3
$\begingroup$

I was interested in knowing if groups with following property have been studied( like what can be said about structure of the group) : "$G$ can be written as disjoint union of a given number of abelian proper subgroups". (this number is not necessarily smallest such number)

$\endgroup$
4
  • 2
    $\begingroup$ Interpreting disjoint as "any two intersect only in the identity",such a group is itself Abelian, as any two such normal subgroups would centralize each other. $\endgroup$ Apr 19, 2016 at 7:29
  • 1
    $\begingroup$ I am sorry, I had just realized that I did not want normal and now I see why :) @GeoffRobinson $\endgroup$
    – user101
    Apr 19, 2016 at 8:11
  • $\begingroup$ See groupprops.subwiki.org/wiki/…. $\endgroup$
    – Stefan Kohl
    Apr 21, 2016 at 16:39
  • $\begingroup$ @StefanKohl That article has no substantive content beyond the definition. $\endgroup$ Apr 24, 2016 at 3:57

2 Answers 2

6
$\begingroup$

A result of mine (appearing as Problem 2.10(b) of my character theory book) says that if $G$ is nonabelian and is a disjoint union of $n$ abelian subgroups, then each of these subgroups has order at most $n-1$ and $|G|$ is at most $(n-1)^2$. As far as I know, there is no non-character proof of this result (though I have not tried very hard to find one).

$\endgroup$
0
$\begingroup$

In the case when the nonabelian $p$-group $G$ admits a partition by cyclic subgroups, an answer is known: either $\exp(G)=p$ or the Hughes subgroup of $G$ is a proper subgroup of $G$. In the general case, when a $p$-group $G$ admits a non-trivial partition, the answer is also known (M. Suzuki). There are papers of R. Baer, O. Kegel, M. Suzuki and other authors devoted to finite groups with partition (see Mathscinet).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.