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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?

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    @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.2012-10-20

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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).

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    I believe the nonabelian analog is called a D-group. Gilbert Baumslag wrote some papers on them a while ago.2012-10-21