TITLE: Product of non-normal subgroups equals the whole group
QUESTION [7 upvotes]: Is there any example of two subgroups $H, K \le G$, none of which normal in $G$, such that $G = HK$?

REPLY [4 votes]: 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 $A_5$.