I like Shakespeare and Greek tragedy, so let me word it as I'm doing: I desperately need J.H.B. Kemperman's 1956 paper On complexes in a semigroup, but the online archive of Indagationes Mathematicae, where it was originally published (Vol. 18, pp. 247-254), goes back until 1990. Are you aware of any comparatively recent reprint or something like that? Thank you much in advance for any help you can provide.
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asked Oct 17, 2012 at 15:19
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$\begingroup$ This is perhaps a stupid comment, but in case the info on your location on the userpage is current, it should not be hard to find a hardcopy in a library near you. $\endgroup$– user9072Commented Oct 17, 2012 at 17:52
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$\begingroup$ Indeed, I tried. But I couldn't find it in the online catalogs, and I really hoped that someone could have a hardcopy in her drawer... $\endgroup$– Salvo TringaliCommented Oct 17, 2012 at 20:53
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1$\begingroup$ Quite possibly you know in more detail than I and there are problems in detail, but sudoc.fr/038722348 and then 'où trouver ce document' suggests it is available at P6 Math-Info (except you should have the bad luck to fall into the unspecfied gaps) and at P6 Stats and at ENS, (and at P11). $\endgroup$– user9072Commented Oct 17, 2012 at 21:26
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$\begingroup$ Thank you, quid. I will try at the ENS (somehow, it didn't appear among the results of my search). $\endgroup$– Salvo TringaliCommented Oct 17, 2012 at 22:30
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2$\begingroup$ I do have it in the drawer in my office. If you have not found it yet, please, let me know, and I will digitize it and send you (by e-mail?) $\endgroup$– SevaCommented Oct 19, 2012 at 9:57
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$\begingroup$ I'm accepting this as an answer (I had completely forgot about the thread), but let me remark that the hand-written note on page 250, line 3 (claiming that the inclusion $A+d+B\subseteq C^\ast$ is "trivially impossible in view of $|A+d+B|=|A+B|=|C|>|C^\ast|$", where $A$ and $B$ are finite non-empty subsets of the base set, $G$, of a fixed cancellative monoid $\mathbb G$, $C$ is the sumset $A+B$, and $d\in A\cap B$) is incorrect. The point is basically that $\mathbb G$ is not abelian, even if Kemperman writes it additively. For a counterexample, let $\mathbb{G}$ be a [...] $\endgroup$ Commented Mar 30, 2014 at 14:20
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$\begingroup$ [...] (multiplicatively written) grp for which there exist $x,y\in G$ s.t. $x^3=1$ and $y^nx\ne xy^m$ for every $m,n\in\mathbf{Z}$ with $m^2+n^2\ne 0$, e.g., a 2-generator 1-relator grp of the form $\langle u,v \mid u^3=1\rangle$. For $n\in{\bf N}$ we let $Y_n := \{y^k: k=0,\ldots,n\}$. Then, we fix $m,n\in{\bf N}$ and let $A:=Y_mx$ and $B:=xY_n$, so that $x\in A\cap B$. We have $$AB=\{y^hx^2y^k: h=0,\ldots,m, k=0,\ldots,n\}$$ and $$AxB=Y_mx^3Y_n=Y_mY_n=Y_{m+n}.$$ Thus, $|AB|=(m+1)(n+1)$ and $|AxB|=m+n+1$, since $y^hx^2y^k=y^rx^2y^s$ for some $h,k,r,s\in \bf Z$ only if $h=r$ and $k=s$. $\endgroup$ Commented Mar 30, 2014 at 14:21