Product of non-normal subgroups equals the whole group Is there any example of two subgroups $H, K \le G$, none of which normal in $G$, such that $G = HK$?
 A: Yeah, there are examples. Take the 5-cycle $\sigma=(1,2,3,4,5) \in A_5$ and let $H= \langle \sigma \rangle \subseteq A_5$. Naturally $A_4$ can be considered as subgroup of $A_5$ (it's the stabilizer of $5$). Then you have
$$A_5 = H A_4.$$
This can be shown by showing the injectivity of the map $H \times A_4 \to A_5$, $(\sigma^k,\tau) \mapsto \sigma^k\tau$: $\sigma^{k_1}\tau_1=\sigma^{k_2}\tau_2$ implies $\sigma^{k_1-k_2} = \tau_2 \tau_1^{-1} \in H \cap A_4 = \{e\}$). Surjectivity follows then by cardinality (you have $|H|=5$, $|A_4|=12$ and $|A_5|=60$).
But since $A_5$ is simple neither $H$ nor $A_4$ is a normal subgroup of $A_5$.
A: The following is a systematic way of constructing such examples: 
Let $A$ be the ring of (strictly) upper triangular matrices of degree $n \geq 3$, with entries in an arbitrary field.  Then, the underlying set $A$ forms a group with respect to the operation $x\circ y=x+y+xy$.  Denote this group by $A^{\circ}$ and the additive group of $A$ by $A^+$.
The group $A^{\circ}$ acts on $A^+$ via $x^g=x+xg$, where $x \in A^+$ and $g \in A^{\circ}$.  Let $G=A^{\circ} \rtimes A^{+}$.
Now, $G$ is a product of the subgroups $H=\{(x,0)\,|\, x\in A^{\circ}\}$, and $K=\{(x,x)\,|\, x\in A\}$. Indeed, any element $(x,y) \in G$ can be written as:
$$(x,y)=(x\circ y^{-1},0)(y,y).$$
I think you can, safely,  complete the assertions without proofs.
Remark. with some trivial exceptions, you can replace $A$ by any nilpotent ring, or more generally by a radical ring (i.e., a ring in which the circle operation defined above is a group law). 
We have $G=LH$, and $G=HK$; where $L=\{(0,x)\,|\, x\in A\}$.  Still there is a third factorization!!
