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Is there a standard name for a (multiplicatively-written) semigroup $(A, \cdot)$ such that, given an arbitrary $a \in A$, the equation $x^n = a$ has at least one solution $x \in A$ for each $n \in \mathbb{N}^+$? And what about the case where such a solution is required to be unique? Unluckily I can find no trace of them in the two volumes of Clifford and Preston's survey: The algebraic theory of semigroups, but this is almost surely for the fact that I don't know what to look for. Thus, I would appreciate much if you could also provide me with references to previous work on the subject.

P.S. If I'm not blinded by the dragon's breath, nothing similar seems to be there in Wiki's article on Special classes of semigroups (click).

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Is $n$ a constant? –  Tara Brough Jun 10 '12 at 12:09
    
Oh sorry! I'm editing once more. –  Salvo Tringali Jun 10 '12 at 12:13
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up vote 5 down vote accepted

These are divisible semigroups and uniquely divisible semigroups respectively. The reason you cannot find anything about them on the internet is that almost nothing is known. Say, every idempotent semigroup is obviously divisible and uniquely divisible. Every divisible infinite semigroup $S$ is not residually finite unless $S$ consists of idempotents. Indeed, if $H$ is a finite homomorphic image of $S$, then for some $n$ $x^n$ is an idempotent in $H$ for every $x$. Now take any $y\in S$. It has a root $a$ of degree $n$. Hence the image of $y$ in $H$ is the image of $a^n$, hence an idempotent. Thus $y$ cannot be separated from $y^2$ in any finite homomorphic image. Thus if $S$ is residually finite, it must consist of idempotents.

There are infinite finitely generated divisible and uniquely divisible groups. The first example was constructed in Guba, V. S. A finitely generated complete group. Izv. Akad. Nauk SSSR Ser. Mat. 50 (1986), no. 5, 883–924.

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Interesting. I will accept this answer in a week, if nobody else has anything to add (as for the part concerning references). Thank you, Mark. –  Salvo Tringali Jun 10 '12 at 12:42
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