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7h
comment What do named “tricks” share?
Another possible entry in the list : the Frattini argument. I thought of it when I read Qiaochu's answer (which I thought was spot on) -- it's a very handy little, um, thingy to know when working with finite groups.
7h
comment What do named “tricks” share?
Just to add to your list -- one of the commenters above has a trick named after him :-) Nikolov and Pyber used a result of Gowers about solving equations in groups to prove a nice fact about the multiplication of (large) sets in groups. The method they used is known as the Gowers trick. I believe it was Pyber who came up with the name -- you can see the name in use in the literature in several places, for instance in Breuillard's Introduction to approximate groups.
1d
comment Sets of matrices which are irreducible but not strongly irreducible
@BenjaminSteinberg, could you expand on your comment a little -- I don't quite understand it. Nice question by the way, Ian.
Feb
4
awarded  Nice Answer
Jan
14
comment long root elements fixed by an automorphism in simple lie type group
Ah, sorry, I misread your question -- you just want to fix elements in a specific long root subgroup. Well, this will just require that you calculate the centralizer (in ${\rm Aut}(G)$) of all the elements in that group, and take the intersection. This seems pretty straightforward.
Jan
14
comment long root elements fixed by an automorphism in simple lie type group
I guess you're wanting to fix some split maximal torus before you start defining root subgroups. After that do you want your automorphism to centralize every long root subgroup, or to normalize every long root subgroup? The former certainly can't be done in the case where there is only one length root (e.g ${\rm PSL_n(q)}$), as the long root subgroups generate the simple group. In the other cases (normalizing and/or more than one root length), I'd have to think a little...
Jan
6
answered embedding of $O_4^-(q)$ in $U_4(q)$
Jan
3
comment embedding of $O_4^-(q)$ in $U_4(q)$
(And, as Derek says, the definitive source on this is Kleidman and Liebeck, Section 4.3.)
Jan
3
comment embedding of $O_4^-(q)$ in $U_4(q)$
Given a $3$-dimensional quadratic form $Q$ over $\mathbb{F}_{q^2}$, any $\mathbb{F}_q$-linear function $L: \mathbb{F}_{q^2}\to \mathbb{F}_q$ will yield a $6$-dimensional quadratic form $L\circ Q$ over $\mathbb{F}_q$ (i.e. you don't need to use the trace form). What is more, you can get all possible $L$ from the trace by adding in some constant, i.e. $L(x)=L(\mu\cdot x)$ for some $\mu \in \mathbb{F}_{q^2}$. If you choose your $\mu$ carefully, the resulting linear form $L$ will give a minus type form $LQ$ for $q\equiv 3\pmod 4$.
Dec
15
answered Dessins d'enfants and absolute Galois group
Dec
15
awarded  Enlightened
Dec
15
awarded  Nice Answer
Dec
8
comment 3-term arithmetic progressions of terms as frequent as primes
Ah, yes, now it's better. I've suffered the curse of the typo many times...
Dec
8
comment 3-term arithmetic progressions of terms as frequent as primes
Perhaps I don't understand, but what is wrong with the sequence $1,2,4,5,7,8,10,11,13,14,\dots$? Isn't this an example of the sort you need that does not contain 3 consecutive members in AP?
Dec
4
comment Maximal torus of Chevalley group $Sp(4)$
I don't quite follow your calculations, but perhaps the problem is that a split torus in $Sp_4(K)$ need not consist of diagonal matrices, it just needs to be conjugate to a set of diagonal matrices. Perhaps you've just chosen a basis for which the $h_\alpha(t)$ are not diagonal? (My guess for how to fix this would be to have $x_\alpha(t)=1+t(E_{12}-E_{34})$ -- i.e. swap the last two vectors in your basis -- but I don't have the wherewithal to do the calculation right now.)
Dec
2
comment Dessins d'enfants and absolute Galois group
Yes! It certainly can't be surjective. It must lie in the Grothendieck-Teichmuller group, $\widehat{GT}$, a proper subgroup of ${\rm Out}(\hat{F_2})$. Guillot discusses this group at the end of his monograph.
Dec
2
comment Dessins d'enfants and absolute Galois group
I strongly recommend this paper by Guillot: arxiv.org/pdf/1309.1968v2.pdf. It develops the theory of dessins from scratch; of course, the faithful action of ${\rm Gal}(\overline{\mathbb Q})/{\mathbb Q})$ on the set of dessins yields the homomorphism you mention. Guillot also proves the remarkable fact that the action of the absolute Galois group on the subset of regular dessins is also faithful, a result that was also proved by Jaikin-Zapirain and Gonzalez-Diaz. Guillot has a follow-up article dealing with explicit computations pertaining to this homomorphism.
Nov
25
comment Strongly real elements of odd order in sporadic finite simple groups
Nice question. Gow has a result in the spirit of your final paragraph: If the Sylow 2-subgroups of $G$ are dihedral or large enough semi-dihedral, then a quaternionic representation exists iff a non-strongly real but real element of odd order exists. See his 1979 J.Algebra paper. (Although I imagine you know of this already...)
Nov
24
answered number of maximal subgroups of the symmetric group
Nov
20
comment When is an erratum necessary?
Jordan Ellenberg wrote a fascinating blog post about an error that was found in one of his preprints: quomodocumque.wordpress.com/2013/11/23/… He was rightly commended by many of his readers for the way he dealt with this error. The post, and the comments that follow, is well worth a read.