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Let $G$ be any group. It's a well-known result that if $H, K$ are subgroups of $G$, then $HK$ is a subgroup itself if and only if $HK = KH$.

Now, I've always wondered about a generalization of this result, something along the lines of:

Theorem: If $H_1, \ldots, H_n$ are subgroups of $G$, then $H_1H_2\dots{H_n}$ is a subgroup if and only if ($\star$) holds, where $(\star)$ is some condition on $H_1, \ldots, H_n$, preferably related to how the smaller products $H_{m_1}\ldots{}H_{m_k}$, for $k < n$, behave.

Question 1: Is there such a theorem?

I do know, and its easy to prove, that if $H_iH_j = H_jH_i$ for every $i, j$, then the big product is a group, but this is not satisfying since it's far from necessary (just take one of the groups to be $G$, and you need no commutativity at all). Also, I've been told that there is no really satisfactory answer; if that is indeed the case, then my question would be why? In particular:

Question 2: Are there really problematic counterexamples where you can see that the behavior of the smaller products has nothing to do with the big product, so that no such a theorem can ever exist?

I would appreciate even an answer for the particular case $n = 3$.

Thanks.

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    Excellent question! I suggest you send a mail to one of the leading experts, prof. dr Bernhard Amberg, at Mainz, Germany, see http://www.mathematik.uni-mainz.de/arbeitsgruppen/gruppentheorie/amberg and browse through his list of publications.2012-09-06
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    Does anyone know if $\prod H_i=\prod H_{\sigma(i)}$ for all $\sigma\in Sym(n)$ is sufficient? That would seem like the first thing to check... Nice question btw, too.2012-09-06
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    Yes, if everything permutes then the product is a subgroup, it can be proved be induction on $n$. It is a bit tedious to write this down properly, but here is the proof for $n=3$: Let $h_i, k_i \in H_i$ for $i=1,2,3$. Then $h_1h_2h_3·k_1k_2k_3 = h_1(h_2h_3k_1)k_2k_3 = h_1k_1'h_3'h_2'k_2k_3 = h_1'h_3'h_2''k_3 = h_1''h_2'''h_3''k_3 = h_1''h_2'''h_3''' \in H_1H_2H_3.$ Further $(h_1h_2h_3)^{-1} = h_3^{-1}h_2^{-1}h_1^{-1} = h_1'h_2'h_3' \in H_1H_2H_3$. Not sure about the converse.2012-09-06
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    @NickyHekster the converse is false. If you take $H = \{1, r\}$ and $K = \{1, \theta{}r\}$ in $D_3$, the dihedral group with $6$ elements, where $\theta$ is the rotation and $r$ the reflection, you can check that $HKH = D_3$, but $HHK = HK$ is not a subgroup.2012-09-07
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    One might begin by requiring that all subgroups involved be distinct. Do you know counterexamples not involving trivial or repeated subgroups?2012-09-08
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    @KevinCarlson I don't know. But requiring all subgroups to be distinct seems a rather unnatural condition to me. The theorem for the $n = 2$ case doesn't require that, and in fact is trivial in this case. It seems that it should be easier to check whether the big product is a subgroup or not when some of the groups coincide, not harder. Besides, we would exclude a lot of cases if we were to assume that.2012-09-08
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    A necessary and sufficient condition is that $\prod H_i=\prod H_{\sigma i}$ for all $\sigma\in C_n$; this can be obtained by applying the $n=2$ case inductively.2012-09-08
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    @anon: as student's example above shows, this is not true. You cannot apply induction to (say) $ABC$, since this may be a subgroup even when $AB$ is not.2012-09-08
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    Hmm. Then "$AB$ is a subgroup if and only if $AB=BA$" is false: take $A=HK$ and $B=H$...2012-09-09
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    @anon if you read carefully, the theorem says that $A$ and $B$ must be subgroups in order for this to hold. In this case, $A = HK$ is not a subgroup.2012-09-09
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    Right. Nevermind me.2012-09-09

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