3,137 reputation
11537
bio website ucs.louisiana.edu/~avm1260
location Lafayette, LA, USA
age 45
visits member for 4 years, 8 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.


2d
reviewed Approve suggested edit on Second Hardy-Littlewood Conjecture theme
Oct
5
reviewed Approve suggested edit on The Convergence of Jacobi and Gauss-Seidel Iteration
Sep
30
awarded  Explainer
Sep
8
reviewed Approve suggested edit on Amenability as a geometric property
Sep
7
reviewed Approve suggested edit on Dimension of totally reflexive modules
Sep
3
reviewed Approve suggested edit on Irreducibility of $x^m-g(y)$
Sep
1
revised Group structure on an arbitrary completely regular topological space that makes $(x,y)\mapsto xy^{-1}$ continuous at $(1,1)$
more accurate title
Aug
23
comment Who defined and who coined “module”?
I would say, in answer to your edit, that if that is your definition of "define" then Dedekind both coined and defined the term (just like he coined the term "ideal", as opposed to "ideal number"). Noether generalized, just like she generalized a lot of the theory of rings and ideals, but according to Stillwell in the work quoted by anon, she was fond of saying "Es steht schon bei Dedekind" ("It is already in Dedekind") when talking about ideals/modules.
Aug
11
comment Self-duality of the subgroup lattice of $G\times H$
I realize the observation above doesn't answer the question; but it now looks more like: if the product of two lattices is self-dual, is one of the lattices self-dual? And that does indeed look suspect.
Aug
10
comment Self-duality of the subgroup lattice of $G\times H$
The condition on the orders of $G$ and $H$ imply that every subgroup of $G\times H$ is of the form $A\times B$ with $A\leq G$ and $B\leq H$. Hence, the lattice of subgroups of $G\times H$ is equal to the product of the lattices of the subgroups of $G$ and of $H$.
Aug
8
revised Combination of two recent problems about finite groups of square orders
fix formatting
Aug
4
reviewed Reject suggested edit on Existence of solutions of a polynomial system
Aug
3
comment Classification of 2-groups with center of index 4
@StefanKohl: Just to nitpick, it's very easy to characterize all groups with center of index 2: they do not exist....
Aug
3
comment On direct product of capable groups
In general, you cannot hope for these kind of converses without conditions; just the fact that the product of two cyclic groups of the same order is always capable, but a nontrivial cyclic group is not, should tell you that this will not in general work; making one of the factors nilpotent does not, in my mind, give you enough leverage. The problems lie much deeper.
Aug
3
comment Existence a finite capable p-group of class two
I do have to ask: why would we care? i.e., where did these conditions come from, and why should we impose them on a group?
Aug
2
revised Existence a finite capable p-group of class two
deleted 1 character in body
Aug
2
revised Existence a finite capable p-group of class two
title was ungrammatical; attempt at a grammatical one.
Aug
2
answered Existence a finite capable p-group of class two
Jul
3
reviewed Reject suggested edit on nontrivial theorems with trivial proofs
Jul
2
awarded  Curious