Note that if $H_1$ and $H_2$ are subgroups of $G$, then $H_1H_2$ is a subgroup if and only if $H_1H_2 = H_2H_1$. So the assumption $G = H_1H_2 = H_2H_1$ is equivalent to $G = H_1H_2$.
One special case where $G = HK$ is when $G$ is the direct product of $H$ and $K$.
Is the direct product of two cyclic groups always cyclic?
0
user38268
On
Consider $G = D_4 = \{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$ and set $H_1 = \{1,y\}$ and $H_2 = \{1,x,x^2,x^3\}$. Then $H_2H_1= G$ and furthermore
$$H_1H_2 = \{1,y,x,yx,x^2,yx^2,x^3,yx^3\} = G.$$
But obviously $G$ is not cyclic.
0
Bumbble Comm
On
The group $\mathbb{Z}_2$ is cyclic but $\mathbb{Z}_2\oplus\mathbb{Z}_2$ is not cyclic.
Note that if $H_1$ and $H_2$ are subgroups of $G$, then $H_1H_2$ is a subgroup if and only if $H_1H_2 = H_2H_1$. So the assumption $G = H_1H_2 = H_2H_1$ is equivalent to $G = H_1H_2$.
One special case where $G = HK$ is when $G$ is the direct product of $H$ and $K$.
Is the direct product of two cyclic groups always cyclic?