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

Question

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!

Answer 1

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.

Answer 2

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.