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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 $n = 3$.


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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 and browse through his list of publications. – Nicky Hekster Sep 6 '12 at 9:07
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. – rschwieb Sep 6 '12 at 12:50
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. – Nicky Hekster Sep 6 '12 at 15:15
@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. – student Sep 7 '12 at 19:03
@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. – student Sep 8 '12 at 17:05

It seems your desired theorem would look like this: If $A, B, C$ are subgroups of $G$, then $A B C$ is a subgroup iff $\Phi_1(A,B)\land\Phi_2(B,C)\land\Phi_3(C,A)$ (where the $\Phi_i$ is some predicate in two variables).

Let's have a look at $\Phi_1$ first. Clearly, "$\Phi_1(A,B):\iff AB\mathrm{\ is\ a\ group}$" would be too strong. Hence $\Phi_1(A,B)$ may be true in cases wheer $AB$ is not a group. But how can $A B$ fail to be a group? Either there is $x\in A B$ with $x^{-1}\notin AB$ or there are $x,y\in AB$ with $xy\notin AB$. For $ABC$ to be a group there must exist a $z\in AB$ such that $x^{-1}=z c$ or $x y = z c$ for some $c\in C$, respectively. All composites in $ABAB$, all inverses are in $BA\subseteq ABAB$, hence we need $ABAB\subseteq ABC$. Unless $ABAB\subseteq AB$ (i.e unless $AB$ is a group), this condition clearly depends on $C$.

Therefore such $\Phi_1$ cannot exist.

The same argument applies to $\Phi_2$ and with slight modifications for $\Phi_3$. One could also argue simply that for subgroups $X, Y$ such that $XY$ is not a group, we have that $XYG$, $XGY$ and $GXY$ are groups, hence all $\Phi_i(X,Y)$ must be true. But $XY1$, $X1Y$ and $1XY$ are not groups, hence some $\Phi_i$ with one argument $=1$ must be false. But that can't be because a product where two factors are the trivial group is a group.

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I'm not sure that your first statement is true. Taking $A = C = H$ and $B = K$, in the notation of my comment above, $ABC$ is a subgroup while $BCA$ and $CAB$ are not. Were you thinking about some special case? – student Sep 8 '12 at 16:55
Hm, maybe I thought things would "gonjugate away". Thus I should rather work with $\Phi_1(A,B)\land \Phi_2(B,C)\land\Phi_3(C,A)$. Then the rest of my post shows that $\Phi_1$ must depend on $C$ ... – Hagen von Eitzen Sep 8 '12 at 22:28

One issue you might have is if one of the subgroups is a normal subgroup, then they commute with the others automatically. You probably need a condition that you have at least one condition involving each pair of subgroups. I am not sure if that is actually enough. If you also include something like that you also have conditions that does not include 1,2,3,etc. subgroups, then you can get the desired answer by induction.

Finally, here is a relatively general example to show that in case of $n=3$ you need all three commutations:

Let $N\subset G$ be a normal subgroup and $H\subset G$ a subgroup such that $N\cap H=1$. Now let $A,B\subset H$ be subgroups such that $A\cap B=1$ and $AB\neq BA$, but $H$ is generated by $A$ and $B$ together. Finally assume that $A,B$ and $N$ are finite.

Since $N$ is normal, $NA=AN$ and $NB=BN$. Since they are all finite and pairwise disjoint, $$|NAB|=|N|\cdot |A| \cdot |B|,$$ but since $AB\neq BA$ it is not a subgroup, and hence, $$|A| \cdot |B|\lneq |H|,$$ so $$|NAB| \lneq |N|\cdot |H| = |NH|.$$ However, since $A$ and $B$ generate $H$ it follows that the subgroup generated by $A,B$ and $N$ is $NH$, which has more elements than $NAB$ does and hence $NAB$ cannot be a subgroup.

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