# Is there a term for abelian groups in which you can divide by natural numbers?

Is there a specific term for an abelian group (or ring) $$G$$ that satisfies the following property?

For every element $$g \in G$$ and natural number $$n$$, there exists a $$q \in G$$ such that $$\underbrace{q + q + \dots + q}_{n\text{ times}} = g$$.

Loosely speaking, this property means that you can "divide by natural numbers", even if you can't divide arbitrary (nonzero) elements as with a field. Obviously any field extension of the rational numbers satisfies this property, but I suspect that there are probably non-field examples as well.

The reason I ask is that it seems that me that the property above gives the minimum amount of algebraic structure necessary to define the arithmetic mean of elements of a set, which seems like kind of a natural thing to study.

• dear tparker, I would caution that your last sentence is perhaps a bit debatable; you write "the property above gives the minimum amount of algebraic structure necessary to define the arithmetic mean of elements of a set", but I would argue that you also need uniqueness of $q$ for that to work ... May 21 at 21:45
• otherwise consider for example the group $\mathbb{Q}/\mathbb{Z}$. what is the arithmetic mean of $0$ and $0$ in this group? on the one hand, it "should" clearly be zero. on the other hand, we also have that $(1/2+\mathbb{Z})+(1/2+\mathbb{Z})$ equals $0$ in this group, even though $1/2+\mathbb{Z}$ is non-zero May 21 at 21:46
• if you do impose uniqueness, then the kinds of groups you are looking at are precisely the divisible torsion-free groups, which are equivalently vector spaces over $\mathbb{Q}$; every such group is isomorphic to $\mathbb{Q}^{\oplus I}$ for some index set $I$ May 21 at 21:47
• @AtticusStonestrom Great point, thanks for pointing that out. So it seems like we can really only take arithmetic means of vectors over $\mathbb{Q}$ (or over extension fields of $\mathbb{Q}$). May 21 at 22:09