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Could you give me an example of a poly-virtually-abelian group that is not virtually-poly-abelian and an example of a virtually-poly-abelian group that is not poly-virtually abelian? (if they exist)

For a definition of virtually look here

Poly-P means that we have a series $1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_n=G$ where $G_{i+1}/G_i$ has property P.

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    *Poly-abelian* is a synonym of *soluble*, which may help with searching.2012-02-28

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In general, if $|G:H|$ is finite, then the core $K$ of $H$ in $G$, which is the intersection of the conjugates of $H$ in $G$, satisfies $K \le H \le G$ with $K \unlhd G$ and $G/K$ finite. Using that, it is clear that virtually-poly-abelian (same as virtually solvable) implies poly-virtually-abelian.

The converse is also true. By using induction on the length of the poly-virtually-abelian series of $G$ we can reduce to the situation where $G$ has a series $1 < H \lhd K \lhd G$ with $H$ and $G/K$ solvable and $K/H$ finite. We may assume that $H$ is the largest normal solvable subgroup of $K$, and so it is characteristic in $K$ and hence normal in $G$. Let $H < C$ with $C/H = C_{G/H}(K/H)$. Then $K/H$ finite implies that $G/C$ is finite. Also $(C \cap K)/H = Z(K/H)$ is abelian and $C/(C \cap K) \cong CK/K$ is solvable, so $C$ is solvable. Hence $G$ is virtually solvable and we are done.

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    In your poly-virtually-abelian series $1 \lhd G_1 \cdots \lhd G_n=G$, by induction $K := G_{n-1}$ is virtually solvable, so has a normal solvable subgroup $H$ of finite index. Now $G/K$ is virtually abelian, but since our aim is to prove that $G$ is virtually solvable, we can ignore the "virtually" and assume that $G/K$ is abelian (and hence solvable).2012-02-28