charged-lepton mass matrix can be explicitly expressed as which indicates that the left-diagonalization matrices V$_{L}$ for the charged-lepton sector is identical as U$_{ω}$ if their mass matrix can be diagonalized as m diag f = V † $_{L}$m$_{f}$ U$_{R}$ = υ Diag . ( y$_{e}$, y$_{µ}$, y$_{τ}$ ) = Diag . ( m$_{e}$, m$_{µ}$, m$_{τ}$ ), and U$_{R}$ is the unit matrix. Taking the scale of A$_{4}$ × Z$_{4}$ symmetry breaking to be above the electroweak scale in our scenario, that is, 〈 χ 〉 , 〈 ϑ 〉 , 〈 ψ 〉 > 〈 Φ $^{0}$〉 , and the required CP-violation can be supplied by the Yukawa interaction f$_{ν}$ Λ ¯ ℓ$_{L}$N$_{R}$ψη , being plus with g$_{ν}$ ¯ ℓ$_{L}$N$_{R}$η , the neutrino Yukawa coupling matrix is given as where < ψ$_{i}$ > = υ$_{ψ}$$_{i}$ ( i = 1 , 2 , 3). The right-handed neutrino Majorana mass terms, being M times the unity matrix plus being driven by 〈 χ 〉 , are given as where < χ$_{i}$ > = υ$_{χ}$$_{i}$ ( i = 1 , 2 , 3) and y$_{R}$ < ϑ > = M . If we assume the vacuum alignment of fields 〈 χ$_{i}$ 〉 and 〈 ψ$_{i}$ 〉 can be chosen as follows A$_{4}$ × Z$_{4}$ symmetry is broken in such a way that while keeping µ − τ symmetry in the right- handed Majorana mass term, Yukawa neutrino sector to be broken $^{2}$. The choice of VEV directions in Eq. (7) and 〈 Φ 〉 require a stable (or at least approximately stable) alignment of the fields χ, ϑ, ψ and Φ, which is displayed in Appendix in which we assume ad hoc 〈 χ$_{1}$ 〉 ≡ υ$_{χ}$ = 0 , 〈 χ$_{2}$ 〉 = 〈 χ$_{3}$ 〉 = 0 , 〈 ψ$_{2}$ 〉 ≡ υ$_{ψ}$ = 0 , 〈 ψ$_{1}$ 〉 = 〈 ψ$_{3}$ 〉 = 0 , (7) M$_{R}$ =    M λ$_{χ}$υ$_{χ}$$_{3}$ λ$_{χ}$υ$_{χ}$$_{2}$ λ$_{χ}$υ$_{χ}$$_{3}$ M λ$_{χ}$υ$_{χ}$$_{1}$ λ$_{χ}$υ$_{χ}$$_{2}$ λ$_{χ}$χ$_{1}$ M    , (6) Y$_{ν}$ =    g$_{ν}$ f$_{ν}$ υ$_{ψ}$$_{3}$ Λ f$_{ν}$ υ$_{ψ}$$_{2}$ Λ f$_{ν}$ υ$_{ψ}$$_{3}$ Λ g$_{ν}$ f$_{ν}$ υ$_{ψ}$$_{1}$ Λ f$_{ν}$ υ$_{ψ}$$_{2}$ Λ f$_{ν}$ υ$_{ψ}$$_{1}$ Λ g$_{ν}$    , (5) 2 It is equivalent to the way of µ − τ symmetry breaking 〈 ψ 3 〉 ≡ υ$_{ψ}$ = 0 , 〈 ψ 1 〉 = 〈 ψ 2 〉 = 0 . 6 m$_{f}$ = $_{U$_{ω}$}$   √$_{3}$$_{y}$$_{e}$$_{υ}$ 0 0 0 √$_{3}$$_{y}$$_{µ}$$_{υ}$ 0 0 0 √$_{3}$$_{y}$$_{τ}$$_{υ}$    , with U$_{ω}$ = 1 $_{√$_{3}$}$   1 1 1 1 ω ω 2 1 ω 2 ω    (4)