3,007 reputation
11535
bio website ucs.louisiana.edu/~avm1260
location Lafayette, LA, USA
age 44
visits member for 4 years, 2 months
seen 1 hour 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.


1h
reviewed Reject suggested edit on Distance between poisson points in two disjoint unit discs
1h
comment Bound for the Frattini subgroup of a $p$-group
Yes, there are $p$-groups that achieve the bound for exponent $p$; namely, the relatively free groups of rank $n$, class $2$, and exponent $p$ have commutator subgroup that is free abelian of rank $\binom{n}{2}$; this group can be realized as $F_n/F_n^p(F_n)_3$, where $F_n$ is the absolutely free group of rank $n$, and $(F_n)_3$ is the third term of the lower central series of $F_n$.
1d
comment SHPS and SPHS inequality using monounary algebra
@GerhardPaseman: Slight correction: I am not currently participating in math.SE, and have not for quite a while.
2d
comment $p$-groups with $\Omega_1(G)\leq\Phi(G)$
$\Phi(G)$ is the Frattini subgroup, which in this context equals $G^pG'$. $\Omega_{\{a\}}(G) = \{g\in G\mid g^{p^a}=1\}$, and $\Omega_a(G) = \langle \Omega_{\{a\}}\rangle$. That said, is this question too open ended?
2d
comment SHPS and SPHS inequality using monounary algebra
@AnuragSharma: I flagged the questions and suggested migrating them to SE.
2d
revised SHPS and SPHS inequality using monounary algebra
Fixed arithmetical error by replacing argument somewhat.
2d
comment SHPS and SPHS inequality using monounary algebra
There was an arithmetical error following the sentence you ask about; I've fixed the argument.
2d
comment SHPS and SPHS inequality using monounary algebra
1. $\Phi$ is a congruence, hence it is both a subalgebra of $A_p\times A_p$, and an equivalence relation; if $(p-k+2,1)\in\Phi$, and $(1,k)\in\Phi$, then we must have $(k,p-k+2)\in\Phi$ by symmetry and transitivity. I applied $f$ to get $(p-k+2,1)$. 2. I showed how: mod out by the congruence $\Phi$ defined in the last paragraph.
Apr
15
comment SHPS and SPHS inequality using monounary algebra
@Gerhard: Fair enough on your final comment; but I'm not sure I understand the first part.
Apr
15
comment SHPS and SPHS inequality using monounary algebra
I know you've said that you don't think you'll get good answers in math.SE. Have you actually tried? These are reasonably basic, though cast in language that a lot of people are not familiar with. It seems the consensus is that you should be trying there (possibly making the questions a bit more accessible if you are afraid people will not be familiar with the concepts on a cold reading).
Apr
15
answered SHPS and SPHS inequality using monounary algebra
Apr
15
comment H S class operator and its equality
I have no idea what you mean by "structure wise". Clearly, there are homomorphic images of $(\mathbb{Z},s)$ that cannot be realized as (isomorphic to) homomorphic images of $(\mathbb{N},s)$, so that means that $\mathbf{H}(\mathbb{Z},s)$ cannot be equal to $\mathbf{H}(\mathbb{N},s)$, whatever "structurewise" is supposed to mean.
Apr
15
comment SHPS and SPHS inequality using monounary algebra
No; S is not just the class of subalgebras, it's the class of algebras that are isomorphic to some subalgebra; likewise, P is the class of algebras that are isomorphic to a product. So "isomorphic but different" does not suffice.
Apr
15
comment SHPS and SPHS inequality using monounary algebra
The only subalgebras of $A_n$ are $\varnothing$ and $A_n$ itself. The empty set does not contribute anything, so you are really down to showing $SHP\neq SPH$ for the class $R$.
Apr
15
comment H S class operator and its equality
@AnuragSharma: For one thing, $(\mathbb{N},s)$ does not have $\mathbb{Z}$ as an image, but $(\mathbb{Z},s)$ certainly does...
Apr
15
comment H S class operator and its equality
So then... does the answer I give below settle it?
Apr
14
answered H S class operator and its equality
Apr
14
comment H S class operator and its equality
Is $Z$ the integers, or an arbitrary set with a "successor" function?
Apr
14
comment Simple groups and words
If you let $\mathfrak{V}$ be the variety generated by $S$, then any word in $\mathfrak{V}(F)$ would fail the property; and as Derek Holt's answer shows, the subgroup $\mathfrak{V}(F)$ does not consist only of power words.
Apr
2
reviewed Approve suggested edit on explicit characterization of the stochastic integrand