Let $a,b$ be two positive integer numbers. A group $G$ is called $a{\times}b$-decomposable if there are subsets $A,B\subset G$ of cardinality $|A|=a$ and $|B|=b$ such that $AB=G$ where $AB=\{xy:x\in A,\;y\in B\}$.

I am looking for an example of a finite group $G$ which is not $a{\times}b$-decomposable for some numbers $a,b$ with $a\cdot b=|G|$.

Remark. It is easy to see that a group $G$ is $a{\times}b$-decomposable if $a\cdot b=|G|$ and $G$ contains a subgroup of order or index equal to $a$ or $b$. Consequently, any Abelian group $G$ is $a{\times}b$-decomposable for any numbers $a,b$ with $a\cdot b=|G|$.

According to the answer of Geoff Robinson to this MO-problem the alternating group $A_9$ contains no subgroups of order or index equal to 35.

Question 1. Is the group $A_9$ $35{\times}5184$-decomposable?

By the comments of @YCor to the same MO-problem,

$\bullet$ the group $PSL_2(11)$ has cardinality $|PSL_2(11)|=15\times 44$ but contains no subgroups of order or index 15;

$\bullet$ the group $PSL_2(13)$ has cardinality $|PSL_2(13)|=21\times 52$ but contains no subgroups of order or index 21.

Question 2. Is the group $PSL_2(11)$ $15{\times}44$-decomposable?

Question 3. Is the group $PSL_2(13)$ $21{\times}52$-decomposable?

  • $\begingroup$ @mathworker21 $A=\{0,1\}$, $B=\{0,2\}$. ($A,B$ are subsets, not necessarily subgroups.) $\endgroup$ – bof Nov 25 '18 at 2:28
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    $\begingroup$ Related question: mathoverflow.net/questions/177747/… $\endgroup$ – Jeremy Rickard Nov 25 '18 at 9:46
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    $\begingroup$ Also: mathoverflow.net/questions/155986/factor-subset-of-finite-group $\endgroup$ – Jeremy Rickard Nov 25 '18 at 9:50
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    $\begingroup$ Questions 1 and 2 have been answered in the affirmative, and the answer to Question 3 is also yes. There is a triple factorization ${\rm PSL}_2(13) = ABC$ into subgroups with $|A|=7$, $|B|=12$, $|C|=13$, with $B \cong A_4$, and you can write $B$ as a product of subgroups of order $3$ and $4$ to get the required $21 \times 52$ factorization. $\endgroup$ – Derek Holt Nov 25 '18 at 20:24
  • $\begingroup$ So, what is the intuitive expectation concerning the general problem? Is each group $a{\times}b$-decomposable? $\endgroup$ – Taras Banakh Nov 25 '18 at 20:59

Answer to question 2: The group $G=\mathrm{PSL}_2(11)$ has an obvious subgroup $H$ of order 11 and (according to Magma) also a subgroup $H'$ of order 60 isomorphic to the alternating group $A_5$. Now $H'$ has a subgroup $K$ of order 4 (the Klein four-group). Letting $T$ be a transversal of $K$ in $H'$, we have $G = HH' = H(KT) = (HK)T$ so that $G$ is $15 \times 44$ decomposable.


Question 1 can be solved in a similar manner to @FrançoisBrunault's answer.

Let $A_{i+1}=L_{i+1}A_i=A_iR_{i+1}$ with $|L_{i+1}|=|R_{i+1}|=i+1$. Then $A_9=A_7R_8R_9=L_7A_6R_8R_9=(L_7L_5)(A_4R_6R_8R_9)$.

  • $\begingroup$ It seems that your argument allows to prove that any alternating group $A_n$ is $a{\times}b$-decomposable for any $a,b$ with $ab=|A_n|$. Right? $\endgroup$ – Taras Banakh Nov 26 '18 at 13:25
  • $\begingroup$ At least, not directly. Check what happens with $A_{35}$ and $a$ being the product of maximal powers of 3 and 5 in $|A_{35}$. $\endgroup$ – Ilya Bogdanov Nov 26 '18 at 14:15
  • $\begingroup$ It seems that the decomposability of the alternating groups $A_n$ can be proved by induction on n. Now I am trying to write down the proof (not very complicated). $\endgroup$ – Taras Banakh Nov 26 '18 at 14:20
  • $\begingroup$ Very interesting! $\endgroup$ – Taras Banakh Nov 26 '18 at 14:30
  • $\begingroup$ No,sorry, I was wrong, as not any finite group is a product of Sylow subgroups (one per prime)... $\endgroup$ – Ilya Bogdanov Nov 26 '18 at 14:40

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