# Find the field by the its multiplicative group

Suppose we have a group G. Is this a multiplicative (or additive) group of some field? I think that аn arbitrary group is not suitable (e.g. in the case of finite fields multiplicative group should be cyclic). What properties must have this group G (it's very interesting in the case of infinite groups)?

## 1 Answer

The additive case is straightforward: every field is a vector space over its prime subfield (the subfield generated by $1$), which is either $\mathbb{Q}$ or $\mathbb{F}_p$ for some $p$, and in both cases fields exist which have every possible dimension as vector spaces. So the problem reduces to characterizing which abelian groups are vector spaces over a prime field (either $\mathbb{Q}$ or $\mathbb{F}_p$ for some $p$). Note that this is a property, not a structure.

The abelian groups which are $\mathbb{Q}$-vector spaces are precisely those which are both divisible and torsion-free, and the abelian groups which are $\mathbb{F}_p$-vector spaces are precisely those for which every element has order (dividing) $p$, what group theorists call the elementary abelian $p$-groups.

In the multiplicative case, every finite subgroup of the multiplicative group of a field is cyclic, which is a pretty strong restriction. For more discussion see this MO question.

• Very thanks! I thought, this theory is very complicated, because my advisor said so. I'm glad that I was wrong) – Michael Galuza Jun 2 '15 at 6:47
• @Michael: well, the MO question I linked to doesn't exactly give a nice characterization of the multiplicative case. I think this is in fact pretty complicated. – Qiaochu Yuan Jun 2 '15 at 6:49
• sorry for stupid question, but it's necessary condition? I mean if abelian group is divisible and torsion-free or elementary $p$-group then it's a additive group of some field? What about multiplication — can we define it by additive group (of a unique way, possibly)? – Michael Galuza Jun 17 '15 at 5:08
• @Michael: yes. To show this is equivalent to showing that $\mathbb{Q}$ and $\mathbb{F}_p$ have extensions of every possible degree, and this is true and not hard to see (the finite degrees are the hardest). – Qiaochu Yuan Jun 17 '15 at 5:24