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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?

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

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