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If $G$ is a finite union of some of its abelian subgroups, then the index of the center of the group is finite

Would I not simply state that by Lagrange's theorem, $Z(G)$ can divide into the abelian subgroups, and the abelian subgroups can divide into $G$? This solution seems too obvious and intuitively wrong. Also I don't know how to approach the question if $G$ was infinite, thank you!

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    $\begingroup$ If $G$ is finite, the index of any subgroup is finite. The hard case is when $G$ is infinite. $\endgroup$ – Ted Mar 28 '14 at 1:59
  • $\begingroup$ If G is infinite, I would have to prove that Z(G) contains more than the identity, because if Z(G) only contains the identity, then the index would be infinite, correct? $\endgroup$ – Ian Mar 28 '14 at 2:09
  • $\begingroup$ @Ian: Yes, but you need to prove much more than that. For $n>1$, the general linear group $GL_n({\bf R})$ has an infinite center (consisting of scalar matrices), but its index is infinite. $\endgroup$ – tomasz Mar 28 '14 at 2:11
  • $\begingroup$ @tomasz I see. I deleted the useless comment. =) $\endgroup$ – Pedro Tamaroff Mar 28 '14 at 2:23
  • $\begingroup$ can I say that since the G is a finite union of abelian subgroups, then Z(G) has finite index in all of those subgroups, which means that the index of Z(G) in the union of those subgroups will just be all the indices of Z(G) in those subgroups added up together? $\endgroup$ – Ian Mar 28 '14 at 3:14
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A sketch of a proof is as follows:

First prove that if $H,A_1,\ldots,A_n\leq G$ are abelian subgroups and $G=H\cup\bigcup_i A_i$, then if $H$ has infinite index, then $G=\bigcup_i A_i$.

This is the hard part (though I would appreciate a simpler explanation), I believe: to show this, choose arbitrary $h\in H$ and:

  1. Notice that there is some $A_i$ such that $A_i/(A_i\cap H)$ is an infinite abelian group, and as such it either contains an infinite cyclic group or an infinite direct sum of cyclic groups.
  2. In the latter case, take a sequence $g_m$ of elements of $A_i$ whose classes generate "orthogonal" cyclic subgroups of $A_i/(A_i\cap H)$.
  3. Then either there is some $m$ with $g_mh\in A_i$ and we are done or else there is some $A_j$ such that there are infinitely many $m$ with $g_m h\in A_j$.
  4. Otherwise without loss of generality we can assume that all $g_mh$ them are in $A_j$ (dropping some $g_m$ if needed). Therefore, for distinct $m$, $g_{2m}g_{2m+1}^{-1}=g_{2m}h(g_{2m+1}h)^{-1}\in A_i\cap A_j$ again generate "orthogonal" cyclic subgroups of each of $A_i/(A_i\cap H),A_j/(A_j\cap H)$.
  5. Repeating this procedure you eventually end up in the "either" case of step 3. (otherwise you get sequences that belong to more and more of the $A_i$ until you get one which is in all of them).
  6. In case where $A_i/(A_i\cap H)$ contains an infinite cyclic subgroup, you do mostly the same thing, but the original $g_m$s need to be picked a little more carefully, and they don't generate "orthogonal" cyclic subgroups, but instead have the property that $g_{m_1}g_{m_2}g_{m_3}\cdot\ldots\cdot g_{m_{2^n}}g_{m_{2^n+1}}^{-1}\cdot\ldots,g_{m_{2^{n+1}}}^{-1}\notin H$ for distinct $m_1,\ldots, m_{2^{n+1}}$, as well all shorter products of similar form. Elements whose classes are distinct power-of-2 multiples of the generator of infinite cyclic subgroup will do, for instance.

Then apply this to $G=\bigcup_i A_i$ (finite union) to deduce that you can assume that each $A_i$ has finite index.

Then under the previous assumption, show that $\bigcap A_i$ is a finite index subgroup of $G$ and that it is contained in the center.

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There is an unpublished result by Reinhold Baer (see Theorem 4.6 in D.J.S. Robinson, Finiteness conditions and generalized soluble groups, Part I, Springer Verlag, New York, 1972.)

Theorem A group is central-by-finite if and only if it is the union of finitely many abelian subgroups.

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  • $\begingroup$ The other direction is rather trivial, really: let $g_1,\ldots,g_n$ be a complete set of representatives of $G/Z(G)$. Then each $\langle Z(G),g_i\rangle$ is abelian and contains $g_i+Z(G)$, so they trivially add up to the entire $G$. $\endgroup$ – tomasz Mar 28 '14 at 10:57
  • $\begingroup$ It also follows easily from the result of B.H. Neumann that if a group is a finite union of subgroups (or even cosets of subgroups), then we can discard any of the (cosets of) subgroups of infinite index, and ti will still be the union of the remaining subgroups (cosets). But that is essentially the argument that tomasz is using in his answer. $\endgroup$ – Derek Holt Mar 28 '14 at 18:47
  • $\begingroup$ Derek, yes indeed. The famous covering theorem of B.H. Neumann! $\endgroup$ – Nicky Hekster Mar 28 '14 at 23:24

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