bio  website  ucs.louisiana.edu/~avm1260 

location  Lafayette, LA, USA  
age  44  
visits  member for  4 years, 2 months 
seen  1 hour ago  
stats  profile views  2,739 
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.
1h

reviewed  Reject suggested edit on Distance between poisson points in two disjoint unit discs 
1h

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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

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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

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$p$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

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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

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SHPS and SPHS inequality using monounary algebra
There was an arithmetical error following the sentence you ask about; I've fixed the argument. 
2d

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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 $(pk+2,1)\in\Phi$, and $(1,k)\in\Phi$, then we must have $(k,pk+2)\in\Phi$ by symmetry and transitivity. I applied $f$ to get $(pk+2,1)$. 2. I showed how: mod out by the congruence $\Phi$ defined in the last paragraph. 
Apr 15 
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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 
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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 
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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 
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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 
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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 
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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 
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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 
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H S class operator and its equality
Is $Z$ the integers, or an arbitrary set with a "successor" function? 
Apr 14 
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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 