bio | website | ucs.louisiana.edu/~avm1260 |
---|---|---|
location | Lafayette, LA, USA | |
age | 45 | |
visits | member for | 5 years, 3 months |
seen | 46 mins ago | |
stats | profile views | 3,261 |
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.
May 21 |
comment |
Primitive elements in a free group
I don't think you are asking what you mean to ask, because as written, you can just take $B$ to be an arbitrary free basis, and $B_0=B$, and $K=\aleph_0$. I suspect you want $B_0$ finite, for one thing... |
May 21 |
revised |
nilpotent of class 2 free product
add comment on general verbal product; add ciitation |
May 21 |
answered | nilpotent of class 2 free product |
May 14 |
comment |
non-intersecting families of subspaces
If $r$ does not divide $m$, then $(k^m-1)/(k^r-1)$ is not an integer (e.g., $k=5$, $m=3$, $r=2$, you get $124/24$), so what would "whose size is equal to..." mean in that case? (P.S., is "size" the cardinality, or the dimension?) |
May 12 |
reviewed | Approve Unordered configuration space of $\mathbb{R}P^1$ |
May 11 |
awarded | Quorum |
May 5 |
reviewed | Approve multidimensional curve fitting (regression) |
Apr 23 |
reviewed | Approve Singular projective variety where the Cartan homomorphism is not an isomorphism? |
Apr 20 |
awarded | Enlightened |
Apr 20 |
awarded | Nice Answer |
Apr 19 |
revised |
amalgamation of locally finite groups
LaTeXifying |
Apr 19 |
comment |
Generating finite groups using subgroups
Of course, this is just the wreath product $K\wr C_2$; you can see examples with arbitrary index for $G$ by taking $K\wr C_n$ instead. |
Apr 18 |
revised |
Generating finite groups using subgroups
clarfiy phrasing |
Apr 18 |
answered | Generating finite groups using subgroups |
Apr 18 |
comment |
Generating finite groups using subgroups
right: it's just that as you phrased it, the subject of the sentence is "$H\leq G_0\leq G$", for which "is a subgroup" makes no sense. |
Apr 18 |
comment |
Generating finite groups using subgroups
"Suppose that $H\leq G_0\leq G$ is a subgroup of index $2$..." You mean "suppose that $G_0$, $H\leq G_0\leq G$, is a subgroup of order $2$", right? |
Apr 13 |
reviewed | Approve Understanding Faltings's Theorem |
Mar 31 |
reviewed | Approve homogeneous polynomials over a finite field |
Mar 30 |
reviewed | Approve (Non)existence of mirrors with more than two foci |
Mar 27 |
revised |
One question about iteration on groups
readability, notation, grammar |