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Background: given an abelian group $G$, we can create a divisible group $\mathrm{Div}(G)$ by defining an operation on the set of $2$-tuples $\{(g,n):g\in G,n\in\Bbb N^+\}$ given by

$$(g,n)+(h,m):=(mg+nh,nm), \tag{a}$$

and quotient by the relation $\sim$ defined by $(g,n)\sim(h,m)\iff mg=nh$. Conceptually, a tuple proxies as a "formal ratio" $g/n$ (this construction idea is essential to creating fields of fractions and Grothendieck groups, too).

How would we construct divisible groups (which we define as those groups $G$ satisfying $nG=G$ for all $n\in\Bbb N^+$, or $G^{[n]}=G$ multiplicatively) out of nonabelian groups?

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I was going to use the feature wherein a user can answer their question and ask it at the same time, but the system disallows users with less than 10 reputation from answering their own question within 7 hours. – blue Jun 21 '12 at 6:12
Note that $G$ is not contained in $\operatorname{Div}(G)$. For instance, take $G=\{0,g\}$ of order 2. Then $(g,n) ~ (0,2n)$ since $n(2g) = n(0)$ and $(0,n) ~ (0,1)$ since $1(0) = n(0)$. Hence $\operatorname{Div}(G)$ has only one element. – Jack Schmidt Jun 21 '12 at 17:39
@JackSchmidt Ah, that's why I originally stipulated torsion-freeness, I forgot though when I read Qiaochu's answer! I will rollback my edit. (Also, $\sim$ is \sim.) – blue Jun 21 '12 at 22:31

You freely adjoin $n^{th}$ roots of the elements of $G$. Of course, since $G$ is noncomutative these $n^{th}$ roots aren't required to commute with anything besides powers of themselves so the result is quite nasty (analogous to what happens when you try to imitate the construction of the Grothendieck group for a noncommutative monoid). More precisely, take the free group on symbols $g^{1/n}, g \in G, n \in \mathbb{N}$ and quotient by the relations $g^1 = (g^{1/n})^n$ as well as any relations between elements $g^1$ of $G$ in $G$. As you correctly indicate in a comment below you of course need to adjoin more roots than this in the non-abelian case.

There is no reason that I can see to require $G$ to be torsion-free. In the abelian case this assumption implies that division has a unique result (if it exists), since $na = nb = g$ then implies $n(a-b) = 0$ which implies $a-b = 0$. But in the non-abelian case, from $a^n = b^n = g$ we cannot conclude $(ab^{-1})^n = 1$ if $a, b$ don't commute, so torsion-freeness doesn't help.

If you just want to construct non-abelian divisible groups, examples include any non-abelian Lie group $G$ for which the exponential map $\mathfrak{g} \to G$ is surjective. If $G$ is compact there is a lot of torsion (think of $\text{SO}(3)$ for example).

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Do expressions like $g^{1/n}h^{1/m}$ necessarily have $k$th roots in this adjunction? I was unsure of this, so what I did was adjoin $n$th roots iteratively (and quotient by the relations you mention) and then take a direct limit of the resulting chain of groups. (But again, I can't post an answer yet.) – blue Jun 21 '12 at 6:19
@sea: ah, yes, that's what I should've done. – Qiaochu Yuan Jun 21 '12 at 6:21
Ah, I hadn't thought about divisible groups with torsion! – blue Jun 21 '12 at 6:23
up vote 1 down vote accepted

We denote by $T(G)$ the set of $2$-tuples $\{(g,n):g\in G,n\in\Bbb N^+\}$ again, but this time we create a free group out of it and quotient by the relation given by $(g,n)^n\sim (e,1)$ as well as $(g,1)(h,1)\sim(gh,1)$ (for all elements $g,h\in G$ and naturals $n$) on top of the equivalence

$$(g,n)\sim(h,m)\iff g^m=h^n, $$

and write $D(G)=F(T(G))/\sim$. This is a group. If commutativity is absent, we no longer can write products of tuples as other tuples in general. Here $(g,n)$ is an avatar for the formal expression $g^{1/n}$.

But $D(G)$ is not necessarily divisible: products of tuples that can't be simplified don't necessarily have roots. Yet it's still true that $G$ embeds into $D(G)^{[n]}$ as a subgroup for any natural $n$. Moreover, we can form a direct system with the embeddings

$$G\hookrightarrow D(G)\hookrightarrow D(D(G))\hookrightarrow D(D(D(G)))\hookrightarrow\cdots.$$

Define $\mathrm{Div}(G)$ to be the direct limit of this system. If $x\in \mathrm{Div}(G)$, then $x\in D^m(G)$ for some $m$, and then $v=(x,n)\in D^{m+1}(G)$ satisfies $v^n=(e_{D^m(G)},1)=e$. It follows that $\mathrm{Div}(G)$ is divisble.

One thing I'm struggling with: is there a sense in which $\mathrm{Div}(G)$ is the smallest divisible group containing $G$? Intuitively this seems right, but I'm unsure of how to formally describe/justify this.

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There is a forgetful functor from divisible groups to groups and ideally what you describe is its left adjoint. This ought to be true but proving it seems annoying. – Qiaochu Yuan Jun 21 '12 at 23:20

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