At first I wondered

Which groups $G$ have the property that $HK$ is a subgroup for all subgroups $H,K<G$?

If a finite group $G$ has this property, I think that $G$ ought to be nilpotent. (I can't prove this though. Would someone help me verify if this is true? What about for infinite groups?)

So, now I am wondering,

Which groups $G$ have the property that $HK$ is a subgroup for all noncyclic subgroups $H,K<G$?

This definitely does not imply nilpotency - for example, $SL(2,3)$ has this property. Is this equivalent to some other condition? Is there a description the class of groups for which this holds?

  • 1
    $\begingroup$ @DonAntonio $S_3$ isn't nilpotent. $\endgroup$ – Alexander Gruber Feb 18 '13 at 6:29

For the first question, when $G$ is a finite group we can indeed show that $G$ must be nilpotent.

If $p$ is a prime and $P$ and $Q$ are two Sylow $p$-subgroups of $G$, then $PQ \leq G$ if and only if $P = Q$. Thus $G$ has a unique Sylow $p$-subgroup. This means that every Sylow subgroup of $G$ is normal, which implies that $G$ is nilpotent.

The converse is not true. The dihedral group $D_8$ is nilpotent, but you can find examples of subgroups $H$ and $K$ of $D_8$ such that $HK$ is not a subgroup.

EDIT: About the second part, I think we can show that such a group is finite, it must be solvable. Let's say that $G$ is a $\mathscr{P}_2$-group if $HK$ is a subgroup for all noncyclic subgroups $H$ and $K$ of $G$. It is straightforward to show that subgroups and quotients of a $\mathscr{P}_2$-group are also $\mathscr{P}_2$-groups, so arguing by induction works here.

Suppose that $G$ is a finite $\mathscr{P}_2$-group. We know that if a group has every Sylow subgroup cyclic, then it must be solvable (this is usually done with Burnside's normal complement theorem). Thus we may assume that $G$ has a noncyclic Sylow $p$-subgroup for some prime $p$. Since $G$ is a $\mathscr{P}_2$-group, it has a unique Sylow $p$-subgroup $P$ (if $Q$ some other Sylow $p$-subgroup, then $Q$ is noncyclic and $PQ \leq G$ implies $P = Q$). Now either $G = P$ and $G$ is solvable as a $p$-group or $G \neq P$ and the solvability of $G$ follows from the solvability of $P$ and $G/P$ by induction. Therefore all finite $\mathscr{P}_2$-groups are solvable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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