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2-sylow subgroups You take a set of permutations that generate a Sylow $2$-subgroup of the symmetric group $S_{2^{r-1}}$.Then make them into permutation matrices of degree $2^r$ that permute $2 \times 2$ block matrices. These permutation matrices, together with a Sylow $2$−subgroup of ${\rm GL}_2(q)$ acting on
one of the $2 \times 2$ blocks, generate a Sylow $2$−subgroup of
${\rm GL}_{2^r}(q)$.
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Apr 17 |
accepted | Classification of Special $p$-Groups |
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Apr 17 |
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Classification of Special $p$-Groups I haven't heard of such a conjecture. Wuth current techniques it seems that there will always be a nonconstant error term in the exponent. So we cannot really justify conjecturing that "almost all $p$-groups are special", although that statement is somehow true logarithmically. |
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Apr 17 |
answered | Classification of Special $p$-Groups |
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Apr 13 |
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wreath product and matrix presentation I am not convinced that $H$ is uniquely defined. The wreath product is an associative operation on permutation groups, but not on abstract groups, and $Z_2$ looks like an abstract group rather than a permutatino group. |
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Apr 12 |
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The automorphisms of a 2-group of nilpotency class 2 I'm impressed! I don't suppose many of us were knowledgeable enough to make that mistake! |
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Apr 11 |
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The automorphisms of a 2-group of nilpotency class 2 In any nilpotent group of class 2, we have $[ab,c]=[a,c][b,c]$ and $[a,b]^{-1} =[b,a]$, so the commutator map is bilinear and alternating, and we get an induced map $G/Z(G) \times G/Z(G) \to Z(G)$. If $|Z(G)|=2$, then $[a^2,b]=1$ for all $a,b$, so $a^2 \in Z(G)$ and $G/Z(G)$ is elementary abelian. So $G$ is extraspecial. |
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Apr 11 |
accepted | The automorphisms of a 2-group of nilpotency class 2 |
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Apr 11 |
answered | The automorphisms of a 2-group of nilpotency class 2 |
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Apr 10 |
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When is Ad(pi) an irreducible representation ? No. It is a subgroup of ${\rm GL}_3({\mathbb R})$, which comes from the adjoint of the 2-dimensional representation of $2.A_5$. |
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Apr 10 |
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When is Ad(pi) an irreducible representation ? I think you want $G=2.A_5 = {\rm SL}_2(5)$ rather than $A_5$. |
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Apr 10 |
answered | When is Ad(pi) an irreducible representation ? |
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Apr 10 |
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Finite subgroups of $PGL(3,K)$ Yes you are right, I had miscalculated the character field! |
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Apr 9 |
answered | Finite subgroups of $PGL(3,K)$ |
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Apr 6 |
accepted | Nilpotency class of a certain finite 2-group |
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Apr 1 |
awarded | ● Enlightened |
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Apr 1 |
accepted | Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$? |
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Mar 31 |
awarded | ● Nice Answer |
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Mar 31 |
answered | Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$? |
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Mar 30 |
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Automorphism classes of the free group You can do this using the Whitehead Algorithm. More generally, it can be decided whether there is an automorphism mapping one $k$-tuple of elements onto another. |
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Mar 29 |
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P-group with abelian centralzer I understand what $cs(G)$ means but I do not understand "has at least three integer". Do you just mean $|cs(G)| \ge 3$. If so, why not write that? |
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Mar 29 |
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P-group with abelian centralzer It is not clear what you mean by "$cs(G)$ has at least three integer". |
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Mar 22 |
awarded | ● Enlightened |
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Mar 22 |
awarded | ● Nice Answer |
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Mar 20 |
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Are all dinilpotent groups solvable, i.e., groups G=AB with nilpotent subgroups A, B ? As far as I know, this is still an open question. Why do you think that there is a counterexample? |
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Mar 17 |
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Subgroups of the general linear groups In general ${\rm GL}_{2n}(q)$ has an abelian subgroup of order $(q-1)q^{n^2}$. |
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Mar 12 |
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What are the known algorithms for computing the inverse of a group automorphism? It's an interesting question for free groups. When the number of generators i not too big (up to about 8), the Whitehead algorithm is quick, but it is not effective when there are more generators. I suspect that the modified coset enumeration procedure might be better. |
