It is well-known that a group cannot be written as the union of two its proper subgroups. Has anybody come across some consequences from this fact? The small one I know is that if H is a proper subgroup of G, then G is generated by the complement G-H.


A consequence is that if a finite group $G$ has only two proper subgroups, then the group itself must be cyclic. This is seen as follows: By the stated result the group has at least one element $g$ that does not belong to either of the proper subgroups. But if there are no other proper subgroups, then the subgroup generated by $g$ cannot be a proper one, and thus must be all of $G$.

This gives an(other) easy proof of the cyclicity of the group of order $pq$, where $p<q$ are primes such that $q\not\equiv 1\pmod p$. By the Sylow theorems there is only one subgroup of order $p$ and only one of order $q$, so the above result applies.


(A very late answer) For a group $G$, let $S_G$ be the set of self-centralizing subgroups, $S_G := \{H \le G \mid H = C_G(H) \}$.

Proposition: There is no group $G$ with $|S_G| = 0$ or $2$.

Proof: This follows from the characterization of self-centralizing subgroups as the maximal (among) abelian subgroups (which exist by Zorn's lemma). Thus $G$ is the union of its maximal abelian subgroups, as this union contains all cyclic subgroups.

Restating, this says that the map of the lattice of subgroups to itself, sending a subgroup to its centralizer, always has at least $3$ fixed points if $G$ is nonabelian (I don't know how useful this is though). One could also easily imagine generalizing such statements to other properties as well.

  • $\begingroup$ Thanks! A surprising application! $\endgroup$ – Nicky Hekster Jan 20 '14 at 17:57

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