# Does such a group exist?

A group with infinite normal subgroups: $G\varsupsetneqq G_{E}\varsupsetneqq G_{1}\varsupsetneqq G_{2}\varsupsetneqq....$ such that there exist a subset $B\subset G$ satisfying: 1) $\forall x,y\in B\,\, x\neq y\Rightarrow x\cdot y^{-1}\notin G_{E}$

2) $\forall n\in\mathbb{N}$ $G/G_{n}=\{{\overline{x}\cdot\overline{y}\cdot\overline{z}\,|\, x,y,z\in B\}}$

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I seriously think you might want to provide some motivation. At the moment this looks random enough to be a homework question. –  Vladimir Dotsenko Jun 15 '12 at 11:57
The notation of 2) makes it look like you want $G_n$ to be a normal subgroup, otherwise the multiplication expression $\bar x \cdot \bar y \cdot \bar z$ is not well-defined. –  Lee Mosher Jun 15 '12 at 12:25
I want to prove that in some specific situation there is elimination of hyperimaginaries in hyper definable group. I did a reduction of this problem to the question I've asked, and i hope someone can help me. –  elad Jun 15 '12 at 14:12

$\mathbb Z \times \mathbb Z$ will do. Take $G_E$ to consist of the elements whose first coordinate is $0$ and $G_n$ to be the subgroup of multiples of $2^n$ inside $G_E$. What $G_n$ are is immaterial, since we will make $G$ itself be covered by $\{x\cdot y\cdot z| x,y,z\in B\}$.
We will do this by building up $B$ step-by-step. At any given time, it contains finitely many elements. Choose some element of $G$ that is not in $\{x\cdot y\cdot z| x,y,z\in B\}$. We will add $3$ elements to $B$ so that it is. We can do this while preserving the property (1), since that just means that the first coordinates of all the elements are different. So choose two very large positive first coordinates, and then a third very large negative first coordinate so that they add up to the desired value. If they are sufficiently large they will not intersect past first coordinates. Then choose second coordinates in any way that adds up to the desired value.
Repeat this process until every element of $G$ is covered. Then clearly $G/G_n$ is covered as well.
This should work any time $G/G_E$ is infinite (of at least the cardinality of $G_E$).