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Embedding a semigroup into a divisible semigroup
Well, I don't know about Šutov's paper (my Russian doesn't go beyond the alphabet), but I read Neumann's, and strictly speaking Neumann doesn't prove, not in the paper referred to in the above answer, that any given sgrp can be embedded into a divisible sgrp. However, he mentions that this can be done (p. 1021), and addresses the reader to [1, Theorem 6.2], where an analogous result is established for groups. But his construction doesn't look very canonical, so let me edit the OP and add another question. References: [1] B.H. Neumann, Adjunction of elements to groups, JLMS, 18 (1943), 4-11.
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Embedding a semigroup into a divisible semigroup
Thanks, Benjamin, I've just retrieved Neumann's paper. For the record, the manuscript comes with a corrigendum (cms.math.ca/cjm/a145888), for "an error in the first proof, p. 1020, of Theorem 3.1", and an addendum addressing the reader to Šutov's work for an alternative proof.
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Non-abelian divisible groups
@Misha. I agree, but my point is that I don't know, among the many things that I don't know, who was the first to address the question explicitly (just to have a trustful reference for all practical purposes). However, I guess that I would be better to give up with this, for it seems hard, and perhaps even pointless, to track back the paternity of the result. Thank you, in any case, for sharing your thoughts.
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On the divisibility of the special linear group of degree $n$ over an algebraically closed field
Clear and very nice. I really wonder if one can cluster all the matrices in ${\rm SL}_n(\mathbb C)$ which don't have a $p$-th root for some prime $p \le n$ in a finite number of conjugacy classes: Your example is still very particular, which tempts me to think that there may be only "few" exceptions.
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On the divisibility of the special linear group of degree $n$ over an algebraically closed field
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Non-abelian divisible groups
OK, I've finally given a look at Niven's 1941 paper. It deals with a more general question, i.e. sort of a fundamental theorem of algebra for polynomials with coefficients in the skew-field of Hamilton's quaternions. And Niven reports a remark of Jacobson according to which this result is a consequence of previous work by Ore, dating back to 1933, on non-commutative polynomials.
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Non-abelian divisible groups
From the paper mentioned in my answer: "The existence of an $m$-th root of a quaternion $a$ is known", and then a note refers the reader to: I. Niven, Equations in quaternions, AMM, Vol. 48, 1941, pp. 654-661. Maybe you're right, but this is the only reference that I've found so far, and it dates back to 1941.
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On the divisibility of the special linear group of degree $n$ over an algebraically closed field
Ehr... Yes, but somehow I missed this "limit case" while working to something slightly more general: If $J = \left[\begin{array}{cc} \lambda & \mu \\ 0 & \lambda^{−1}\end{array}\right]$ for some $\lambda \in \mathbb C^\times\setminus\{−1\}$ and $\mu \in \mathbb C$, then $\left[\begin{array}{cc} a & c \\ 0 & b\end{array}\right]^2=J$ for $a=|\lambda|^{1/2}e^{i\frac{\theta}{2}}$, $b=a^{−1}$ and $c=(a+b)^{−1} \mu$, where $\theta$ is (the principal value of) the complex argument of $\lambda$. I edited the OP, fixed my mistake, and updated Q2. Thanks!
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Non-abelian divisible groups
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On the divisibility of the special linear group of degree $n$ over an algebraically closed field
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Are there results in "Digit Theory"?
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Non-abelian divisible groups
To clarify: This doesn't count as an answer to the specific question in the OP, but it is an attempt to suggest a direction, and it was too long for a comment.
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All and the only algebraically closed fields s.t. any regular n-by-n matrix has a k-th root for every k
I had a look at Higham's Functions of Matrices - Theory and Applications, before asking. And even if the 7th chapter of the book is entirely focused on $k$-th roots of complex matrices, nothing like your (absolutely brilliant) proof seems to be there. And the 3rd edition of the classical Matrix Computations by Golub and Van Loan spends no word for arbitrary $k$-th roots, while square roots are just mentioned in relation to the Cholesky and polar decompositions (and again, there's nothing in the lines of your slick argument).
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All and the only algebraically closed fields s.t. any regular n-by-n matrix has a k-th root for every k
You're absolutely right! I had completely overlooked the Lefschetz principle.