11.43 (1990)

No, it is not true in general: let $G = \langle a, b \mid a^{30} = b^2 = 1, a^b = a^{-1} \rangle \cong D_{60}$ be the dihedral group of order 60, and let $A = \langle a^{10}, ba \rangle \cong D_6$ and $B = \langle a^6, b \rangle \cong D_{10}$. Then $G = AB$, but $G$ has 18 conjugacy classes and $A$ and $B$ only 3 and 4, respectively. From (P. Gallagher, Math. Z., 118 (1970), 175–179) a positive answer follows if $A$ or $B$ is normal. The problem remains open in the case where $A$ and $B$ have coprime orders, see new problem 14.44. (J. Sangroniz, Letter of December, 17, 1998.)