Suppose $G$ is a group such that $G=H_1H_2$, then one can easily see that $G=H_2H_1$, one can switch the order of the subgroups: for a general element $g$ one writes $g^{-1}$ as $h_1h_2$ with $h_1\in H_1$ and $h_2\in H_2$ and then $g=h_2^{-1}h_1^{-1}$.

The question is the analogous for the general case with more subgroups: suppose $G$ is a group and $H_1$, $H_2$, ... ,$H_n$ are subgroups such that $G=H_1H_2\cdots H_n$. Is it always possible to permute the factors and still obtain the full group?

If one has the property that $H_1H_2\cdots H_i$ is a subgroup of $G$ for every $i$ then one can use induction and the case with two subgroups to show that it is always possible, but this condition may not always hold.

  • 2
    $\begingroup$ Note that the existence of a counterexample was easily predictable, since otherwise this would imply that if a finite group $G$ is generated by subgroups $H,K$, so that $G=HKHKHK\dots HK$ for enough many factors, then by permuting we get $G=H\dots HK\dots K$ and thus $G=HK$. Thus it's enough to find the smallest possible examples, which, no surprisingly, is the smallest non-abelian group. This is why I voted to migrate on MathSE, it's not of research level. $\endgroup$
    – YCor
    Nov 3 '14 at 15:23


Take $G=\textrm{Sym}(3)$ and $H_1=H_3=\langle (12)\rangle$ and $H_2=\langle (23)\rangle$. Then $H_1H_2H_3=G$ but $H_1H_3H_2=H_1H_2\subsetneq G$.


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.