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bio website boolesrings.org/nickgill
location San Jose, Costa Rica
age 37
visits member for 5 years, 1 month
seen 10 hours ago

I'm a visiting professor at the Universidad de Costa Rica.


11h
comment abelian p- subgroups of E_6(q)
Jim, one query. I don't have GLS3 in front of me but, from memory, their results describe the *maximal possible rank of an elementary abelian $p$-subgroup, as opposed to the possible ranks of a maximal elementary abelian $p$-subgroup. (In theory there could be different maximal elementary abelian $p$-subgroups with different ranks, couldn't there?)
1d
comment Can we say that $A$ is a complement for a group $G$?
Thanks, Geoff, that is very interesting.
1d
comment Finding an overgroup or a subgroup in PGL
Tomo, I think $H'$ does have an infinite number of overgroups - you just need to caclulate its normalizer to see this... But I haven't time to do the details... If you restrict to connected overgroups then the answer might be different...
1d
comment abelian p- subgroups of E_6(q)
You can use Chevalley's commutator formula to figure out which of the root groups commute - it's conceivable I suppose that all maximal abelian unipotents are conjugates of these.... Also, the abelian unipotent subgroups of maximal order were classified by Mal'cev I believe. See the references in this paper: math.nsc.ru/~vdovin/evdavg.ps
1d
comment Can we say that $A$ is a complement for a group $G$?
Geoff, my profound ignorance of this type of thing prompts the following questions: Is it conceivable that this is the only counter-example? Could one imagine a classification of characters of this form? Would such a classification be interesting?!
1d
comment On the size of centralizers in a non-abelian finite simple group
Finally: regarding my comments about asymptotics above. I think $\varepsilon$ would need to depend on the rank $n$ in the Lie type case, so my guess is not true in general (take $G=PSL_2(q)$ and let $q\to\infty$ for instance).
1d
comment On the size of centralizers in a non-abelian finite simple group
An aside: Marty Isaacs has proved that if $G$ is solvable, then there is a $g$ such that $|G|<|C_G(g)|^2$. The relevant paper is: Isaacs, I. M. Solvable groups contain large centralizers. Israel J. Math. 55 (1986), no. 1, 58–64.
1d
comment On the size of centralizers in a non-abelian finite simple group
It seems pretty clear that the claim is true provided $|G|$ is large enough. For alternating groups, take $g$ a $3$-cycle, for groups of Lie type of rank $n$ (most of the time you could) take $g$ central in a Levi subgroup of rank $n-1$. Modulo a load of details I would think that asymptotically you could get $|G|<|C_G(g)|^{1+\varepsilon}$ for any $\varepsilon>0$.
Nov
25
comment Finding an overgroup or a subgroup in PGL
For an overgroup take the normalizer of the connected subgroup of dimension $1$ that I describe above.
Nov
24
comment Finding an overgroup or a subgroup in PGL
It might be worth mentioning in the question that $H$ is abelian in general.
Nov
24
comment Finding an overgroup or a subgroup in PGL
OK, for a connected subgroup of dimension 1 just take $y=w=0$. The resulting set is is an abelian unipotent subgroup (in particular it has exponent $2$). In fact, if I am not mistaken, over $\overline{k}$ it looks like a root group for $PSp_4(k)$. (Although the question over $\overline{k}$ needs clarification because in this context one cannot choose $a$ to be a non-square.)
Nov
21
comment Finding an overgroup or a subgroup in PGL
Being pedantic: I guess the trivial subgroup answers the second part of your question.
Nov
21
comment Finding an overgroup or a subgroup in PGL
Do $w$ and $z$ vary across $k$ too?
Nov
20
comment Rational conjugation of elements of a finite group
@FriederLadisch, make this an answer!
Nov
20
awarded  Nice Answer
Nov
13
comment Vertex-primitive graphs with two vertices having almost the same neighbourhood
@BrendanMcKay Thanks. I get it now!
Nov
12
comment Vertex-primitive graphs with two vertices having almost the same neighbourhood
Sorry for the dumb question about 2-transitivity. Re the other question: I don't immediately see why $(uv)$ is in the group. Am I being dumb again?
Nov
11
comment A lemma on verbal conjugacy classes in groups
That each conjugate of $H$ has finite index is more-or-less what you show in the fourth line of the proof, I think. The only caveat is that that line concerns elements of $G_w$ so one needs to iterate to get the same result for elements of $w(G)$. Since $E$ contains $w(G)$ you are done.
Nov
11
comment Vertex-primitive graphs with two vertices having almost the same neighbourhood
Two questions about what you know already: (1) is the statement true if $G$ is vertex 2-transitive? (2) it seems like whether $v$ is in $N(u)$ or not naturally splits the problem into cases - has the statement been proved in either of these?
Nov
11
comment Vertex-primitive graphs with two vertices having almost the same neighbourhood
+1 - great question. Except that when I saw the title, I thought "oh, if anyone can answer that it will be verret..." :-)