3,787 reputation
12042
bio website ucs.louisiana.edu/~avm1260
location Lafayette, LA, USA
age 46
visits member for 5 years, 5 months
seen 3 hours ago

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 low-level) 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 e-mail 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 e-mail, 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 $x^3-y^2=2$
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 2-generated group (and I would guess one should be able to extend to proving that every finite group can be embedded into a finite 2-generated 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), 399-400, 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 all-LaTeX title. On the other hand, ASCII art and pseudo-latex ( k^x )seem like a bad idea when you are talking about such a small part of the title. This is not a LaTeX-heavy 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.