# Non-abelian divisible groups

I recently stumbled over the example in http://ysharifi.wordpress.com/2010/03/09/a-uniquely-divisible-non-abelian-group/ of a non-abelian group $G$ with the property that for all natural numbers $n$ and elements $x\in G$ there is $y\in G$ such that $x=y^n$. (In this particular example $y$ is unique but I don't care about that.) I will call such groups divisible, although I'm not sure if this term is used for abelian groups exclusively.

I wonder wether there are examples of non-abelian divisible groups, that satisfy additional assumptions. I am in particular interested in non-abelian divisible

$\bullet$ simple groups,

$\bullet$ finitely generated groups,

$\bullet$ finitely presented groups,

$\bullet$ groups satisfying all or some of the above properties at the same time.

Unfortunately I don't manage to find examples, but I'd appreciate any help.

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Example of nilpotent divisible groups are the radicable subgroups of $U_n(\mathbb{Q})$, the group of upper triangular matrices $n\times n$ matrices with 1s in the diagonal. They can be obtained as $\exp(\mathfrak{g})$ where $\mathfrak{g}$ is a subalgebra of $\mathfrak{u}(n,\mathbb{Q})$, the Lie algebra of strictly upper triangular matrices over $\mathbb{Q}$. These groups do not satisfy your particular conditions by they are of finite rank, that is, for $G$ unipotent and radicable there exists $r$ such that every finitely generated subgroup of $G$ can be generated by $r$ elements. – Diego Sulca Mar 7 '12 at 12:32
Most of these issues have already been discussed in the following questions: mathoverflow.net/questions/56128/… mathoverflow.net/questions/85236/examples-of-monster-groups – Benjamin Steinberg Mar 7 '12 at 16:09

See:

V. S. Guba, Finitely generated complete groups, Izv. Akad. Nauk SSSR Ser. Mat. 50 (1986), 883-924.

for an interesting 2-generated example. (Furthermore, roots in Guba's examples are unique.) Note that a group $G$ is called complete if for any non-trivial word $u(x_1,\cdots,x_m)$ and every $g\in G$, the equation $u(x_1,\cdots,x_m)=g$ has a solution. In your case, the words $u$ are of the form $x^n$. I do not know if Guba's group is simple, but it does have a (nontrivial) simple quotient, which is necessarily verbal, and, in particular, divisible.

None of these groups is finitely-presented and, I think, the problem of existence of fp divisible groups is open. The philosophical reason why is the following. Call an infinite finitely-generated group "exotic" if it satisfies some bizarre, seemingly impossible property, e.g., being a torsion group, containing very few conjugacy classes, being divisible, etc. The most common method for constructing exotic groups $G$ is by direct limit of a sequence of (relatively) hyperbolic groups $G_k$ which are quotients of a single group $G_0$. If $G$ were finitely presented, it would be isomorphic to one of the groups $G_k$ and, hence, non-exotic.

Mark Sapir will probably have more comments on this.

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Any connected, compact Lie group has this property. To get convinced about this, take an element $g \in G$ where $G$ is a compact Lie group, then $g$ can be put inside a torus $T$ in $G$, and a torus being a direct product of finitely many copies of circle, $S^1$, the result now follows.

Caution: One needs to prove the result that every element of $G$ has an $n$-th root to show that every $g \in G$ can be put in a torus. The proof uses degree argument of a smooth map between manifolds.

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You do not need any of these, just the fact that exponential map for every compact Lie group $G$ equals Riemannian exponential map for the bi-invariant Riemannian metric on $G$. – Misha Mar 21 '13 at 16:58
Yes Misha, that will also do. But I thought that the knowledge of (even existence of) Riemannian metric is difficult than the basic cohomology theory of compact manifolds. :-). – Shripad Mar 26 '13 at 3:56

One often reads that divisible groups are important since they help us understanding the structure of abelian groups, for they are all and the only injectives in the usual category of abelian groups (which is, of course, undeniable). Yet, I find that the non-abelian case is, if possible, even more interesting. A 'natural' example of a non-commutative divisible group is the group of units of Hamilton's quaternions; afak, the result is due to I. Niven [1]. On another hand, it was recently proved on this forum that the general linear group of degree $n$ over an algebraically closed field $\mathbb K$ is divisible iff $\mathbb K$ has zero characteristic (see here), and I've just posed a similar question for ${\rm SL}_n(\mathbb K)$ (see here)

[1] I. Niven, The Roots of a Quaternion, The Amer. Math. Monthly, Vol. 49, No. 6 (Jun. - Jul., 1942), pp. 386-388.

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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. – Salvo Tringali Mar 21 '13 at 11:17
Divisibility of SO(3) was known by mid-19th century (since every element of SO(3) is a rotation). Divisibility of SU(2) is also quite trivial (use normal form for unitary matrices). I do not think Niven should be credited with this. – Misha Mar 21 '13 at 16:56
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. – Salvo Tringali Mar 21 '13 at 18:01
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. – Salvo Tringali Mar 21 '13 at 18:58
Salvo: You just need to know is that if $A\in U(n)$, then eigenvectors with distinct eigenvalues are orthogonal (this was probably known to Hermite or Jordan). It then follows that $A$ is unitarily conjugate to a diagonal matrix $Diag(e^{it_1},...,e^{it_n})$, which gives you all the roots you need. I assume, Ore and Niven were doing something more advanced. – Misha Mar 21 '13 at 22:45