On Decompositions of Finite Group Any finite non-cyclic abelian group $G$ can be written as product $HK$ of two proper subgroups. Here $HK=\{ hk\colon h\in H, k\in K\}$. A step further, if $G$ is a finite group such that the commutator subgroup $[G,G]$ is proper subgroup of $G$, then $G$ has a decomposition $HK$ for some proper subgroups $H,K$, since


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*if $G/[G,G]$ is non-cyclic then  we can pull back the decomposition for $G/[G,G]$;

*if $G/[G,G]$ is cyclic, then we have the decomposition $G=HK$ where $H=[G,G]$, and $K=\langle x\rangle$ is a subgroup such that $G/[G,G]=\langle x[G,G]\rangle$. 
The question I would like to ask is the natural one:
Q. Does every finite group admits a decompositon $G=HK$ where $H,K$ are proper subgroups?
By initial observations, it is sufficient to visit the question for groups $G$ such that $[G,G]=G$ (such groups are called perfect groups
 A: If it were really true that all finite groups (other than cyclic $p$-groups) possessed a factorization, then it would suffice to prove this for non-abelian simple groups: if $G/N = (H/N) (K/N)$ then $G=HK$. A quick check reveals that if $G$ is a finite group with a top composition factor of order less than $1000$, then $G$ can be factored as $G=HK$. However, $G=\operatorname{PSL}(2,13)$ has no such factorization: the only possibilities for $H,K$ by order considerations are a Borel subgroup and a non-split torus. However, their intersection is always of size 2, which is too large, $|HK| = |G|/2$.
There are large families where there are never factorizations: finite simple groups of exceptional Lie types $E_6$, $E_7$, $E_8$ or twisted types ${}^2G_2$, ${}^3D_4$, ${}^2F_4$, and ${}^2E_6$ are not factorizable. This is shown in Hering–Liebeck–Saxl (1987).


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*Hering, Christoph; Liebeck, Martin W.; Saxl, Jan.
“The factorizations of the finite exceptional groups of Lie type.”
J. Algebra 106 (1987), no. 2, 517–527.
MR880974
DOI:10.1016/0021-8693(87)90013-5
