A finite group that has no decomposition of given cardinality

Question

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?

Answer 1

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.

Answer 2

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 $A_9=A_7R_8R_9=L_7A_6R_8R_9=(L_7L_5)(A_4R_6R_8R_9)$.