Let $G$ be a finite group such that $HK=KH$ for any subgroups $H,K$ of $G$ . Then is every subgroup of $G$ normal ?

[Edit] (to anyone thinking this question is not worth keeping: Here is the response from the OP to a comment from Tobias Kildetoft stating that the condition implies uniqueness of all Sylow subgroups, and hence that $G$ is nilpotent, JL)

If for a fixed prime $p$, $H,K$ are two different Sylow-p subgroups then $|H \cap K| < |H|=|K|$ and then the subgroup $HK$ is a $p$-subgroup with order $|H||K|/|H \cap K| > |H|$ , impossible ! Hence for a given prime $p$ , there is a unique Sylow-p subgroup . But I have no idea whether these line of arguments passes to all subgroups or not ...


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    $\begingroup$ @Shuri2060 : I think the definition of $HK$ is pretty common , $HK:=\{hk : h \in H , k\in K\}$ . groupprops.subwiki.org/wiki/Product_of_subgroups $\endgroup$ – user456828 Jul 10 '17 at 19:03
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    $\begingroup$ Do you see that if this is the case then all Sylow subgroups are normal? $\endgroup$ – Tobias Kildetoft Jul 10 '17 at 19:09
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    $\begingroup$ @lulu Those are equivalent. $\endgroup$ – Tobias Kildetoft Jul 10 '17 at 19:11
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    $\begingroup$ No. Groups satisfying this property are sometimes called M-groups, or Iwasawa groups, and they are not necessarily Hamiltonian. This is discussed in Section 2.3 of the book Subgroup Lattices of Groups, by R. Schmidt. There, he gives the following example, for odd primes $p$: $$\langle a,x\mid a^{p^3} = 1, x^{p^{3}} = a^{p^{2}}, a^x = a^{1+p}\rangle.$$ $\endgroup$ – James Jul 10 '17 at 19:38
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    $\begingroup$ @amWhy Do you think the material I lifted from the OP's comment demonstrates a satisfactory research effort from their part? We apparently don't agree whether such material is necessary, but that's ok. For my part: if answering a group theory question requires Derek Holt to show up, then it is both sufficiently interesting and sufficiently difficult to be worth keeping. But, I am seeking for a harmonious coexistence of our points of view :-) $\endgroup$ – Jyrki Lahtonen Jul 12 '17 at 10:24

The group $\langle x,y \mid x^{p^2} = y^p=1, y^{-1}xy=x^{p+1} \rangle$ of order $p^3$ is a counterexample for all odd primes $p$.

The only non-normal subgroups are the $p$ conjugates of $\langle y \rangle$, and the product of any two of these is the subgroup $\langle x^p,y \rangle$.

Added: Sorry, I can see now that this example has already been mentioned in comments by James, but it is completely buried in zillions of comments, so it should really be an answer.

  • $\begingroup$ It looks similar, but it's not quite the same example. Yours is smaller and, it seems, easier to verify has the required properties. $\endgroup$ – James Jul 10 '17 at 21:21

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