A finite group that has no decomposition of given cardinality

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?

• @mathworker21 $A=\{0,1\}$, $B=\{0,2\}$. ($A,B$ are subsets, not necessarily subgroups.) – bof Nov 25 '18 at 2:28
• Related question: mathoverflow.net/questions/177747/… – Jeremy Rickard Nov 25 '18 at 9:46
• – Jeremy Rickard Nov 25 '18 at 9:50
• 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. – Derek Holt Nov 25 '18 at 20:24
• So, what is the intuitive expectation concerning the general problem? Is each group $a{\times}b$-decomposable? – 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.
Let $$A_{i+1}=L_{i+1}A_i=A_iR_{i+1}$$ with $$|L_{i+1}|=|R_{i+1}|=i+1$$. Then $$A_9=A_7R_8R_9=L_7A_6R_8R_9=(L_7L_5)(A_4R_6R_8R_9)$$.
• 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? – Taras Banakh Nov 26 '18 at 13:25
• At least, not directly. Check what happens with $A_{35}$ and $a$ being the product of maximal powers of 3 and 5 in $|A_{35}$. – Ilya Bogdanov Nov 26 '18 at 14:15
• 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). – Taras Banakh Nov 26 '18 at 14:20