bio  website  ucs.louisiana.edu/~avm1260 

location  Lafayette, LA, USA  
age  46  
visits  member for  5 years, 5 months 
seen  3 hours ago  
stats  profile views  3,351 
With the move of MathOverflow into the SE network, this account is now associated with dormant accounts in math.SE and other sites in the network. While I plan to continue my (generally lowlevel) participation in MO, my current plans do not include restarting my participation in those other sites. Therefore, I will be ignoring any comments or pings that reach me from those sites, unless and until I resume my active participation there.
I remain "gone for the foreseeable future" from math.SE, tex.SE, and meta.SE.
Please do not send me private email to call my attention to comments, questions, or other matters related to those sites. Thank you. Also, as I no longer participate in those sites, I do not wish to be sent, by private email, questions that you can just as well ask on those sites. I would have thought it was obvious, but apparently I need to say so explicitly.
2d

comment 
When did people know that all real polynomials of degree greater than 2 are reducible?
As I understand it, one of the reasons proving the FTA was important was to ensure that partial fractions would always work, at least in principle; I do not believe that it was a known fact before then. 
Jul 29 
reviewed  Approve Fermat's proof for $x^3y^2=2$ 
Jul 28 
comment 
Is there a structure theorem or group law for finite groups generated by two elements?
@shane.orourke: D'oh. Of course. That's what I get for posting past 11pm... 
Jul 28 
comment 
Is there a structure theorem or group law for finite groups generated by two elements?
@TT_: As you say, it's not going to make things any easier; it still tells you that every finite group is embeddable into a 2generated group (and I would guess one should be able to extend to proving that every finite group can be embedded into a finite 2generated group). 
Jul 28 
awarded  Enlightened 
Jul 27 
awarded  Nice Answer 
Jul 27 
answered  Is there a structure theorem or group law for finite groups generated by two elements? 
Jul 20 
awarded  Informed 
Jul 20 
comment 
Why can't a nonabelian group be 75% abelian?
On the other hand, note that the probability that two semigroup elements commute can be any rational number, as shown by Givens and by Ponomarenko and Selinski (Givens, B. The probability that two semigroup elements commute can be almost anything, College Math J. 39 (5), 399400, 2008; and also the paper by Michelle Soule). 
Jul 9 
awarded  Guru 
Jun 20 
revised 
Representation of finite group
add tags 
Jun 12 
reviewed  Reject Is regularity closed under products? 
Jun 7 
revised 
Min number of primes up to n
fix some latex 
Jun 4 
comment 
For a ring $k$ and a set $X$, what are the $k$algebra homomorphisms $k^X \to k$?
@TomLeinster: Yes, you should keep LaTeX to a minimum in titles, and you definitely don't want to have an allLaTeX title. On the other hand, ASCII art and pseudolatex ( k^x )seem like a bad idea when you are talking about such a small part of the title. This is not a LaTeXheavy title... 
Jun 4 
revised 
For a ring $k$ and a set $X$, what are the $k$algebra homomorphisms $k^X \to k$?
latex title 
Jun 1 
reviewed  Edit Quintic Equation 
Jun 1 
revised 
Quintic Equation
format edited 
Jun 1 
comment 
Quotients of finitely generated nilpotent groups
@YCor: I think yo umean "torsionfree" where you say "finite", but yes, I see there would be issues because the factors for $H/N_3$ (e.g., $(H/N_3)^{\rm ab}$) could be strictly proper factors of the corresponding ones for $H$, so we are not sure that $H$ itself has torsionfree factors. Thanks. 
Jun 1 
comment 
Quotients of finitely generated nilpotent groups
@YCor: Wouldn't we be able to do the same thing in general by induction on the class? First, go to a torsionfree subgroup of finite index $N$, so we may assume $N$ Is torsionfree. Then find a subgroup $H$ of finite index that contains $N_{k+1}$ such that $H/N_{k+1}$ has the desired property. But since $N_{k+1}\subseteq Z(N)\cap H\subseteq Z(H)$, then the commutator subgroup of $H/N_{k+1}$ is "essentially" the same as that of $H$ because the center is marginal in the commutator bracket, so the fact that $H/N_{k+1}$ has torsionfree quotients shoudl give $H$ does as well... 
May 31 
comment 
Quotients of finitely generated nilpotent groups
@YCor: Thanks! I confess to not knowing it before. 