\rangle_{2} \\ 0 & 0 & \omega Y_{S2}\langle S_{3} \rangle_{2} & \omega^2Y_{S2}\langle S_{3} \rangle_{2} & Y_{S2}\langle S_{3} \rangle_{3} & Y_{S2}(V_{S} + \langle S_{1} \rangle) & Y_{S2}\langle S_{3} \rangle_{1} \\
0 & 0 & \omega^2Y_{S2}\langle S_{3}\rangle_{3} & \omega Y_{S2}\langle S_{3}\rangle_{3} & Y_{S2}\langle S_{3} \rangle_{2} & Y_{S2}\langle S_{3} \rangle_{1} & Y_{S2}(V_{S} + \langle S_{1} \rangle) \end{array}\right)\nonumber \\\end{aligned}$$ written in the left-handed and right-handed charged leptons bases given by $(L_{L1}, L_{L3}= (L_{L3_1}, L_{L3_2}, L_{L3_3}),l^c_{R3}=(l^c_{R3_1},l^c_{R3_{2}},l^c_{R3_{3}}))$ and $(l^{(1)}_{R1}, l^{(2)}_{R1}, l_{R1'}, l_{R1''},l_{R3_{1}},l_{R3_{2}},l_{R3_{3}})^T$ respectively. Note that the first two columns of the mass matrix are proportional to each other, hence the determinant of this matrix is zero, Det$(M_{l}) = 0$, which implies it has a zero eigenvalue to be identified as the electron mass. We thus predicts the electron is massless at tree level in our model. This is because the two $A_{5}$ singlet right-handed charged leptons (see Eq. (\[leprt\])) have the same Yukawa couplings structure due to the $A_5$ multiplication rules and this result is independent of the values of VEVs. The construction provides another example of electron-muon universality proposed in Ref [@Barr:1976bk], where a class of models was devised to evaluate the small ratio $m_{e}/m_{\mu}$. Under the assumption that the $A_{5}$ breaking scale is much 