I have a group $G$, and for all $g \in G$ and $a,b \in \mathbb{Z}$ it makes sense to talk about the element $g^{a \over b} \in G$.

To get some intuition, I've been thinking about what it would mean to do this to the integers over addition: it's not a group unless we extend the set to all rational numbers. Likewise if we consider the rationals over multiplication, it's not a group unless we extend the set to all real numbers. But I would like to see some formal treatments of this structure and am wondering where to look.

Is it true that this could be an alternative definition for Abelian groups? or is this something else entirely?

  • 3
    $\begingroup$ Divisble groups or $R$-modules ($\Bbb Q\subseteq R$ a subring) would probably be your best bet. $\endgroup$
    – anon
    Oct 19, 2012 at 23:10
  • 1
    $\begingroup$ So in other words, you have a group with a mapping (or maybe partial map? Do all elements have roots?) $R_n$ on $G$ ($n \in \mathbb{Z}$) with $R_n(\underbrace{g\cdot\ldots\cdot g}_{n\text{ times}}) = g$ and $R_n(ab)=R_n(a)R_n(b)$. If that mapping is defined on the whole group (i.e., not a partial mapping), then it's an automorphism, no? Maybe you can describe those groups as groups with special kinds of automorphisms... $\endgroup$
    – fgp
    Oct 20, 2012 at 0:25
  • $\begingroup$ Oh, and btw, you only need to extend $\mathbb{Q}$ to (at most) the set of all algebraic numbers, not to $\mathbb{R}$, no? And not even to all those, I think - not all algebraic numbers are representable by radicals as far as I remember - that is guaranteed only for those defined by polynomials of order $\leq 4$ I think. $\endgroup$
    – fgp
    Oct 20, 2012 at 0:31
  • $\begingroup$ @fgp I'll have to think a little more about your first comment and its implications, but that seems true. For your second comment, I think you're right. It did seem weird to me that in the first case I went from a countable set to a countable set and in the second I went from countable to uncountable. $\endgroup$ Oct 20, 2012 at 1:15

1 Answer 1


I'll assume that $\frac{a}{b}$ is in lowest terms and that $b \neq 0$. Whatever $g^{a/b}$ is, it ought to be some element $h$ such that

$$h^b = g^a.$$

Such an element need not exist in general, nor will it be unique in general.

Abelian groups such that such elements always exist are said to be divisible. Divisibility, however, says nothing about uniqueness. Abelian groups such that such elements both always exist and are always unique are $\mathbb{Q}$-vector spaces (exercise).

  • $\begingroup$ I believe the nonabelian analog is called a D-group. Gilbert Baumslag wrote some papers on them a while ago. $\endgroup$
    – user641
    Oct 21, 2012 at 4:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .