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Let $G$ be a group. Consider it as a structure in the language of groups $L=(e,\cdot, (-)^{-1})$. Suppose that $G$ eliminates quantifiers in this language. Is it true that $G$ must be Abelian then?

My thinking was that either the group is divisible, or there exists a number $N>1$ such that for any element $x$ the maximal number by which $x$ is "divisible" is $n$, i.e. for any $x$ if $k$ is the maximal number for which the formula $\exists y\ y^k=x$ has a solution then $k=N$. The latter is clearly impossible in the Abelian case, and I don't know how one can make a non-Abelian group that satisfies such condition. Divisible non-Abelian groups also look quite nasty, so I thought that maybe there are none among them that eliminate quantifiers.

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I haven't thought through the details carefully, but I bet you can eliminate quantifiers in $(\mathbb{Z}/2) \ltimes \mathbb{R}$. –  David Speyer Aug 7 '12 at 13:23
I believe the integers have quantifier elimination, and they are equivalent to $\langle 1/p\vert p\textrm{ prime}\rangle$, which is not divisible, but where every element is divisible by arbitrarily large (if not arbitrary) numbers, so the premise is false. –  tomasz Aug 7 '12 at 19:37
Tomasz, I am not sure what you mean here. Surely $(\mathbb Z,+)$ doesn't have QE in the group language, the sets of $p$-divisible elements have to be added to the language . –  Dima Sustretov Aug 7 '12 at 20:50
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